Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn. Viết đề như thế này gây khó đọc.

Ta có : 1+\(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}\) (đkxđ x\(\ne\pm2\) )

\(\Leftrightarrow\) \(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}-1\)

\(\Leftrightarrow\)\(\dfrac{1}{x+2}=\dfrac{12-\left(8-x^3\right)}{8-x^3}\)

\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{x^3+4}{8-x^3}\)

\(\Leftrightarrow8-x^3=\left(x+2\right)\left(x^3+4\right)\)

\(\Leftrightarrow8-x^3=x^4+4x+2x^3+8\)

\(\Leftrightarrow-x^3-x^4-4x-2x^3=8-8\)

\(\Leftrightarrow-x^4-3x^3-4x=0\)

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

\(\Rightarrow-x=0\)\(\Rightarrow x=0\) (TM x\(\ne\pm2\))

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

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

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

a) \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(\Leftrightarrow3x-9+5x-5-x-6=0\)

\(\Leftrightarrow7x=20\)

\(\Rightarrow x=\frac{20}{7}\)

b) \(\left|4x-2\right|=8\Leftrightarrow\orbr{\begin{cases}4x-2=8\\4x-2=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=10\\4x=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

c) \(-3\left|6x+1\right|=-12\)

\(\Leftrightarrow\left|6x+1\right|=4\Leftrightarrow\orbr{\begin{cases}6x+1=4\\6x+1=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}6x=3\\6x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{6}\end{cases}}\)

                                                   Bài giải

a, \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(3x-9+5x-5-x-6=0\)

\(7x-20=0\)

\(7x=20\)

\(x=\frac{20}{7}\)

b, \(\left|4x-2\right|=8\)

\(4x-2=\pm8\)

\(\Rightarrow\orbr{\begin{cases}4x-2=-8\\4x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-6\\4x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy \(x\in\left\{-3\text{ ; }2\right\}\)

c, \(-3\left|6x+1\right|=-12\)

\(\left|6x+1\right|=4\)

\(6x+1=\pm4\)

\(\Rightarrow\orbr{\begin{cases}6x+1=-4\\6x+1=4\end{cases}}\Rightarrow\orbr{\begin{cases}6x=-5\\6x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{6}\\x=\frac{1}{2}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{-\frac{5}{6}\text{ ; }\frac{1}{2}\right\}\)

Sửa lại đề nha:

 \(\dfrac{1}{x^2+9x+12}thành\dfrac{1}{x^2+9x+20}\)

⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)

⇔ \(x^2+8x+12=32\)

⇔ \(x^2+8x-20=0\)

⇔ \(\left(x-2\right)\left(x+10\right)=0\)

⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

2: =>2x^2-8x+4=x^2-4x+4 và x>=2

=>x^2-4x=0 và x>=2

=>x=4

3: \(\sqrt{x^2+x-12}=8-x\)

=>x<=8 và x^2+x-12=x^2-16x+64

=>x<=8 và x-12=-16x+64

=>17x=76 và x<=8

=>x=76/17

4: \(\sqrt{x^2-3x-2}=\sqrt{x-3}\)

=>x^2-3x-2=x-3 và x>=3

=>x^2-4x+1=0 và x>=3

=>\(x=2+\sqrt{3}\)

6:

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

=>\(\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=-2\)

=>\(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}+1+2=\sqrt{x-1}+3\)

=>1-căn x-1=căn x-1+3 hoặc căn x-1-1=căn x-1+3(loại)

=>-2*căn x-1=2

=>căn x-1=-1(loại)

=>PTVN

1) ĐK: \(x\ge\dfrac{5}{2}\)

pt <=> \(x-4=\sqrt{2x-5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left(x-4\right)^2=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-8x+16=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-10x+21=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left(x-3\right)\left(x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left[{}\begin{matrix}x=3\left(l\right)\\x=7\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=7

2) ĐK: \(2x^2-8x+4\ge0\)

pt <=> \(\left\{{}\begin{matrix}x\ge2\\2x^2-8x+4=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\left(x-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=0\left(l\right)\\x=4\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=4

3) ĐK: \(x\ge3\)

pt <=> \(\left\{{}\begin{matrix}x\le8\\x^2+x-12=x^2-16x+64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\17x=76\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x=\dfrac{76}{17}\left(n\right)\end{matrix}\right.\) 

Vậy, pt có nghiệm duy nhất là \(x=\dfrac{76}{17}\)\(\)

4) ĐK: \(x\ge3\)

pt <=> \(x^2-3x-2=x-3\)

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

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

a, (x-3)(2x+2) - (2x+1)(x-3)+12 =0

(x-3)(2x +2-2x-1) +12 = 0

(x-3) . 1 +12=0

x - 3 +12 =0

x = 9