a) \(=\left(x-y\right)\left(3+5x\right)\)

b) \(=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)

c) \(=5\left(x^2-xy-2x+2y\right)=5\left[x\left(x-y\right)-2\left(x-y\right)\right]=5\left(x-y\right)\left(x-2\right)\)

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

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

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

xy + 2x + y + 2 = y(x + 1) + 2(x + 1) = (x + 1).(y + 2)

x(x - 1) + x(x + 3) = x(x - 1 + x + 3) = x. ( 2x + 2) = 2x.(x + 1)

\(-4x^2+8x-4=-4\left(x^2-2x+1\right)=-4\left(x-1\right)^2\)

c: \(-4x^2+8x-4\)

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

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

a) \(xy+2x+y+2\)

\(=\left(xy+y\right)+\left(2x+2\right)\)

\(=y\left(x+1\right)+2\left(x+1\right)\)

\(=\left(y+2\right)\left(x+1\right)\)

b) \(x\left(x-1\right)+x\left(x+3\right)\)

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

\(=2x^2+2x\)

\(=2x\left(x+1\right)\)

c) \(-4x^2+8x-4\)

\(=-\left[\left(2x\right)^2-2.2x.2+2^2\right]\)

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

a: =>(3x+1)(3x-1)-(3x+1)(2x-3)=0

=>(3x+1)(3x-1-2x+3)=0

=>(3x+1)(x+2)=0

=>x=-1/3 hoặc x=-2

b: =>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>x=-1/3 hoặc x=-4/5

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)

=13/54

Đề trước đó: 

(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x

<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x

<=>x^2-11x-6=0

<=>x^2-2x. 11/2 + 121/4-145/4=0

<=>(x-11/2)^2=145/4

<=>|x-11/2|=căn(145)/2

<=>x=[11+-căn(145)]/2

2: \(ax+ay+bx+by\)

\(=a\left(x+y\right)+b\left(x+y\right)\)

\(=\left(x+y\right)\left(a+b\right)\)

3: \(x\left(x-2y\right)-x+2y\)

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

\(=\left(x-2y\right)\left(x-1\right)\)