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20 tháng 2 2021
Bài 35:
b) ĐKXĐ: \(x\notin\left\{5;2\right\}\)
Ta có: \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\)
\(\Leftrightarrow\dfrac{x+2}{x-5}+3-\dfrac{6}{2-x}=0\)
\(\Leftrightarrow\dfrac{x+2}{x-5}+3+\dfrac{6}{x-2}=0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\dfrac{3\left(x-5\right)\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\dfrac{6\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=0\)
Suy ra: \(x^2-4+3\left(x^2-7x+10\right)+6x-30=0\)
\(\Leftrightarrow x^2-4+3x^2-21x+30+6x-30=0\)
\(\Leftrightarrow4x^2-15x-4=0\)
\(\Leftrightarrow4x^2-16x+x-4=0\)
\(\Leftrightarrow4x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\4x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{4;-\dfrac{1}{4}\right\}\)
Bài 36:
a) Ta có: \(\left(3x^2-5x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(3x^2-5x+1\right)=0\)
mà \(3x^2-5x+1>0\forall x\)
nên (x-2)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: S={2;-2}
HD
1
\(a,\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x-1}{\sqrt{x}}.\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{x-1}\)
\(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\sqrt{x}}\)
\(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\frac{2x+2}{\sqrt{x}}\)
\(A=\frac{2\sqrt{x}}{\sqrt{x}}+\frac{2x+2}{\sqrt{x}}\)
\(A=\frac{2\sqrt{x}+2x+2}{\sqrt{x}}=\frac{2\left(x+\sqrt{x}+1\right)}{\sqrt{x}}\)
\(b,A=\frac{2\sqrt{x}+2x+2}{\sqrt{x}}=6\)
\(2\sqrt{x}+2x+2=6\sqrt{x}\)
\(2x-4\sqrt{x}+2=0\)
\(\left(\sqrt{2}\sqrt{x}\right)^2-4\sqrt{x}+\sqrt{2}^2=0\)
\(\left(\sqrt{2x}-\sqrt{2}\right)^2=0\)
\(\sqrt{2x}=\sqrt{2}\)
\(x=1\)