|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:

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

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

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

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

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

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

=>-5x<9

hay x>-9/5

a: Ta có: \(\sqrt{x^2-x+3}+7=10\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)

\(\Leftrightarrow x^2-4x+8=4\)

\(\Leftrightarrow x-2=0\)

hay x=2

(x+1)(x+2)(x+4)(x+8)=28x²

\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+9x+8\right)=28x^2\)(1)

Thấy x=0 không là nghiệm của (1). CHia 2 vế (1) cho x² ta đc:

\(\left(1\right)\Leftrightarrow\left(x+\frac{8}{x}+6\right)\left(x+\frac{8}{9}+9\right)=28\)

Đặt \(t=x+\frac{8}{x}\)ta có:

\(\left(1\right)\Rightarrow\left(t+6\right)\left(t+9\right)=28\)

\(\Leftrightarrow t^2+15t+26=0\Leftrightarrow\orbr{\begin{cases}t=-2\\t=-13\end{cases}}\)

\(\Delta=13^2-4\left(1.8\right)=137\)\(\Rightarrow x_{1,2}=\frac{-13\pm\sqrt{137}}{2}\)(thỏa mãn)

Vậy...

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{8}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{5}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{5}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+3\left(\dfrac{1}{8}-\dfrac{1}{y}\right)=\dfrac{5}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+\dfrac{3}{8}-\dfrac{3}{x}=\dfrac{5}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{1}{x}=\dfrac{1}{24}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\x=24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{12}\\x=24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=24\end{matrix}\right.\)

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}

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

\(\frac{1}{x+1+\frac{1}{x}}+\frac{2}{x+2+\frac{1}{x}}=\frac{8}{15}\)

Đặt \(x+1+\frac{1}{x}=a\)

\(\frac{1}{a}+\frac{2}{a+1}=\frac{8}{15}\)

\(\Leftrightarrow a+1+2a=\frac{8}{15}a\left(a+1\right)\)

\(\Leftrightarrow8a^2-37a-15=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-\frac{3}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1+\frac{1}{x}=5\\x+1+\frac{1}{x}=-\frac{3}{8}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+1=0\\x^2+\frac{11}{8}x+1=0\end{matrix}\right.\)