|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:

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

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

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

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}

Lời giải:

ĐKXĐ: $x\neq -1; x\neq -2$

PT \(\Leftrightarrow 2(2x^2+6x+4)+2-\frac{10}{x^2+3x+2}=5\)

\(\Leftrightarrow 4(x^2+3x+2)-\frac{10}{x^2+3x+2}-3=0\)

Đặt \(x^2+3x+2=a\). Khi đó PT trở thành:

\(4a-\frac{10}{a}-3=0\)

\(\Rightarrow 4a^2-3a-10=0\)

\(\Leftrightarrow (a-2)(4a+5)=0\Rightarrow \left[\begin{matrix} a-2=0\\ 4a+5=0\end{matrix}\right.\)

Nếu \(a-2=0\Leftrightarrow x^2+3x+2-2=0\Leftrightarrow x^2+3x=0\)

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

Nếu \(4a+5=0\Leftrightarrow 4(x^2+3x+2)+5=0\)

\(\Leftrightarrow 4x^2+12x+13=0\)

\(\Leftrightarrow (2x+3)^2=-4< 0\) (vô lý- loại)

Vậy.........

Đặt \(t=2x^2+3x-1\) thì pt trở thành :

\(t\left(t-5\right)=-4\) \(\Leftrightarrow t^2-5t+4=0\)

\(\Leftrightarrow t^2-t-4t+4=0\)

\(\Leftrightarrow\left(t-1\right)\left(t-4\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=4\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)\left(x+2\right)=0\\\left(2x+5\right)\left(x-1\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-2\\x=-\frac{5}{2}\\x=1\end{matrix}\right.\) ( TM )

\(\Leftrightarrow\left(2x^2+3x-1\right)^2-5\left(2x^2+3x-1\right)+4=0\)

\(\Leftrightarrow\left(2x^2+3x-1-1\right)\left(2x^2+3x-1-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-2=0\\2x^2+3x-5=0\end{matrix}\right.\)

Bấm máy...

bài nó dàiiiiiiii , khôg hiểu chỗ nèo hỏi lại mình hen

\(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)

\(\Leftrightarrow\left(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{\left(3x+2\right)\left(x+1\right)}\right)=1\)

\(\Leftrightarrow\dfrac{2x\left(3x+2\right)\left(x+1\right)-\left(7x.\left(3x^2-x+2\right)\right)}{\left(3x^2-x+2\right).\left(3x+2\right)\left(x+1\right)}=\dfrac{-15x^3+17x^2-10x}{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}\)

\(\Leftrightarrow\dfrac{-15x^3+17^2-10x }{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}-1=0\)

rồi quy đồng tùm lum từa lưa nữa được như này:

\(\Leftrightarrow\dfrac{-9x^4-27x^3+10x^2-18x-4}{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}=0\)

\(\Leftrightarrow-9x^4-27x^3+10x^2-18x-4=0\)

\(\Leftrightarrow x^2+\dfrac{5}{3}.x+\dfrac{25}{26}=0\)

\(\Leftrightarrow x+\left(\dfrac{5}{6}\right)^2=\dfrac{1}{36}\)

Sử dụng công thức bậc 2 hen:

\(\Leftrightarrow x=\dfrac{-5\pm\sqrt{1}}{6}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{1}}{6}\\x_2=\dfrac{-5-\sqrt{1}}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{2}{3}\\x_2=-1\end{matrix}\right.\)