Question

bài 1: giải phương trình

a) ( 2x - 10) * 95x + 3 ) = 0

b) ( 2 / x + 1 ) - ( 1 / x - 2 ) = ( 3x - 11 / x2 - x - 2 )

Answer 1

Bài 1:

a) ĐKXĐ: \(x\notin\left\{-1;2\right\}\)

Ta có: \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{x^2-x-2}\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x-2\right)\left(x+1\right)}\)

Suy ra: \(2x-4-x-1=3x-11\)

\(\Leftrightarrow x-5-3x+11=0\)

\(\Leftrightarrow-2x+6=0\)

\(\Leftrightarrow-2x=-6\)

hay x=3(thỏa mãn đkxđ)

Vậy: S={3}

Answer 2

a) (2x - 10)(95x + 3 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x-10=0\\95x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\95x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\frac{3}{95}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{5;-\frac{3}{95}\right\}\)

b) ĐKXĐ: \(x\ne-1;x\ne2\)

\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{x^2-x-2}\)

\(\Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\)

\(\Leftrightarrow2x-4-x-1=3x-11\)

\(\Leftrightarrow x-5=3x-11\)

\(\Leftrightarrow-2x=-6\)

\(\Leftrightarrow x=3\left(TM\right)\)

Vậy phương trình có tập nghiệm là: S ={3}

Answer 3

;

Answer 4

o