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a) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)
đk: x >/ 0
(*) \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\) \(\Leftrightarrow\sqrt{2x}=\dfrac{28}{13}\Leftrightarrow2x=\left(\dfrac{28}{13}\right)^2\Leftrightarrow x=\dfrac{392}{169}\left(N\right)\)
Kl: \(x=\dfrac{392}{169}\)
b) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) (*)
đk: x >/ 5
(*) \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(N\right)\)
Kl: x=9
c) \(\sqrt{\dfrac{3x-2}{x+1}}=2\) (*)
Đk: \(\left[{}\begin{matrix}x< -1\\x\ge\dfrac{2}{3}\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{3x-2}{x+1}=4\Leftrightarrow3x-2=4x+4\Leftrightarrow x=-6\left(N\right)\)
Kl: x=-6
d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (*)
Đk: \(x\ge\dfrac{4}{5}\)
(*) \(\Leftrightarrow\sqrt{5x-4}=2\sqrt{x+2}\Leftrightarrow5x-4=4x+8\Leftrightarrow x=12\left(N\right)\)
Kl: x=12
\(a,x-3\sqrt{x}+2\)
\(=x-3\sqrt{x}+\frac{9}{4}-\frac{1}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2=\left(x+2\right)\left(x-2\right)\)
câu a mình nhìn nhầm :
\(=\left(x-1\right)\left(x+2\right)\)
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
ĐKXĐ:...
\(M=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)
\(N=\frac{x\sqrt{x}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}=\frac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}=x-1\)
Để \(M=N\Leftrightarrow x-1=2\sqrt{x}+1\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\left(\sqrt{3}+1\right)^2=4+2\sqrt{3}\)
ĐKXĐ:...
\(A=\left(\frac{\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a}+1}\right).\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}=\frac{1}{a}\)
\(C=\left(\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(=\left(\frac{\left(\sqrt{x}+1\right)}{-\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)}.\frac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)}\)
\(=\left(-1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\sqrt{x}=\left(\frac{-x-\sqrt{x}-1+x+\sqrt{x}}{x+\sqrt{x}+1}\right)\sqrt{x}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}\)
Bài 2 xét x=0 => A =0
xét x>0 thì \(A=\frac{1}{x-2+\frac{2}{\sqrt{x}}}\)
để A nguyên thì \(x-2+\frac{2}{\sqrt{x}}\inƯ\left(1\right)\)
=>cho \(x-2+\frac{2}{\sqrt{x}}\)bằng 1 và -1 rồi giải ra =>x=?
1,Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)
=> \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(a+2=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(b+2=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)\)
\(c+2=\left(\sqrt{c}+\sqrt{b}\right)\left(\sqrt{c}+\sqrt{a}\right)\)
=> \(\frac{\sqrt{a}}{a+2}+\frac{\sqrt{b}}{b+2}+\frac{\sqrt{c}}{c+2}=\frac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\frac{\sqrt{b}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)}+...\)
=> \(\frac{\sqrt{a}}{a+2}+...=\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\frac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
=> M=0
Vậy M=0
Lời giải:
\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{2(2+\sqrt{2})}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{(2+\sqrt{2})(2-\sqrt{2})}=\sqrt{2}.\sqrt{2^2-2}=2\)
\(y=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{\frac{2}{3}(9\sqrt{2}-6\sqrt{3}+3\sqrt{5})}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\frac{2}{3}\)
Do đó:
\(E=\frac{1+xy}{x+y}-\frac{1-xy}{x-y}=\frac{1+\frac{4}{3}}{2+\frac{2}{3}}-\frac{1-\frac{4}{3}}{2-\frac{2}{3}}=\frac{9}{8}\)
a, Điều kiện x ∉ {\(\frac{5}{3};\frac{1}{7}\)}
\(\sqrt{3x-5}=\sqrt{7x-1}\)
\(\left(\sqrt{3x-5}\right)^2=\left(\sqrt{7x-1}\right)^2\)
\(\left|3x-5\right|=\left|7x-1\right|\)
\(3x-5=7x-1\)
\(-4x=4\) => x = -1