# Phương trình bậc nhất một ẩn

a,3(x+1) - (4x-3)=5

b, (2x+1)(x-1)= x2-2x +1

c,|-5x|=3x-16

d, $\frac{x-2}{x-2}$- $\frac{2}{x-2}$= $\frac{5x+2}{4-x^2}$

giải các pt trên

giúp mk vs ạ

Thiếu tướng -
28 tháng 6 2020 lúc 9:18

a) Ta có: 3(x+1)-(4x-3)=5

$\Leftrightarrow3x+3-4x+3-5=0$

$\Leftrightarrow-x+1=0$

hay x=1

Vậy: S={1}

b) Ta có: $\left(2x+1\right)\left(x-1\right)=x^2-2x+1$

$\Leftrightarrow\left(2x+1\right)\left(x-1\right)=\left(x-1\right)^2$

$\Leftrightarrow\left(2x+1\right)\left(x-1\right)-\left(x-1\right)^2=0$

$\Leftrightarrow\left(x-1\right)\left(2x+1-x+1\right)=0$

$\Leftrightarrow\left(x-1\right)\left(x+2\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.$

Vậy: S={1;-2}

c) Ta có: |-5x|=3x-16(*)

Trường hợp 1: $-5x\ge0\Leftrightarrow x\le0$

(*)$\Leftrightarrow-5x=3x-16$

$\Leftrightarrow-5x-3x=-16$

$\Leftrightarrow-8x=-16$

hay x=2(loại)

Trường hợp 2: $-5x< 0\Leftrightarrow x>0$

(*)$\Leftrightarrow5x=3x-16$

$\Leftrightarrow5x-3x=-16$

$\Leftrightarrow2x=-16$

hay x=-8(loại)

Vậy: $S=\varnothing$

d) ĐKXĐ: $x\notin\left\{2;-2\right\}$

Ta có: $\frac{x-2}{x-2}-\frac{2}{x-2}=\frac{5x+2}{4-x^2}$

$\Leftrightarrow\frac{x-4}{x-2}=\frac{5x+2}{4-x^2}$

$\Leftrightarrow\frac{\left(x-4\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{5x+2}{\left(x-2\right)\left(x+2\right)}=0$

$\Leftrightarrow\frac{x^2-2x-8+5x+2}{\left(x-2\right)\left(x+2\right)}=0$

Suy ra: $x^2+3x-6=0$

$\Leftrightarrow x^2+2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{33}{4}=0$

$\Leftrightarrow\left(x+\frac{3}{2}\right)^2=\frac{33}{4}$

$\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{2}=\frac{\sqrt{33}}{2}\\x+\frac{3}{2}=-\frac{\sqrt{33}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{33}-3}{2}\left(tm\right)\\x=\frac{-\sqrt{33}-3}{2}\left(tm\right)\end{matrix}\right.$

Vậy: $S=\left\{\frac{\sqrt{33}-3}{2};\frac{-\sqrt{33}-3}{2}\right\}$

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