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}

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

a) Thay a = -1 vào phương trình

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

\(\Rightarrow\dfrac{x^2-1+x^2-9}{\left(x+3\right)\left(x+1\right)}=2\)

\(\Rightarrow2x^2-10=2\left(x+3\right)\left(x+1\right)=2x^2+8x+6\)

\(\Rightarrow2x^2+8x+6-2x^{10}+10=0\)

\(\Rightarrow8x+16=0\Rightarrow x=-2\)

b, c Làm tương tự như câu a

d)

Phương trình nhận x = 1 làm nghiệm

=> \(\dfrac{1+a}{1+3}+\dfrac{1-3}{1-a}=2\)

\(\Rightarrow\dfrac{a+1}{4}+\dfrac{2}{a-1}=2\)

\(\Rightarrow\dfrac{a^2-1+8}{4\left(a-1\right)}=2\)

\(\Rightarrow a^2+7=2\left(4a-1\right)=8a-2\)

\(\Rightarrow a^2-8x+9=0\)

\(\Rightarrow\left[{}\begin{matrix}a=4+\sqrt{7}\\a=4-\sqrt{7}\end{matrix}\right.\)

Bài 1 :

a )Thế \(m=1\) vào phương trình ta được :

\(2x^2-3x-2=0\)

\(\Leftrightarrow2x^2+x-4x-2=0\)

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

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

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

Vậy \(S=\left\{-\frac{1}{2};2\right\}\)

b ) Theo hệ thức vi-et ta có :

\(\left\{{}\begin{matrix}x_1+x_2=\frac{6m-3}{2}\\x_1x_2=\frac{-3m+1}{2}\end{matrix}\right.\)

\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(\frac{6m-3}{2}\right)^2-\frac{2\left(-3m+1\right)}{2}\)

\(=\frac{36m^2-36m+9}{4}+3m-1\)

\(=\frac{36m^2-36m+9+12m-4}{4}\)

\(=\frac{36m^2-24m+5}{4}\)

\(=\frac{36m^2-24m+4+1}{4}\)

\(=\frac{\left(6m-2\right)^2+1}{4}\ge\frac{1}{4}\)

Vậy GTNN của A là \(\frac{1}{4}\) . Dấu bằng xảy ra khi \(x=\frac{1}{3}\)

\(pt:x^2-\left(2m-3\right)x-1=0\)

\(Thay\cdot m=1:pt\Leftrightarrow x^2+x-1=0\\ \Delta=1^2-4.\left(-1\right).1=5>0\\ \Rightarrow\left\{{}\begin{matrix}x_1=\frac{-1+\sqrt{5}}{2}\\x_2=\frac{-1-\sqrt{5}}{2}\end{matrix}\right.\)

 `x^3+1/x^3=6(x+1/x)(x ne 0)`

`<=>(x+1/x)(x^2-1+1/x^2)=6(x+1/x)`

`<=>(x+1/x)(x^2-1+1/x^2-6)=0`

`<=>((x^2+1)/x)(x^2+1/x^2-7)=0`

`(x^2+1)/x ne 0(AA x)`

`=>x^2+1/x^2-7=0`

`=>x^2+2+1/x^2-9=0`

`<=>(x+1/x)^2-3=0`

`<=>(x+1/x+3)(x+1/x-3)=0`

`+)x+1/x+3=0`

`<=>(x^2+3x+1)/x=0`

`<=>x^2+3x+1=0`

`<=>x^2+3x+9/4=5/4`

`<=>(x+3/2)^2=5/4`

`<=>x=(+-\sqrt{5}-3)/2`

`+)x+1/x-3=0`

`<=>(x^2-3x+1)/x=0`

`<=>x^2-3x+1=0`

`<=>x^2-3x+9/4=5/4`

`<=>(x-3/2)^2=5/4`

`<=>x=(+-\sqrt{5}+3)/2`

Vậy `S={(\sqrt{5}-3)/2,(-\sqrt{5}-3)/2,(\sqrt{5}+3)/2,(-\sqrt{5}+3)/2}`

ĐK: \(x\ne0\)

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

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=0\\x^2+\dfrac{1}{x^2}-7=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(loai\right)\\x^4-7x^2+1=0\left(1\right)\end{matrix}\right.\)

Giải (1): \(\Leftrightarrow x^2=\dfrac{7\pm3\sqrt{5}}{2}\) \(\Rightarrow x=\pm\sqrt{\dfrac{7\pm3\sqrt{5}}{2}}\)