|x-9|=2x+5

Xét 3 TH

TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)

TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)

TH3: x=9 =>0=23(L)

Vậy  x= 4/3

Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)

\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)

\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)

Ta có:

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\)

\(\dfrac{2\left(x+3\right)+3\left(x-3\right)}{x^2-9}=\dfrac{3x+5}{x^2-9}\)

\(5x-4=3x+5\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\)

\(x^2-2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b.\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(ĐK:x\ne\pm2\)

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

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

\(\Leftrightarrow x^2+3x+2-5x+10=12+x^2-4\)

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

\(\Leftrightarrow x=2\left(ktm\right)\)

Vậy pt vô nghiệm

\(a,x^2-2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\dfrac{x+1}{x-2}-\dfrac{5}{x-2}=\dfrac{12}{x^2-4}+1\) (ĐKXĐ : x ≠ 2 ; x ≠ -2)

\(\Rightarrow\left(x+1\right)\left(x+2\right)-5\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x^2+3x+2-5x-10=12+x^2+2x-2x+4\)

\(\Leftrightarrow2x=24\)

\(\Leftrightarrow x=12\left(N\right)\)

câu c chưa học :vv

a)

<=> x (x-2 ) = 0

<=> x =0 

x = 2

b)

đkxđ : x khác 2 , x khác -2

<=> \(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{12}{x^2-4}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\dfrac{x^2+3x+2}{....}-\dfrac{5x-10}{....}-\dfrac{12}{...}+\dfrac{x^2-4}{....}=0\)

<=> \(x^2+3x+2-5x+10-12+x^2-4=0\)

<=> \(2x^2-2x-4=0\)

<=> x =2 (ktm)

Vậy..

Chọn C

\(\Leftrightarrow16-3\left(x+1\right)< 24+2\left(x-1\right)\)

=>16-3x-3<24+2x-2

=>-3x+13<2x+22

=>-5x<9

hay x>-9/5

a) \(\frac{2-x}{3}< \frac{3-2x}{5}\)

<=> \(10-5x< 9-6x\)

<=> x < - 1 

Vậy S = { x| x < -1 }

b)

 

câu a bài 1:(2x+1)(3x-2)=(5x-8)(2x+1)

<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(6-2x)=0

bước sau tự làm nốt nha !

câu b:gợi ý: tách 4x^2-1thành (2x-1)(2x+1) rồi làm như câu a

Bài 2: 

a: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

\(\Leftrightarrow x-3=5\left(2x-3\right)=10x-15\)

=>-9x=-12

hay x=4/3

b: \(\Leftrightarrow x\left(x+2\right)-x+2=2\)

=>x²+2x-x+2=2

=>x²+x=0

=>x=0(loại) hoặc x=-1(nhận)

c: \(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow x^2+3x+2+x^2-3x+2=2x^2+4\)

=>4=4(luôn đúng)

Vậy: S={x|x<>2; x<>-2}

\(\Leftrightarrow\dfrac{3\left(x+7\right)}{15}+\dfrac{5\left(4x+5\right)}{15}\ge0\)

\(\Leftrightarrow3\left(x+7\right)+5\left(4x+5\right)\ge0\)

\(\Leftrightarrow23x+46\ge0\)

\(\Leftrightarrow23x\ge-46\)

\(\Leftrightarrow x\ge-2\)

Lời giải:

$\frac{x+7}{5}+\frac{4x+5}{3}\geq 0$

$\Leftrightarrow \frac{x}{5}+\frac{4x}{3}+\frac{7}{5}+\frac{5}{3}\geq 0$

$\Leftrightarrow \frac{23}{15}x+\frac{46}{15}\geq 0$

$\Leftrightarrow 23x+46\geq 0$

$\Leftrightarrow 23x\geq -46$

$\Leftrightarrow x\geq -2$

ta có: x⁴-4x³-2x²+12x+9 < x⁴-4x³-2x²+15x-3

=> x⁴-4x³-2x²+15x-3 - (x⁴-4x³-2x²+12x+9) > 0

=> 3x+6>0

(đề bài có cho điều kiện của x thì chứng minh 3x+6>0 là xong ạ)

Ta có: \(\left(x^2-2x-3\right)^2< x^2\left(x^2-4x-2\right)+3\left(5x-1\right)\)

\(\Leftrightarrow x^4+4x^2+9-4x^3-6x^2+12x< x^4-4x^3-2x^2+15x-3\)

\(\Leftrightarrow3x-12>0\)

\(\Leftrightarrow x-4>0\Rightarrow x>4\)

Vậy x > 4

hình như bn @ÀKhôngLỗiChín mới đúng á...mình h lộn