\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(4x-7\right)}\left(dkxd:x\ne-2;x\ne\dfrac{7}{4}\right)\)

\(=\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x^2-4\right)}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x-2\right)}{4x-7}\)

\(=\dfrac{4x-8}{4x-7}\)

===============================

\(\dfrac{x}{x-2}+\dfrac{2}{2-x}\left(dkxd:x\ne2\right)\)

\(=\dfrac{x}{x-2}-\dfrac{2}{x-2}\)

\(=\dfrac{x-2}{x-2}\)

\(=1\)

\(a,đk:x\ne0;4;1\)

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

\(đk:x\ne-2;1\)

\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{x\left(7x-7\right)}{\left(x+2\right)\left(7x-7\right)}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{7x^2-7x+7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{7x^2-16}{\left(x+2\right)\left(7x-7\right)}\)

a)

\(\dfrac{x-1}{x^2-5x+4}-\dfrac{4}{x^2-4x}\) \(ĐKXĐ:x\ne0;x\ne4;x\ne1\)

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

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

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

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

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

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

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

b)

\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\)  \(ĐKXĐ:x\ne-2;x\ne1\)

\(=\dfrac{x\left(7x-7\right)}{\left(x+2\right)\left(7x-7\right)}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\)

\(=\dfrac{7x^2-7x+7x-16}{\left(x+2\right)\left(7x-7\right)}\)

\(=\dfrac{7x^2-16}{\left(x+2\right)\left(7x-7\right)}\)

1, bạn xem lại đề 

2, 15(x-3) + 8x-21 = 12(x+1) +120 

<=> 23x - 66 = 12x + 132 

<=> 11x = 198 <=> x = 198/11 

3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4 

<=> 30x + 10 - 95 = 18x -12

<=> 12x = 73 <=> x = 73/12 

\(\dfrac{5x-1+x+1}{3x^2y}=\dfrac{6x}{3x^2y}=\dfrac{2}{xy}\)

\(\dfrac{21x^2+22y}{36x^3y^2}\)

\(\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x-8}{4x-7}=1-\dfrac{1}{4x-7}\)

a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

=>3x-9-10x+2=-4

=>-7x-7=-4

=>-7x=3

=>x=-3/7

b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)

=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)

=>10-2x+7x-14=4x-4+x

=>5x-4=5x-4

=>0x=0(luôn đúng)

Vậy: S=R\{0;2}

a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)

b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)

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

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

\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)

c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)

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

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

\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)

a: =>x^2+4x+4-x^2+4x-4>8x-2

=>8x>8x-2

=>0>-2(luôn đúng)

b: =>15x-10(7x+5)-24x>-240

=>15x-70x-50-24x>-240

=>-79x-50>-240

=>-79x>-190

=>x<190/79