a.(x+10) /(4*x)-8* 4 -(2*x)/x+2

-(127*x-10)/(4*x)

(5/2-127*x/4)/x

Câu a

a) \(4\left(2-x\right)^2+xy-2y\)

\(=4\left(x-2\right)^2+\left(xy-2y\right)\)

\(=4\left(x-2\right)\left(x-2\right)+y\left(x-2\right)\)

\(=\left(x-2\right)\left(4x-8\right)+y\left(x-2\right)\)

\(=\left(x-2\right)\left(4x-8+x-2\right)\)

\(=\left(x-2\right)\left(5x-10\right)\)

\(=5\left(x-2\right)^2\)

a, \(=4\left(x-2\right)^2+y\left(x-2\right)=\left(x-2\right)\left(4x-8+y\right)\)

b, \(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-xy+y^2-y^2\right]=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)=x\left(x-y\right)\left(x^2-2xy+y^2-y\right)\)

c, \(=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\)

d, không phân tích được

c, x²y - xy² - 3x + 3y
= xy(x-y) - 3(x-y)
= (x-y)(x-3)

Đề bài yêu cầu gì?

tìm nghiệm hay j

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

cho A(x) = 0

\(=>-3x+1=0=>-3x=-1=>x=\dfrac{1}{3}\)

cho B(x) = 0

\(=>6-\dfrac{1}{3}x=0=>\dfrac{1}{3}x=6=>x=6\times3=18\)

Cho C(x)=0

\(=>x^2+2x=0=>x\left(x+2\right)=0=>\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

cho D(x) =0

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

cho E(x)=0

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

Bài 4:

a: \(=7xy\left(2-3-4\right)=-35xy\)

b: \(=\left(x-5\right)\left(x+y\right)\)

c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)

d: \(=\left(x+y\right)^3-\left(x+y\right)\)

=(x+y)(x+y+1)(x+y-1)

e: =x^2+8x-x-8

=(x+8)(x-1)

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

g: =-5x^2+15x+x-3

=(x-3)(-5x+1)

h: =x^2-3xy+xy-3y^2

=x(x-3y)+y(x-3y)

=(x-3y)*(x+y)

Bài 4:

a: \(=7xy\left(2-3-4\right)=-35xy\)

b: \(=\left(x-5\right)\left(x+y\right)\)

c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)

d: \(=\left(x+y\right)^3-\left(x+y\right)\)

=(x+y)(x+y+1)(x+y-1)

e: =x^2+8x-x-8

=(x+8)(x-1)

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

g: =-5x^2+15x+x-3

=(x-3)(-5x+1)

h: =x^2-3xy+xy-3y^2

=x(x-3y)+y(x-3y)

=(x-3y)*(x+y)

\(3x^3+6x^2-3x\Rightarrow ChonD\)

D

\(\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}\)

\(=\dfrac{4\left(x-3\right)+x+7}{x^2-9}\)

\(=\dfrac{4x-12+x+7}{x^2-9}\)

\(=\dfrac{5x-5}{x^2-9}\)

Bài 1:
\(a,\left(x+4\right)\left(x+3\right)-7x=x^2+4x+3x+12-7x=x^2+12\\ b,\left(x+4\right)^2+x-16=x^2+8x+16+x-16=x^2+9x\\ c,\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}=\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x-12+x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{5x-5}{\left(x+3\right)\left(x-3\right)}\)

Bài 2:
\(7a-7b=7\left(a-b\right)\\ b,x^2-8x+16=\left(x-4\right)^2\\ c,ax-ay+3x-3y=a\left(x-y\right)+3\left(x-y\right)=\left(a+3\right)\left(x-y\right)\\ d,x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

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

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

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

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

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

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)