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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, 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
2. \(\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}=\dfrac{7}{\sqrt{x-3}}\) (2)
\(\Leftrightarrow\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}-\dfrac{7}{\sqrt{x-3}}=0\)
\(\Leftrightarrow\dfrac{\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7}{\sqrt{x-3}}=0\)
\(\Leftrightarrow\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7=0\)
\(\Leftrightarrow\left|x\right|-16+\sqrt{x^2-9}-7=0\)
\(\Leftrightarrow\left|x\right|-23+\sqrt{x^2-9}=0\)
\(\Leftrightarrow\sqrt{x^2-9}=-\left|x\right|+23\)
\(\Leftrightarrow x^2-9=-\left(-\left|x\right|+23\right)^2\)
\(\Leftrightarrow x^2-9=-\left(-\left|x\right|\right)^2-46\cdot\left|x\right|+529\)
\(\Leftrightarrow x^2-9=\left|x\right|^2-46+\left|x\right|+529\)
\(\Leftrightarrow x^2-9=x^2-46\cdot\left|x\right|+529\)
\(\Leftrightarrow-9=-46\cdot\left|x\right|+529\)
\(\Leftrightarrow46\cdot\left|x\right|=529+9\)
\(\Leftrightarrow49\cdot\left|x\right|=538\)
\(\Leftrightarrow\left|x\right|=\dfrac{269}{23}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{269}{23}\\x=-\dfrac{269}{23}\end{matrix}\right.\)
Sau khi dùng phép thử ta nhận thấy \(x\ne-\dfrac{269}{23}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{269}{23}\right\}\)
3. sửa đề: \(\sqrt{14-x}=\sqrt{x-4}\sqrt{x-1}\) (3)
\(\Leftrightarrow\sqrt{14-x}=\sqrt{\left(x-4\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-x-4x+4}\)
\(\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-5x+4}\)
\(\Leftrightarrow14-x=x^2-5x+4\)
\(\Leftrightarrow14-x-x^2+5x-4=0\)
\(\Leftrightarrow10+4x-x^2=0\)
\(\Leftrightarrow-x^2+4x+10=0\)
\(\Leftrightarrow x^2-4x-10=0\)
\(\Leftrightarrow x=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\cdot1\cdot\left(-10\right)}}{2\cdot1}\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{16+40}}{2}\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{56}}{2}\)
\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{14}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4-2\sqrt{14}}{2}\\x=\dfrac{4+2\sqrt{14}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{14}\\x=2-\sqrt{14}\end{matrix}\right.\)
sau khi dùng phép thử ta nhận thấy \(x\ne2-\sqrt{14}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{2+\sqrt{14}\right\}\)
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow x-2+\sqrt{2x}-\sqrt[4]{5x+6}=0\)
\(\Leftrightarrow x-2+\frac{\left(2x\right)^2-\left(5x+6\right)}{\left(\sqrt{2x}+\sqrt[4]{5x+6}\right)\left(2x+\sqrt{5x+6}\right)}=0\)
\(\Leftrightarrow x-2+\frac{\left(x-2\right)\left(4x+3\right)}{\left(\sqrt{2x}+\sqrt[4]{5x+6}\right)\left(2x+\sqrt{5x+6}\right)}=0\)
\(\Leftrightarrow\left(x-2\right)\left(1+\frac{4x+3}{\left(\sqrt{2x}+\sqrt[4]{5x+6}\right)\left(2x+\sqrt{5x+6}\right)}\right)=0\)
\(\Leftrightarrow x-2=0\Rightarrow x=2\)
\(a,x-3\sqrt{x}+2\)
\(=x-3\sqrt{x}+\frac{9}{4}-\frac{1}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2=\left(x+2\right)\left(x-2\right)\)
câu a mình nhìn nhầm :
\(=\left(x-1\right)\left(x+2\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\x\ge-3\\x\ge-4\\x\ge-7\end{matrix}\right.\Leftrightarrow}x\ge-2\)
\(\sqrt{x+2}-\sqrt{x+3}=\sqrt{x+4}-\sqrt{x+7}\)
\(\Leftrightarrow x+2-2\sqrt{\left(x+2\right)\left(x+3\right)}+x+3=x+4-2\sqrt{\left(x+4\right)\left(x+7\right)}+x+7\)
\(\Leftrightarrow-2\sqrt{\left(x+2\right)\left(x+3\right)}+2\sqrt{\left(x+4\right)\left(x+7\right)}=6\)
\(\Leftrightarrow2\left[\sqrt{\left(x+4\right)\left(x+7\right)}-\sqrt{\left(x+2\right)\left(x+3\right)}\right]=6\)
\(\Leftrightarrow\sqrt{\left(x+4\right)\left(x+7\right)}-\sqrt{\left(x+2\right)\left(x+3\right)}=3\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)-2\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}+\left(x+2\right)\left(x+3\right)=9\)
\(\Leftrightarrow-2\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}=-2x^2-16x-8\)
\(\Leftrightarrow\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}=x^2+8x+4\)
Có lẽ làm sai ở đâu đó, mk lười :V
ĐKXĐ: \(x\ge-2\)
\(\Leftrightarrow\sqrt{x+2}+\sqrt{x+7}=\sqrt{x+3}+\sqrt{x+4}\)
\(\Leftrightarrow2x+9+2\sqrt{x^2+9x+14}=2x+7+2\sqrt{x^2+7x+12}=0\)
\(\Leftrightarrow\sqrt{x^2+9x+14}+1=\sqrt{x^2+7x+12}\)
\(\Leftrightarrow x^2+9x+15+2\sqrt{x^2+9x+14}=x^2+7x+12\)
\(\Leftrightarrow2\sqrt{x^2+9x+14}=-2x-3\) (\(x\le-\frac{3}{2}\))
\(\Leftrightarrow4\left(x^2+9x+14\right)=4x^2+12x+9\)
\(\Leftrightarrow24x=-47\)
\(\Leftrightarrow x=-\frac{47}{24}\)