a,(2x+y)²

b,(3m-n)²

c,(4a+5b)²

d,(x-\(\dfrac{1}{2}\))²

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

=\(\dfrac{x}{1+x}+\dfrac{1}{1+x}+\dfrac{1-2x}{x-1}+1\)

=\(1+1+\dfrac{1-2x}{x-1}\)

Thay x=5 ta có \(1+1+\dfrac{1-10}{10-1}\)

=2-1=1

\(\dfrac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}\)

n(2n-3)-2n(n+1)=2n²-3n+2n²-2n=-5n \(⋮\) 5 với mọi n

có phải M=\(\dfrac{x+3}{3x}+\dfrac{2}{x+1}-3:\dfrac{2-4x}{x+1}-3x-x^2+\dfrac{1}{3x}\)

ko bạn

a, Ta có a(a-1)-(a+3)(a+2)

= a²-a-a²-5a-6

= -6a-6

= -6(a+1) chia hết cho 6 (đpcm)

b,Ta có a(a+2)-(a-7)(a-5)

= a²+2a-a²+12a+35

= 14a+35

= 7(a+5) chia hết cho 7 (đpcm0

a,x²-100-(x+10)=0 b,x²(x²+2)-x²-2=0

(x-10)(x+10)-(x+10)=0 x²(x²+2)-(x²+2)=0

(x+10)(x-10-1)=0 (x²-1)(x²+2)=0

(x+10)(x-11)=0 (x-1)(x+1)(x²+2)=0

<=>1)x+10=0 <=>1)x=-10 <=>1)x-1=0 <=>1)x=1

2)x-11=0 2)x=11 2)x+1=0 2)x= -1

Vậy... 3)x²+2=0 3)x²= -2(vô lí)

Vậy....

c,(3x-1)²-(9x²-1)=0

(3x-1)²-(3x-1)(3x+1)=0

(3x-1)(3x-1-3x-1)=0

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

<=>3x-1=0

<=>x=\(\dfrac{1}{3}\)

Vậy ....

a)x²-4x-5=x²+x-5x-5=(x²+x)-(5x+5)=x(x+1)-5(x+1)=(x-5)(x+1)

b)x²-4xy+4y²-36=[x²-2x2y+(2y)²]-36=(x-2y)²-36=(x-2y-6)(x-2y+6)

c)x²-9x+20=x²-4x-5x+20=(x²-4x)-(5x-20)=x(x-4)-5(x-4)=(x-5)(x-4)

d)(9x²-36)-(3x-6)(2x+5)=(3x-6)(3x+6)-(3x-6)(2x+5)=(3x-6)(3x+6-2x-5)=3(x-2)(x+1)

a) Ta có VT = (a+b)(a²-ab+b²)+(a-b)(a²+ab+b²) = a³+b³+a³-b³=2a³=VP(đpcm)

b)Mk chỉ cm đc a³+b³=(a+b)[(a-b)²+ab] thôi

Ta có VP = (a+b)(a²-2ab+b²+ab) = (a+b)(a²-ab+b²) = a³+b³ = VT(đpcm)

c)Ta có VP = a²c²+2acbd+b²d²+a²d²-2adbc+b²c² = a²c²+b²d²+a²d²+b²c²

=a²(c²+d²)+b²(c²+d²)

=(a²+b²)(c²+d²)=VT(đpcm)