\(\left(2x+4\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)

\(2\left(x+2\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)

\(\left(x+2\right)\left(2x-6-x+4\right)=x\left(x+5\right)\)

\(\left(x+2\right)\left(x-2\right)-x^2-5x=0\)

\(x^2-2x+2x-4-x^2-5x=0\)

\(-5x-4=0\)

\(-5x=4\)

\(\Rightarrow\)\(x=\frac{-4}{5}\)

\(\left(x-2\right)^2=\left(2x-4\right)\left(x+5\right)\)

\(\left(x-2\right)^2-2\left(x-2\right)\left(x+5\right)=0\)

\(\left(x-2\right)\left(x-2-2x-10\right)=0\)

\(\left(x-2\right)\left(-x-12\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\-x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-12\end{cases}}}\)

Bạn tự kết luận 2 câu nhé

làm hệ PT nghĩ ngay đến xét hàm

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

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)

TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)

TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)

b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)

c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)

\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)

Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

* 2x - 1 = -(x + 2)

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....

giúp mình nha mn

giúp mk với

Đặt a = 2x - 5; b = 3x - 4 => x + 1 = b - a

Khi đó ta có pt: a³ - b³ + (b - a)³ = 0

<=> (b - a)³ - (b³ - a³) = 0

<=> b³ - a³ - b³ + a³ - 3ab(b - a)= 0

<=> 3ab(b - a) = 0 <=> a = 0 hoặc b = 0 hoặc a= b

* Với a = 0, ta có: 2x - 5 = 0 <=> x = \(\frac{5}{2}\)

* Với b = 0, ta có: 3x - 4 = 0 <=> x = \(\frac{4}{3}\)

* Với a = b, ta có: 2x - 5 = 3x - 4 <=> x = -1

Vậy S = {-1;\(\frac{4}{3}\);\(\frac{5}{2}\)}

=0 bạn nha

giúp với trí ơi

giúp nốt trí ui

a) \(4x^2-2x-y^2-y=\left(4x^2-y^2\right)-\left(2x+y\right)\)

\(=\left(2x+y\right)\left(2x-y\right)-\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\)

b) \(9x^2-25y^2-6x+10y=\left(3x-5y\right)\left(3x-5y\right)-2\left(3x-5y\right)\)

\(=\left(3x-5y\right)\left(3x+5y-2\right)\)

c) \(x^3-2x^2+2x-1=x^3-1-2x^2+2x\)

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

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

d) \(x^4+2x^3-4x-4=x^4-4+2x^3-4x\)

\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)=\left(x^2-2\right)\left(x^2+2+2x\right)\)

\(\left(2x-2\right)^2=\left(x+1\right)^2+3.\left(x-2\right)\left(x+5\right)\)

\(\left(2x-2\right)^2-\left(x+1\right)^2=3.\left(x-2\right)\left(x+5\right)\)

\(\left(2x-2-x-1\right)\left(2x-2+x+1\right)=3.\left(x-2\right)\left(x+5\right)\)

\(\left(x-3\right)\left(3x-1\right)=3.\left(x-2\right)\left(x+5\right)\)

\(3x^2-x-9x+3=\left(3x-6\right)\left(x+5\right)\)

\(3x^2-10x+3=3x^2+15x-6x-30\)

\(3x^2-3x^2-10x+6x-15x+3+30=0\)

\(-19x+33=0\)

\(-19x=-33\)

\(x=\frac{33}{19}\)

\(\Leftrightarrow\left(2x-2\right)^2-\left(x+1\right)^2-3.\left(x-2\right).\left(x+5\right)=0\)

\(\Leftrightarrow4x^2-8x+4-\left(x^2+2x+1\right)-\left(3x-6\right).\left(x+5\right)=0\)

\(\Leftrightarrow4x^2-8x+4-x^2-2x-1-\left(3x^2+15x-6x-30\right)=0\)

\(\Leftrightarrow4x^2-8x+4-x^2-2x-1-3x^2-15x+6x+30=0\)

\(\Leftrightarrow-19x+33=0\)

\(\Leftrightarrow-19x=-33\)

\(\Leftrightarrow x=\frac{33}{19}\)

Vậy...............