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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
Bài 1:
a/ \(\sqrt{\dfrac{2x^2-4x+2}{6}}=1\) .
\(\Leftrightarrow\dfrac{2\left(x^2-2x+1\right)}{6}=1\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{3}=1\)
\(\Leftrightarrow\left(x-1\right)^2=3\) \(\Rightarrow\left[{}\begin{matrix}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{matrix}\right.\)
vậy tập nghiệm của phương trình S=\(\left\{1-\sqrt{3};\sqrt{3}+1\right\}\)
b/ ta có: \(\dfrac{6}{x-4}=\sqrt{2}\Leftrightarrow\sqrt{2}\left(x-4\right)=6\)
\(\Leftrightarrow x\sqrt{2}-4\sqrt{2}=6\)
\(\Leftrightarrow x\sqrt{2}=6+4\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{6+4\sqrt{2}}{2}=4+3\sqrt{2}\)
vậy \(x=4+3\sqrt{2}\) là nghiệm của phương trình
c/ \(\sqrt{\dfrac{20}{2x^2-8x+8}}=\sqrt{5}\)
\(\Leftrightarrow\left(\sqrt{\dfrac{20}{2x^2-8x+8}}\right)^2=\left(\sqrt{5}\right)^2\)
\(\Leftrightarrow\dfrac{20}{2\left(x^2-4x+4\right)}=5\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)^2}=\dfrac{10}{2}\)
\(\Rightarrow\left(x-2\right)^2=2\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{2}\\x-2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{2}\\x=2-\sqrt{2}\end{matrix}\right.\)
vậy tập nghiệm của phương trình \(S=\left\{2+\sqrt{2};2-\sqrt{2}\right\}\)
Bài 2:
a/ đặt A= \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow A^2=3+\sqrt{5}+3-\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(\Leftrightarrow A^2=6-2\sqrt{9-5}\)
\(\Leftrightarrow A^2=6-2\sqrt{4}=6-4=2\)
\(\Rightarrow A=\sqrt{2}\)
\(\Rightarrow\)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\) = \(\sqrt{2}\)
\(\Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
b/ \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\dfrac{\sqrt{12}}{\sqrt{15}}+\dfrac{\sqrt{75}}{\sqrt{15}}+\dfrac{\sqrt{27}}{\sqrt{15}}=\sqrt{\dfrac{12}{15}}+\sqrt{\dfrac{75}{15}}+\sqrt{\dfrac{27}{15}}\)
\(=\dfrac{2\sqrt{5}}{5}+\sqrt{5}+\dfrac{3\sqrt{5}}{5}=\left(\dfrac{2\sqrt{5}}{5}+\dfrac{3\sqrt{5}}{5}\right)+\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)
c/ \(\left(12\sqrt{20}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(24\sqrt{5}-80\sqrt{2}+105\sqrt{2}\right):\sqrt{10}\)
\(=\left(24\sqrt{5}+25\sqrt{2}\right):\sqrt{10}=\dfrac{24\sqrt{5}}{\sqrt{10}}+\dfrac{25\sqrt{2}}{\sqrt{10}}\)
\(=12\sqrt{2}+5\sqrt{5}\)
\(1a.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{548}=2\sqrt{16.5\sqrt{3}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=8\sqrt{\sqrt{75}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=6\sqrt{\sqrt{75}}-6\sqrt{137}\) \(b.\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right).\dfrac{1}{\sqrt{15}}=10\sqrt{3}.\dfrac{1}{\sqrt{3}.\sqrt{5}}=2\sqrt{5}\) \(d.\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}=\left(60\sqrt{2}-80\sqrt{2}+105\sqrt{2}\right).\dfrac{1}{\sqrt{10}}=85\sqrt{2}.\dfrac{1}{\sqrt{2}.\sqrt{5}}=17\sqrt{5}\) \(e.\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right):\sqrt{7}=\left(\sqrt{\dfrac{1}{7}}-4\sqrt{\dfrac{1}{7}}+3\sqrt{\dfrac{1}{7}}\right).\dfrac{1}{\sqrt{7}}=0\) \(2a.A=\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}=\sqrt{9-5-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\) \(b.B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{4-2}=2\)
Bài 4:
a: ĐKXĐ: x>=0; x<>1
b: \(P=\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\)
\(=\dfrac{2a^2+4}{-\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{-\sqrt{a}+1+\sqrt{a}+1}{a-1}\)
\(=\dfrac{-2a^2-4}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{2}{a-1}\)
\(=\dfrac{-2a^2-4+2a^2+2a+2}{\left(a-1\right)\left(a^2+a+1\right)}=\dfrac{2a+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
a)\(\left(\sqrt{10}-\sqrt{15}+3\sqrt{3}\right)\sqrt{5}-\sqrt{72}\)
\(=\sqrt{15}-\sqrt{15}+15-6\sqrt{2}\)
\(15-6\sqrt{2}\)
b)\(\dfrac{\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right)}{8\sqrt{10}}\)
\(=\dfrac{\left(15.5\sqrt{2}+5.10\sqrt{2}-3.15\sqrt{2}\right)}{8\sqrt{10}}\)
\(=\dfrac{\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right)}{8\sqrt{10}}\)
\(=\dfrac{80\sqrt{2}}{8\sqrt{10}}=\dfrac{10\sqrt{2}}{\sqrt{10}}=\sqrt{20}=2\sqrt{5}\)
Bài 3:
a: \(=\left(4\sqrt{2}-6\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}=-2\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=-2\)
b: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
Bài 1 :
a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)
= \(2\left|a-3\right|+2\left|a+2\right|\)
\(=2.\left(-a+3\right)+2\left(-a-2\right)\)
b) có sai đề ko ?
c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
1:
a: \(\sqrt{36}-\sqrt{100}=6-10=-4\)
b: Để \(\sqrt{\dfrac{2}{2x-1}}\) có nghĩa thì \(\dfrac{2}{2x-1}>=0\)
=>2x-1>0
=>x>1/2
2:
a: \(A=\dfrac{\left(15\sqrt{180}-5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}\)
\(=15\sqrt{\dfrac{180}{10}}-5\sqrt{\dfrac{200}{10}}-3\sqrt{\dfrac{450}{10}}\)
\(=15\sqrt{18}-5\sqrt{20}-3\sqrt{45}\)
\(=45\sqrt{2}-10\sqrt{5}-9\sqrt{5}\)
\(=45\sqrt{2}-19\sqrt{5}\)
b: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)
\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)
\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)