Sửa đề: thêm (...) phần mẫu :

\(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\\ \)

ĐK: \(x^2-3x+3\ne0\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\left(3-\frac{9}{4}\right)\ne0\) có (3-9/4)>0 vậy các mẫu khác không với mọi x

Đặt x^2-3x+4=t => t>=(4-9/4)=7/4

\(\Leftrightarrow\frac{1}{t-1}+\frac{2}{t}=\frac{6}{t+1}\Leftrightarrow\frac{t\left(t+1\right)}{t\left(t-1\right)\left(t+1\right)}+\frac{2\left(t^2-1\right)}{t\left(t-1\right)\left(t+1\right)}=\frac{6t\left(t-1\right)}{t\left(t-1\right)\left(t+1\right)}\)

\(\Leftrightarrow\left(t^2+t\right)+\left(2t^2-2\right)=6t^2-6t\)\(\Leftrightarrow3t^2-7t=-2\)

\(\Leftrightarrow t^2-2.\frac{7}{6}t+\left(\frac{7}{6}\right)^2=\frac{49}{36}-\frac{2}{3}=\frac{3.49-2.36}{3.36}=\frac{49-2.12}{36}=\frac{25}{36}=\left(\frac{5}{6}\right)^2\)

\(\Leftrightarrow\left(t-\frac{7}{6}\right)^2=\left(\frac{5}{6}\right)^2\Rightarrow\left\{\begin{matrix}t=\frac{7+5}{6}=2\\t=\frac{7-5}{6}=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\) 7/4<2 loại luôn

Kết luận vô nghiệm

Nhầm 7/4<2 có nghiệm

tiếp:

x^2-3x+4=2<=>x^2-3x+2=0 {a+b+c=0}

x=1 hoạc x=2

Kết luận: pt có nghiệm x=1 hoạc x=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}

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

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

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

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..

Viết đề mà ko ai đọc được vậy :v

a) \(3x^2+2x+3=\left(3x+1\right)\sqrt{x^2+3}\)

\(\Leftrightarrow3x^2+2x+3-3x\sqrt{x^2+3}-\sqrt{x^2+3}=0\)

\(\Leftrightarrow x^2+3-x\sqrt{x^2+3}-\sqrt{x^2+3}-2x\sqrt{x^2+3}+2x^2+2x=0\)

\(\Leftrightarrow\sqrt{x^2+3}\cdot\left(\sqrt{x^2+3}-x-1\right)-2x\cdot\left(\sqrt{x^2+3}-x-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x^2+3}-x-1\right)\left(\sqrt{x^2+3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)

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

Vậy...

\(\left(4x-1\right)\sqrt{x^2+1}=2x^2+2x+1\) (1)

<=>\(\left(4x-1\right)\left[\sqrt{x^2+1}-\left(3-x\right)\right]=6x^2-11x+4\)

Xét \(\sqrt{x^2+1}+3-x=0\)

<=> \(x^2+1=x^2-6x+9\) <=>\(x=\frac{4}{3}\)(tm phương trình (1))

Xét \(\sqrt{x^2+1}+3-x\ne0\)

pt <=>\(\frac{\left(4x-1\right)\left(x^2+1-x^2+6x-9\right)}{\sqrt{x^2+1}+3-x}=\left(3x-4\right)\left(2x-1\right)\)

<=> \(\frac{\left(4x-1\right)\left(6x-8\right)}{\sqrt{x^2+1}+3-x}-\left(3x-4\right)\left(2x-1\right)=0\)

<=>\(\left(3x-4\right)\left(\frac{2\left(4x-1\right)}{\sqrt{x^2+1}+3-x}-2x+1\right)=0\)

<=>\(\left[{}\begin{matrix}x=\frac{4}{3}\left(tm\right)\\\frac{8x-2}{\sqrt{x^2+1}+3-x}-2x+1=0\left(2\right)\end{matrix}\right.\)

pt (2) <=>\(8x-2=\left(2x-1\right)\sqrt{x^2+1}-2x^2+7x-3\)

<=>\(2x^2+x+1=\left(2x-1\right)\sqrt{x^2+1}\)( đk: \(x\ge\frac{1}{2}\))

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

<=>\(4x^4+4x^3+5x^2+2x+1=4x^4-4x^3+5x^2-4x+1\)

<=>\(8x^3+6x=0\) <=> \(x\left(8x^2+6\right)=0\) <=>x=0 (do 8x²+6>0) (không t/m (2))

=>(2) vô nghiệm

Vậy pt có tập nghiệm \(S=\left\{\frac{4}{3}\right\}\)

P/s: Hơi dài :)

Mấy anh chị khác god phân tích lắm nên em đành làm cách khác:(

\(2x^2+2x+1=\left(4x-1\right)\sqrt{x^2+1}\)

Đặt \(\sqrt{x^2+1}=a\ge1\)

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

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

Xét (1): Do \(a\ge1\rightarrow a+1\ge2\Rightarrow x\ge1\)

(1) \(\Leftrightarrow2x=\sqrt{x^2+1}+1\)

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

\(\Leftrightarrow\left(x-\frac{4}{3}\right)\left[\frac{\frac{3}{16}\left(3x+4\right)}{\frac{5}{4}x+\sqrt{x^2+1}}+\frac{3}{4}\right]=0\)

\(\Leftrightarrow x=\frac{4}{3}\) (vì cái ngoặc to luôn > 0 với mọi \(x\ge1\))

Vậy...

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

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

\(ĐK:x\ne\pm1\)

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

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

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

\(\Leftrightarrow x^2+6x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

=>[3x−2=02x+5=0=>[3x=22x=−5=>⎡⎢ ⎢⎣x=23x=−52