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a/ ĐKXĐ: \(x\ne-1\)
\(\Leftrightarrow4\left(3-7x\right)=x+1\)
\(\Leftrightarrow12-28x=x+1\)
\(\Rightarrow29x=11\Rightarrow x=\frac{11}{29}\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(\Leftrightarrow1-\left(\sqrt{x}-2\right)=3-\sqrt{x}\)
\(\Leftrightarrow3=3\) (luôn đúng)
Vậy nghiệm của pt là \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne7\)
\(\Leftrightarrow8-x-8\left(x-7\right)=1\)
\(\Leftrightarrow8-x-8x+56=1\)
\(\Leftrightarrow-9x=-63\Rightarrow x=7\left(ktm\right)\)
Vậy pt vô nghiệm
d/ ĐKXĐ: \(x\ne4\)
\(\Leftrightarrow\frac{28}{6\left(x-4\right)}-\frac{6\left(x+2\right)}{6\left(x-4\right)}=\frac{-9}{6\left(x-4\right)}-\frac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
\(\Rightarrow x=5\)
e/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow3x=-15\Rightarrow x=-5\)
a/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow3x=-15\Rightarrow x=-5\)
b/ ĐKXĐ: \(x\ne\left\{-\frac{4}{3};1\right\}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+37x+28=12x^2-7x-5\)
\(\Leftrightarrow44x=-33\Rightarrow x=-\frac{3}{4}\)
c/ ĐKXĐ: \(x\ne\left\{-\frac{1}{4};0\right\}\)
\(\Leftrightarrow\frac{3\left(x^2-1\right)}{4x+1}+\frac{2\left(1-x^2\right)}{x}-\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{3}{4x+1}-\frac{2}{x}-1\right)=0\)
TH1: \(x^2-1=0\Rightarrow x=\pm1\)
TH2: \(\frac{3}{4x+1}-\frac{2}{x}-1=0\Leftrightarrow3x-2\left(4x+1\right)-x\left(4x+1\right)=0\)
\(\Leftrightarrow4x^2+6x+2=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{2}\end{matrix}\right.\)
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\x\ne2\\x\ne\frac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)
Đặt \(x^2-x-1=a\) ta được:
\(\frac{4}{a-1}+\frac{2}{a}=5\Leftrightarrow4a+2\left(a-1\right)=5a\left(a-1\right)\)
\(\Leftrightarrow5a^2-11a+2=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x-1=2\\x^2-x-1=\frac{1}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3=0\\5x^2-5x-6=0\end{matrix}\right.\) (bấm máy)
b/ ĐKXĐ: \(x>2\)
Đặt \(\sqrt{x-2}=a>0\)
\(\frac{4}{a+1}-\frac{1}{a}=1\Leftrightarrow4a-\left(a+1\right)=a\left(a+1\right)\)
\(\Leftrightarrow a^2-2a+1=0\Rightarrow a=1\)
\(\Rightarrow\sqrt{x-2}=1\Rightarrow x=3\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\frac{4}{9}\end{matrix}\right.\)
\(\Leftrightarrow4\left(2-3\sqrt{x}\right)-\left(\sqrt{x}+1\right)=3\left(\sqrt{x}+1\right)\left(2-3\sqrt{x}\right)\)
\(\Leftrightarrow9x-10\sqrt{x}+1=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=\frac{1}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{81}\end{matrix}\right.\)
Cảm ơn bn nhiều :>