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:

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

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

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}

|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:

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

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

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

a) 2x + 1 = 15 - 5x

<=> 2x + 5x = 15 - 1 

<=> 7x         = 14

<=>   x         = 2

Vậy phương trình có nghiệm duy nhất là x = 2 

b) 3x - 2 = 2x + 5

<=> 3x - 2x = 5 + 2

<=> x          = 7 

Vậy phương trình có nghiệm duy nhất là x = 7 

c) x ( 2x + 1 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

Vậy _______

d) 7 ( x - 2 ) = 5 ( 3x + 1 )

<=> 7x - 14 = 15x + 5

<=> 7x - 15x = 5 + 14

<=> -8x         = 19

<=> \(x=-\frac{19}{8}\)

Vậy ______

\(\frac{2x-8}{6}-\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\)

\(\Leftrightarrow\frac{4\left(2x-8\right)}{24}-\frac{6\left(3x+1\right)}{24}=\frac{3\left(9x-2\right)}{24}+\frac{2\left(3x-1\right)}{24}\)

\(\Leftrightarrow\frac{8x-32}{24}-\frac{18x+6}{24}=\frac{27x-6}{24}+\frac{6x-2}{24}\)

\(\Leftrightarrow8x-32-18x-6=27x-6+6x-2\)

\(\Leftrightarrow8x-18x-27x-6x=-6-2+32+6\)

\(\Leftrightarrow-42x=30\)

\(\Leftrightarrow x=-\frac{5}{7}\)

Nhận thấy \(x=0\) không phải nghiệm, chia cả tử và mẫu vế trái cho x:

\(\frac{2}{3x-5+\frac{2}{x}}+\frac{13}{3x+1+\frac{2}{x}}=6\)

Đặt \(3x-5+\frac{2}{x}=a\)

\(\frac{2}{a}+\frac{13}{a+6}=6\)

\(\Leftrightarrow6a\left(a+6\right)=2\left(a+6\right)+13a\)

\(\Leftrightarrow6a^2+34a-12=0\Rightarrow\left[{}\begin{matrix}a=\frac{1}{3}\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x-5+\frac{2}{x}=\frac{1}{3}\\3x-5+\frac{2}{x}=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x^2-\frac{16}{3}x+2=0\\3x^2+x+2=0\end{matrix}\right.\)