Bài 10: Ôn tập chương I

Hướng dẫn giải:

a) $\sqrt{\dfrac{25}{81} \cdot \dfrac{16}{49} \cdot \dfrac{196}{9}}$

$=\sqrt{\dfrac{25}{81}} \cdot \sqrt{\dfrac{16}{49}} \cdot \sqrt{\dfrac{196}{9}}$

$=\sqrt{\left(\dfrac{5}{9}\right)^{2}} \cdot \sqrt{\left(\dfrac{4}{7}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{3}\right)^{2}}$

$=\dfrac{5}{9} \cdot \dfrac{4}{7} \cdot \dfrac{14}{3}=\dfrac{40}{27}$.

b) $\sqrt{3 \dfrac{1}{16} \cdot 2 \dfrac{14}{25} \cdot 2 \dfrac{34}{81}}$

$=\sqrt{\dfrac{49}{16} \cdot \dfrac{64}{25} \cdot \dfrac{196}{81}}$

$=\sqrt{\dfrac{49}{16}} \cdot \sqrt{\dfrac{64}{25}} \cdot \sqrt{\dfrac{196}{81}}$

$=\sqrt{\left(\dfrac{7}{4}\right)^{2}} \cdot \sqrt{\left(\dfrac{8}{5}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{9}\right)^{2}}$

$=\dfrac{7}{4} \cdot \dfrac{8}{5} \cdot \dfrac{14}{9}=\dfrac{196}{45}$.

c) $\dfrac{\sqrt{640} \cdot \sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.343}{567}}$

$=\sqrt{\dfrac{64.49 .7}{81.7}}=\sqrt{\dfrac{64.49}{81}}$

$=\dfrac{\sqrt{64} \cdot \sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}$

$=\dfrac{56}{9}$.

d) $\sqrt{21,6} \cdot \sqrt{810} \cdot \sqrt{11^{2}-5^{2}}$

$=\sqrt{21,6.810 \cdot\left(11^{2}-5^{2}\right)}$

$=\sqrt{216.81 .(11+5)(11-5)}$

$=\sqrt{36.6 .9^{2} \cdot 4^{2} .6}$

$=\sqrt{36^{2} .9^{2} \cdot 4^{2}}=36.9 .4=1296$.

Hướng dẫn giải:

a) $(\sqrt{8}-3 \cdot \sqrt{2}+\sqrt{10}) \sqrt{2}-\sqrt{5}$

$=\sqrt{8} \cdot \sqrt{2}-3 \cdot \sqrt{2} \cdot \sqrt{2}+\sqrt{10} \cdot \sqrt{2}-\sqrt{5}$

$=\sqrt{16}-3.2+\sqrt{20}-\sqrt{5}$

$=\sqrt{4^{2}}-6+\sqrt{2^{2} \cdot 5}-\sqrt{5}$

$=4-6+2 \sqrt{5}-\sqrt{5}=-2+\sqrt{5}$.

b) $0,2 \sqrt{(-10)^{2} \cdot 3}+2 \sqrt{(\sqrt{3}-\sqrt{5})^{2}}$

$=0,2|-10| \sqrt{3}+2|\sqrt{3}-\sqrt{5}|$

$=0,2.10 \cdot \sqrt{3}+2(\sqrt{5}-\sqrt{3})$

$=2 \sqrt{3}+2 \sqrt{5}-2 \sqrt{3}=2 \sqrt{5}$.

c) $\left(\dfrac{1}{2} \cdot \sqrt{\dfrac{1}{2}}-\dfrac{3}{2} \cdot \sqrt{2}+\dfrac{4}{5} \cdot \sqrt{200}\right): \dfrac{1}{8}$

$=\left(\dfrac{1}{2} \sqrt{\dfrac{2}{2^{2}}}-\dfrac{3}{2} \sqrt{2}+\dfrac{4}{5} \sqrt{10^{2} \cdot 2}\right): \dfrac{1}{8}$

$=\left(\dfrac{1}{2}.\dfrac{\sqrt{2}}{2}-\dfrac{3}{2} \sqrt{2}+\dfrac{4}{5} \cdot 10 \sqrt{2}\right): \dfrac{1}{8}$

$=\left(\dfrac{1}{4} \sqrt{2}-\dfrac{3}{2} \sqrt{2}+8 \sqrt{2}\right): \dfrac{1}{8}$.

$=\left(\dfrac{1}{4}-\dfrac{3}{2}+8\right) \cdot \sqrt{2}: \dfrac{1}{8}$

$=\dfrac{27}{4} \sqrt{2} .8=54 \sqrt{2}$.

d) $2 \sqrt{(\sqrt{2}-3)^{2}}+\sqrt{2 \cdot(-3)^{2}}-5 \sqrt{(-1)^{4}}$

$=2|\sqrt{2}-3|+|-3| \sqrt{2}-5 \cdot(-1)^{2}$

$=2(3-\sqrt{2})+3 \sqrt{2}-5$

$=6-2 \sqrt{2}+3 \sqrt{2}-5=1+\sqrt{2}$

Hướng dẫn giải:

a) $x y-y \sqrt{x}+\sqrt{x}-1$

$=y \cdot \sqrt{x} \cdot \sqrt{x}-y \sqrt{x}+\sqrt{x}-1$

$=y \sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y \sqrt{x}+1)$.

b) $\sqrt{a x}-\sqrt{b y}+\sqrt{b x}-\sqrt{a y}$

$=(\sqrt{a x}+\sqrt{b x})-(\sqrt{a y}+\sqrt{b y})$

$=(\sqrt{a} \cdot \sqrt{x}+\sqrt{b} \cdot \sqrt{x})-(\sqrt{a} \cdot \sqrt{y}+\sqrt{b} \cdot \sqrt{y})$

$=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})$

$=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})$.

c) $\sqrt{a+b}+\sqrt{a^{2}-b^{2}}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b} \cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

d) $12-\sqrt{x}-x $

$=12-4 \sqrt{x}+3 \sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.

Hướng dẫn giải:

a) $\sqrt{-9a}-\sqrt{9+12 a+4 a^{2}}$

$=\sqrt{-9 a}-\sqrt{3^{2}+2.3 .2 a+(2 a)^{2}}$

$=\sqrt{3^{2} \cdot(-a)}-\sqrt{(3+2 a)^{2}}$

$=3 \sqrt{-a}-|3+2 a|$

Thay $a=-9$ ta được:

$3 \sqrt{9}-|3+2 \cdot(-9)|=3.3-15=-6$.

b) Điều kiện: $m \neq 2$

$1+\dfrac{3 m}{m-2} \sqrt{m^{2}-4 m+4}$

$=1+\dfrac{3 m}{m-2} \sqrt{m^{2}-2.2 \cdot m+2^{2}}$

$=1+\dfrac{3 m}{m-2} \sqrt{(m-2)^{2}}$

$=1+\dfrac{3 m|m-2|}{m-2}$

+) $m>2$, ta được: $1+\dfrac{3 m}{m-2} \sqrt{m^{2}-4 m+4}=1+3 m$. $(1)$

+) $m<2$, ta được: $1+\dfrac{3 m}{m-2} \sqrt{m^{2}-4 m+4}=1-3 m$. $(2)$

Với $m=1,5<2$. Thay vào biểu thức $(2)$ ta có: $1-3 m=1-3.1,5=-3,5$

Vậy giá trị biểu thức tại $m=1,5$ là $-3,5$.

c) $\sqrt{1-10 a+25 a^{2}}-4a$

$=\sqrt{1-2.1 .5 a+(5 a)^{2}}-4 a$

$=\sqrt{(1-5a)^{2}}-4 a$

$=|1-5 a|-4 a$

+) Với $a <\dfrac{1}{5}$, ta được: $1-5a-4 a=1-9a$. $(3)$

+) Với $a \ge \dfrac{1}{5}$, ta được: $5 a-1-4 a=a-1$. $(4)$

Vì $a=\sqrt{2}>\dfrac{1}{5}$. Thay vào biểu thức $(4)$ ta có: $a-1=\sqrt{2}-1$.

Vậy giá trị của biểu thức tại $a=\sqrt{2}$ là $\sqrt{2}-1$.

d) $4 x-\sqrt{9 x^{2}+6 x+1}$

$=4 x-\sqrt{(3 x)^{2}+2.3 x+1}=4 x-\sqrt{(3 x+1)^{2}}$

$=4 x-|3x+1|$

+) Với $3x+1 \geq 0$ $\Leftrightarrow$ $x \ge -\dfrac{1}{3}$, ta có: $4 x-(3x+1)=4 x-3 x-1 =x-1$. $(5)$

+) Với $3x+1<0$ $\Leftrightarrow$ $x <-\dfrac{1}{3}$, ta có: $4 x+(3 x+1)=4 x+3x+1=7x+1$. $(6)$

Vì $x=-\sqrt{3}<-\dfrac{1}{3}$. Thay vào biểu thức $(6)$, ta có: $7 x+1=7 .(-\sqrt{3})+1=-7 \sqrt{3}+1$.

Giá trị của biểu thức tại $x=-\sqrt{3}$ là $-7 \sqrt{3}+1$.

Hướng dẫn giải:

a) $\sqrt{(2x-1)^{2}}=3$

$\Rightarrow$ $|2 x-1|=3$

$\Leftrightarrow$ $2x-1=\pm 3$

+) TH1: $2x-1=3$

$\Rightarrow$ $2 x=4$

$\Rightarrow$ $x=2$
+) TH2: $2x-1=-3$

$\Rightarrow$ $2 x=-2$
$\Rightarrow$ $x=-1$
Vậy  $x=-1$ ; $x=2$ .
b) Điều kiện: $x \geq 0$

$\dfrac{5}{3} \sqrt{15 x}-\sqrt{15x}-2=\dfrac{1}{3} \sqrt{15 x}$

$\Leftrightarrow$ $\dfrac{5}{3} \sqrt{15 x}-\sqrt{15 x}-\dfrac{1}{3} \sqrt{15 x}=2$

$\Leftrightarrow$ $\left(\dfrac{5}{3}-1-\dfrac{1}{3}\right) \sqrt{15} x=2$

$\Leftrightarrow$ $\dfrac{1}{3} \sqrt{15 x}=2$

$\Leftrightarrow$ $\sqrt{15 x}=6$

$\Leftrightarrow$ $15 x=36$

$\Leftrightarrow$ $x=\dfrac{12}{5}$

Vậy $x=\dfrac{12}{5}$ .

Hướng dẫn giải:

a) $VT=\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}-\sqrt{6}}{\sqrt{2^{2} \cdot 2}-2}-\dfrac{\sqrt{6^{2} .6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{\sqrt{2} \cdot \sqrt{6}-\sqrt{6}}{2 \sqrt{2}-2}-\dfrac{6 . \sqrt{6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left[\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}-\dfrac{6 \sqrt{6}}{3}\right] \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{\sqrt{6}}{2}-2 \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4 \sqrt{6}}{2}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=\left(\dfrac{-3}{2} \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}$

$=-\dfrac{3}{2}=-1,5=V P$.
b) $VT=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}$

$=\left(\dfrac{\sqrt{7} \cdot \sqrt{2}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{5} \cdot \sqrt{3}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}$

$=\left[\dfrac{\sqrt{7}(\sqrt{2}-1)}{1-\sqrt{2}}+\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}\right]: \dfrac{1}{\sqrt{7}-\sqrt{5}}$

$=(-\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})$

$=-(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})$

$=-(7-5)=-2=VP$.

c) $V T=\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}$

$=\dfrac{\sqrt{a} \cdot \sqrt{a} \cdot \sqrt{b}+\sqrt{b} \cdot \sqrt{b} \cdot \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}$

$=\dfrac{\sqrt{a} \cdot \sqrt{a b}+\sqrt{b} \cdot \sqrt{a b}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}$

$=\dfrac{\sqrt{a b}(\sqrt{a}+\sqrt{b})}{\sqrt{a b}} \cdot(\sqrt{a}-\sqrt{b})$

$=(\sqrt{a}+\sqrt{b}) \cdot(\sqrt{a}-\sqrt{b})$

$=a-b=V P$.

d) $VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)$

$=\left(1+\dfrac{\sqrt{a} \cdot \sqrt{a}+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a} \cdot \sqrt{a}-\sqrt{a}}{\sqrt{a}-1}\right)$

$=\left[1+\dfrac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}\right]\left[1-\dfrac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}\right]$

$=(1+\sqrt{a})(1-\sqrt{a})$

$=1-(\sqrt{a})^{2}=1-a=V P$

Hướng dẫn giải:

a) $\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\left(1+\dfrac{a}{\sqrt{a^{2}-b^{2}}}\right): \dfrac{b}{a-\sqrt{a^{2}-b^{2}}}$

$=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}-b^{2}}} \cdot \dfrac{a-\sqrt{a^{2}-b^{2}}}{b} $

$=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(\sqrt{a^{2}-b^{2}}\right)^{2}}{b \sqrt{a^{2}-b^{2}}}$

$=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(a^{2}-b^{2}\right)}{b \sqrt{a^{2}-b^{2}}}$

$=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b^{2}}{b \cdot \sqrt{a^{2}-b^{2}}}$

$=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b}{\sqrt{a^{2}-b^{2}}}$

$=\dfrac{a-b}{\sqrt{a^{2}-b^{2}}}$

$=\dfrac{\sqrt{a-b} \cdot \sqrt{a-b}}{\sqrt{a-b} \cdot \sqrt{a+b}}$ (do $a>b>0$)$

$=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}$

Vậy $Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}$.
b) Thay $a=3b$ vào $Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}$, ta được:

$Q=\dfrac{\sqrt{3 b-b}}{\sqrt{3 b+b}}=\dfrac{\sqrt{2 b}}{\sqrt{4 b}}$

$=\dfrac{\sqrt{2 b}}{\sqrt{2} \cdot \sqrt{2 b}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}$.