g, $\frac{x}{x-1{}}$=$\frac{4-3x}{x(x-1)}$
h, $\frac{2x-1}{x+2}$-$\frac{x}{x-2}$=$\frac{5x-2}{4-x^2}$
k, $\frac{x+2}{x-5}$+3=$\frac{6}{2-x}$
n, 3$x$-|4$x$-3|= 2
g, $\frac{x}{x-1{}}$=$\frac{4-3x}{x(x-1)}$
h, $\frac{2x-1}{x+2}$-$\frac{x}{x-2}$=$\frac{5x-2}{4-x^2}$
k, $\frac{x+2}{x-5}$+3=$\frac{6}{2-x}$
n, 3$x$-|4$x$-3|= 2
Đáp án + giải thích các bước giải:
`g) x\ne1;0`
`x/(x-1)=(4-3x)/(x(x-1))`
`->(x^2-4+3x)/(x(x-1))=0`
`->x^2+3x-4=0`
`->x^2-x+4x-4=0`
`->x(x-1)+4(x-1)=0`
`->(x+4)(x-1)=0`
`->`\(\left[ \begin{array}{l}x+4=0\\x-1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x-4\\x=1(KTM)\end{array} \right.\)
`h)x\ne±2`
`(2x-1)/(x+2)-x/(x-2)=(5x-2)/(4-x^2)`
`->((2x-1)(x-2)-x(x+2))/((x-2)(x+2))+(5x-2)/(x^2-4)=0`
`->(2x^2-5x+2-x^2-2x+5x-2)/((x-2)(x+2))=0`
`->x^2+2x=0`
`->x(x+2)=0`
`->`\(\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=0\\x=-2(KTM)\end{array} \right.\)
`k)x\ne5;2`
`(x+2)/(x-5)+3=6/(2-x)`
`->((x+2)(2-x)+3(x-5)(2-x)-6(x-5))/((x-5)(2-x))=0`
`->4-x^2+3(2x-10-x^2+5x)-6x+30=0`
`->4-x^2+6x-30-3x^2+15x-6x+30=0`
`->-4x^2+15x+4=0`
`->4x^2-15x-4=0`
`->4x^2+2.2x.15/4+225/16=289/16`
`->(2x+15/4)^2=289/16`
`->`\(\left[ \begin{array}{l}2x+\dfrac{15}{4}=\dfrac{17}{4}\\2x+\dfrac{15}{4}=\dfrac{-17}{4}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}2x=\dfrac{1}{2}\\2x=-8\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\dfrac{1}{4}\\x=-4\end{array} \right.\)
n) `3x-|4x-3|=2`
`->|4x-3|=3x-2 (3x-2>=0->x>=3/2)`
`->`\(\left[ \begin{array}{l}4x-3=3x-2\\4x-3=2-3x\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1\\7x=5\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1(KTM)\\x=\dfrac{5}{7}(KTM)\end{array} \right.\)
Đáp án:chúc bạn học tốt.
bài làm mình làm dưới ảnh nhé