1:

a: A=(x+4)^3=10^3=1000

b: B=(x-2)^3=20^3=8000

1

a) \(A=x^3+3.x^2.4+3x.4^2+4^3=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

b) \(B=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

2

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

\(a,=5\left(x^2+2xy+y^2\right)-10y^2+5=5\left(x+y\right)^2-10y^2+5\\ =5\left(1+2\right)^2-10\cdot4+5=45-40+5=10\\ b,=7\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(7-x+y\right)\\ =\left(2-2\right)\left(7-2+2\right)=0\)

b: \(=7\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(7-x+y\right)=0\)

\(a,\)\(A\left(x\right)+B\left(x\right)=x^2-2x+1+x^2+2x+1=2x^2+2\)

\(b,\)\(A\left(x\right)-B\left(x\right)=x^2-2x+1-x^2-2x-1=-4x\)

\(c,\)Thay \(x=1\) vào \(A\left(x\right)\) ta được

\(A\left(x\right)=1^2-2.1+1=0\)

a, Thay x=2 vào A, ta được:

\(A\left(2\right)=3.2^3+5-6.2+5.2^2=37\)

Vậy A= 37 khi x=2.

b,

 \(A\left(x\right)+B\left(x\right)=\left(3x^3+5-6x+5x^2\right)+\left(4x^2+6x-2x^7-9\right)\\ =-2x^7+3x^3+9x^2-4\)

Đề phải là x^4-12x^3+12x^2-12x+111 tại x=11.

x=11

=>x+1=12

thay x+1=12 vào x^4-12x^3+12x^2-12x+111 ta được:

x⁴-(x+1)x³+(x+1)x²-(x+1)x+111

=x⁴-x⁴-x³+x³+x²-x²-x+111

=111-x

=111-12

=99

x+1=12

thay 'x+1=12 vào x^4-12x^3+12x^2-12x+111 ta có

x^4-(x+1)x^3+(x+1)x^2-(x+1)x+111

=x^4-x^4-x^3+x^3+x^2-x^2-x+111

=111-x

=111-11

=100

ban kia lam dung roi do

k tui nha

thanks

x=11

nên x+1=12

\(x^4-12x^3+12x^2-12x+111\)

\(=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+111\)

\(=x^4-x^4-x^3+x^3+x^2-x^2-x+111\)

=111-x

=111-11=100

Tại x=11

\(\Rightarrow f\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-...+\left(x+1\right)x-1\)

\(f\left(x\right)=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-...+x^2+x-1\)

\(f\left(x\right)=x-1\)

\(f\left(x\right)=10\)

\(x=11\Leftrightarrow12=x+1\)

Mà \(f\left(x\right)=x^{17}-12x^{16}+12x^{15}-12x^{14}+........+12x-1\)

\(\Leftrightarrow f\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-.......+\left(x+1\right)x-1\)

\(\Leftrightarrow f\left(x\right)=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-.....+x^2+x-1\)

\(\Leftrightarrow f\left(x\right)=x-1\)

Mà \(x=11\)

\(\Leftrightarrow f\left(11\right)=11-1=10\)

Vậy \(f\left(11\right)=10\)

a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)

\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)

b: Ta có: |2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>x=-2

Thay x=-2 vào A, ta được:

\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)

c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)

=>x+3=24x+12

=>24x+12=x+3

=>23x=-9

hay x=-9/23

d: Để A<0 thì x+3<0

hay x<-3