Cần gấp ^^

\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)

=>x+100=0

=>x=-100

1.\(13.87+13.12+13\)

\(=13\left(87+12+1\right)\)

\(=13.100=1300\)

2.Đề sai à ???

3.\(x\left(x+4\right)-x\left(x-6\right)\)

\(=x^2+4x-x^2+6x\)

\(=10x\)

\(=10.123=1230\)

1, \(13.87+13.12+13=13\left(87+12+1\right)=13.100=1300\)

2, bổ sung \(\left(x-3\right)2x+\left(x-3\right)y=\left(x-3\right)\left(2x+y\right)\)

Thay x = 13 ; y = 4 ta được : \(\left(13-3\right)\left(26+4\right)=10.30=300\)

3, \(x\left(x+4\right)-x\left(x-6\right)=x\left(x+4-x+6\right)=10x\)

Thay x = 123 ta được \(1230\)

Bài 1: 

a: \(M=3\left[\left(x+y\right)^2-2xy\right]-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+1\)

\(=3\left(4-2xy\right)-\left[8-6xy\right]+1\)

\(=12-6xy-8+6xy+1=5\)

b: \(N=\left(2x-y\right)^3+3\left(2x-y\right)^2+3\left(2x-y\right)+11\)

\(=9^3+3\cdot9^2+3\cdot9+11\)

=729+243+27+11

=729+270+11=1010

\(\dfrac{1}{6\left(1-3x\right)}-\dfrac{1}{\left(3x-1\right)^2}+\dfrac{1}{6\left(3x+11\right)}+\dfrac{3}{\left(3x+11\right)^2}=0\)

\(\Leftrightarrow\dfrac{-1}{6\left(3x-1\right)}-\dfrac{1}{\left(3x-1\right)^2}=\dfrac{-1}{6\left(3x+1\right)}-\dfrac{3}{\left(3x+11\right)^2}\)

\(\Leftrightarrow-\left(3x-1\right)-6=-1\left(3x+11\right)-18\)

=>-3x+1-6=-3x-11-18

=>-3x-5=-3x-19

=>-5=-19(vô lý)

a: A=(-x)^3+3*(-x)^2*2+3*(-x)*2^2+2^3=(-x+2)^3

=(28+2)^3=30^3=27000

b: \(C=\left(x+2y-2\right)^3=\left(20+2\cdot9-2\right)^3\)

=36^3

c: 11^3-1

=(11-1)(11^2+11+1)

=10*(121+12)

=1330

d: x^3-y^3=(x-y)^3+3xy(x-y)

=6^3+3*6*9

=216+162

=378