a,\(\left|-5x\right|\)=3x-16

\(\Leftrightarrow\)\(\left[{}\begin{matrix}-5x=3x-16\\-5x=-3x+16\end{matrix}\right.\)            \(\Leftrightarrow\)\(\left[{}\begin{matrix}-8x=-16\\-2x=16\end{matrix}\right.\)                 \(\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

a:

ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)

 \(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)

=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)

=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)

b:

ĐKXĐ: x<>-3

 \(y=\left(x+3\right)+\dfrac{4}{x+3}\)

=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)

\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)

=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)

y'=0

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

=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)

=>(x+5)(x+1)=0

=>x=-5 hoặc x=-1

c:

ĐKXĐ: x<>-2

 \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)

=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)

=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)

\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)

d: 

ĐKXĐ: x<>2

\(y=x-2+\dfrac{9}{x-2}\)

=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)

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

=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)

y'=0

=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)

=>\(\left(x-2\right)^2-9=0\)

=>(x-2-3)(x-2+3)=0

=>(x-5)(x+1)=0

=>x=5 hoặc x=-1

`(x^2-x+1)^4+4x^4=5x^2(x^2-x+1)^2`

Đặt `a=(x^2-x+1)^2,b=x^2`

`pt<=>a^2+4b^2=5ab`

`<=>a^2-5ab+4b^2=0`

`<=>a^2-ab-4ab+4b^2=0`

`<=>a(a-b)-4b(a-b)=0`

`<=>(a-b)(a-4b)=0`

`<=>` $\left[ \begin{array}{l}a=b\\a=4b\end{array} \right.$

`+)a=b`

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

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

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

`<=>x=1`

`+)a=4b`

`<=>x^2=4(x^2-x+1)^2`

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

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

`+)2x^2-x+1=0`

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

`<=>(x-1/4)^2+7/16=0` vô lý

`+)2x^2-3x+1=0`

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

`<=>2x(x-1)-(x-1)=0`

`<=>(x-1)(2x-1)=0`

`<=>` $\left[ \begin{array}{l}x=1\\x=\dfrac{1}{2}\end{array} \right.$

Vậy `S={1,1/2}`

\(Đk:-1\le x\le3\)

Đặt: \(\hept{\begin{cases}u=\sqrt{x+1}\\v=\sqrt{3-x}\end{cases}}\) Ta suy ra:

\(u^2=x+1\)

\(3u^2-2v^2=5x-3\)

\(4u^2-v^2=5x+1\)

\(u^2+v^2=4\)

Pt đã cho trở thành:

\(2\left(3u^2-2v^2\right)+5uv^2=3\left(4u^2-v^2\right)\Leftrightarrow6u^2\left(2-u\right)=v^2\left(u+3\right)\)

Thay \(v^2=4-u\) ta thu được pt:

\(2\left(3u^2-2v^2\right)+5uv^2=3\left(4u^2-v^2\right)\)

\(\Leftrightarrow6u^2\left(2-u\right)=\left(4-u^2\right)\left(u+3\right)\Leftrightarrow\orbr{\begin{cases}u=2\\u=\frac{5+\sqrt{145}}{10}\end{cases}}\)

Từ đó tìm đc các nghiệm của pt là: \(\orbr{\begin{cases}x=3\\x=\frac{7+\sqrt{145}}{10}\end{cases}}\)

Sai r bn ơi 

Sao thay vào lại đc 5uv^2 vậy ạ phải là 5u^2v chứ