\(a,4x^2-6x=2x\left(2x-3\right)\\ b,9x^4y^3+3x^2y^4=3x^2y^3\left(2x^2+y\right)\\ c,x^3-2x^2+5x=x\left(x^2-2x+5\right)\\ d,3x\left(x-1\right)+5\left(x-1\right)=\left(3x+5\right)\left(x-1\right)\\ e,2x^2\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(2x^2+4\right)=2\left(x+1\right)\left(x^2+2\right)\\ f,2x^2y-4xy^2+6xy=2xy\left(x-y+3\right)\\ g,4x^3+4x^2+4x=4x\left(x^2+x+1\right)\\ h,x^3+x^2-3x-27=x^3-3x^2+4x^2-12x+9x-27=x^2\left(x-3\right)+4x\left(x-3\right)+9\left(x-3\right)=\left(x^2+4x+9\right)\left(x-3\right)\\ i,4x^2-12x+9=\left(2x-3\right)^2\\ k,8x^3-27=\left(2x\right)^3-3^3=\left(2x-3\right)\left(4x^2+6x+9\right)\\ l,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

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Đây nè bạn đã cố gắng ko làm tắt rồi nhé bạn

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)

c: \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)

d: \(3x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(3x+5\right)\)

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

f: \(2x^2y-4xy^2+6xy=2xy\left(x-2y+3\right)\)

a: Ta có: \(2x+3=x+1\)

\(\Leftrightarrow2x-x=1-3\)

hay x=-2

b: Ta có: \(2x\left(2x-1\right)-\left(2x+3\right)^2=5\)

\(\Leftrightarrow4x^2-2x-4x^2-12x-9=5\)

\(\Leftrightarrow-14x=14\)

hay x=-1

c: Ta có: \(4x^2-25\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(2x-5x-10\right)\left(2x+5x+10\right)=0\)

\(\Leftrightarrow\left(-3x-10\right)\left(7x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=-\dfrac{10}{7}\end{matrix}\right.\)

d: Ta có: \(2x^2+7x+5=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e: Ta có: \(4x^2-4x=-1\)

\(\Leftrightarrow4x^2-4x+1=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

f: Ta có: \(\dfrac{1}{9}x^3-x=0\)

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

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

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

g: Ta có: \(x^3+3x^2+3x=7\)

\(\Leftrightarrow\left(x+1\right)^3=8\)

\(\Leftrightarrow x+1=2\)

hay x=1

a, x = -2

b, x = -1

c, x = -10/3 hoặc -10/7

b: Ta có: \(x\left(x+1\right)-\left(2x+3\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

d: Ta có: \(\left(x-1\right)^2-4\left(x+2\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)

b cần bài nào thế

Chụp rõ hơn tí đi cậu, mờ quá!

a: (x-4)(x+5)>0

=>x-4>0 hoặc x+5<0

=>x>4 hoặc x<-5

b: (2x+1)(x-3)<0

=>2x+1>0 và x-3<0

=>-1/2<x<3

c: (x-7)(3-x)<0

=>(x-7)(x-3)>0

=>x>7 hoặc x<3

d: x^2+6x-16<0

=>(x+8)(x-2)<0

=>-8<x<2

e: 3x^2+7x+4<0

=>3x^2+3x+4x+4<0

=>(x+1)(3x+4)<0

=>3x+4>0 và x+1<0

=>-4/3<x<-1

f: 5x^2-9x+4>0

=>(x-1)(5x-4)>0

=>x>1 hoặc x<4/5

g: x^2+6x-16<0

=>(x+8)(x-2)<0

=>-8<x<2

h: x^2+4x-21>0

=>(x+7)(x-3)>0

=>x>3 hoặc x<-7

i: x^2-9x-22<0

=>(x-11)(x+2)<0

=>-2<x<11

l: 16x^2+40x+25<0

=>(2x+5)^2<0(loại)

m: 3x^2-4x-4>=0

=>3x^2-6x+2x-4>=0

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

=>x>=2 hoặc x<=-2/3

a: \(x^2-4x-5=\left(x-5\right)\left(x+1\right)\)

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

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

k: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)