cái thứ nhất bạn dùng phương pháp đổi biến,đặt x^2+3x+2=a rùi thay vào và ptdt thành nhân tử thui

còn cái thứ 2 bạn nhân x+1 với x+4;x+2 với x+3 rùi lại dùng phương pháp đổi biến la ra thui

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}

a) Thay m=3 vào phương trình, ta được:

\(x^2-2x+3^2-3\cdot3+5=0\)

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

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

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

Vậy: Khi m=3 thì phương trình vô nghiệm

1)

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right).\left(x+2\right)\left(x+4\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x+5\right).\left(x^2+6x+8\right)-40=0\)

Đặt \(a=x^2+6x+6\) ta có:

\(\Leftrightarrow\left(a-1\right)\left(a+2\right)-40=0\)

\(\Leftrightarrow a^2+a-2-40=0\)

\(\Leftrightarrow a^2-6x+7x-42=0\)

\(\Leftrightarrow a\left(a-6\right)+7\left(a-6\right)=0\)

\(\Leftrightarrow\left(a-6\right)\left(a+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=6\\a=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+6=6\\x^2+6x+6=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)

(\(x^2+6x+13=\left(x+3\right)^2+4>0\left(loại\right)\))

Vậy.................

3)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=7\end{matrix}\right.\)

Vậy..........

2)\(x^3-7x^2+15x-25=0\)

\(\Leftrightarrow x^3-5x^2-2x^2+10x+5x-25=0\)

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

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

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

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

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+5\right)=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+5=5\\x^2+5x+5=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+5\right)=0\\\left(x+\frac{5}{2}\right)^2=-\frac{15}{4}\left(VL\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\) ( TM )

ĐKXĐ: \(x\ne\pm2\)

Ta có: \(\frac{x-2}{x+2}-\frac{x+2}{x-2}=\frac{24}{4-x^2}\)

\(\Leftrightarrow\frac{x-2}{x+2}-\frac{x+2}{x-2}=\frac{-24}{x^2-4}\)

\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\frac{-24}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2-4x+4-\left(x^2+4x+4\right)=-24\)

\(\Leftrightarrow x^2-4x+4-x^2-4x-4+24=0\)

\(\Leftrightarrow24-8x=0\)

\(\Leftrightarrow8x=24\)

hay x=3(tm)

Vậy: Tập nghiệm S={3}

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

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

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

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..