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\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
Bài 2:
a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)
b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)
a: \(=2ab\cdot\dfrac{-15}{b^2a}=\dfrac{-30}{b}\)
b: \(=\dfrac{2}{3}\cdot\left(1-a\right)=\dfrac{2}{3}-\dfrac{2}{3}a\)
c: \(=\dfrac{\left|3a-1\right|}{\left|b\right|}=\dfrac{3a-1}{b}\)
d: \(=\left(a-2\right)\cdot\dfrac{a}{-\left(a-2\right)}=-a\)
a) \(\sqrt{\left(\sqrt{7-2}\right)^2}=\sqrt{5}\)
b)\(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-3\sqrt{2}\right)^2}\)
=\(\sqrt{2}-1-2+3\sqrt{2}=4\sqrt{2}-3\)
c)\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=2\sqrt{3}\)
d) hình như bn ghi sai
e)\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}\right):\sqrt{2}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{3}+1}\right):\sqrt{2}\)
=\(\dfrac{\sqrt{2+\sqrt{3}}\left(\sqrt{3}+1\right)+\sqrt{2-\sqrt{3}}\left(\sqrt{3}-1\right)}{2\sqrt{2}}\)
=\(\dfrac{\sqrt{6+3}+\sqrt{2+\sqrt{3}}+\sqrt{6-3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{2+\sqrt{3}}+\sqrt{3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{3}}{2\sqrt{2}}\)
f) \(\sqrt{9a^2}+3a-7=-3a+3a-7=-7\)
g)\(\dfrac{\sqrt{4x^2-4x+1}}{4x-2}+3x+2\)
=\(\dfrac{\sqrt{\left(2x-1\right)^2}}{4x-2}+3x+2=\dfrac{2x-1}{2\left(2x-1\right)}+3x+2\)
=\(\dfrac{1}{2}+3x+2=\dfrac{5}{2}+3x\)
h)\(\sqrt{\left(5a-1\right)^2}+2a-3\)
nếu a<0 :\(-5a+1+2a-3=-3a-2\)
nếu a>0 : \(5a-1+2a-3=7a-4\)
i)\(\sqrt{\dfrac{2a}{5}}.\sqrt{\dfrac{5a}{18}}+2\left(a-1\right)\)
=\(\sqrt{\dfrac{10a^2}{90}}+2a-2=\sqrt{\dfrac{a^2}{9}}+2a-2\)
=\(\dfrac{a}{3}+2a-2=\dfrac{7a}{3}-2\)
a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)
b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)
\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)
c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
\(1a.\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=21-2\sqrt{21}+2\sqrt{21}=21\) \(b.\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}=11+2\sqrt{30}-2\sqrt{30}=11\)
\(2a.\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}.b^2}+\sqrt{\dfrac{a^2}{b^2}.\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+b\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}}=\left(2+b\right)\sqrt{\dfrac{a}{b}}\) \(b.\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{\left(x-1\right)^2}}.\sqrt{\dfrac{\left(2\sqrt{m}x-2\sqrt{m}\right)^2}{81}}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{\text{|}2\sqrt{m}x-2\sqrt{m}\text{|}}{9}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{2\sqrt{m}\text{|}x-1\text{|}}{9}=\dfrac{2m}{9}\) \(3a.VP=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1=VT\)
KL : Vậy đẳng thức được chứng minh.
\(b.VP=\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{\text{|}a+b\text{|}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{a+b}=\text{|}a\text{|}=VT\)
KL : Vậy đẳng thức được chứng minh .
P/s : Dài v ~
a, Vì trong dấu căn là số âm nên biểu thức này vô nghĩa. b)\(\sqrt{\dfrac{1}{200}}=\dfrac{1}{\sqrt{200}}=\dfrac{1}{10\sqrt{2}}=\dfrac{\sqrt{2}}{10\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{2}}{20}\)
c,\(\sqrt{\dfrac{7}{500}}=\dfrac{\sqrt{7}}{\sqrt{500}}=\dfrac{\sqrt{7}}{10\sqrt{5}}=\dfrac{\sqrt{7}.\sqrt{5}}{10\sqrt{5}.\sqrt{5}}=\dfrac{\sqrt{35}}{50}\)