1: \(=x^3-6x^2+12x-8-8x^3-36x^2-54x-27+7\left(x-1\right)^3\)

\(=-7x^3-42x^2-42x-35+7x^3-21x^2+21x-7\)

\(=-63x^2-21x-42\)

2: \(=x^3+125-\left(x^3-8\right)=125+8=133\)

3: \(=8x^3-27-8x^3-12x^2-6x-1=-12x^2-6x-28\)

\(A=\left(2x-3\right).\left(3x^2+2x-1\right)-\left(4x+1\right)\cdot\left(x-1\right)\)

\(A=6x^3+4x^2-2x-9x^2-6x+3-\left(4x^2-4x+x-1\right)\)

\(A=6x^3+4x^2-2x-9x^2-6x+3-4x^2+4x-x+1\)

\(A=6x^3-9x^2-5x+4\)

Với \(x=\frac{1}{2}\).Ta có : 

\(A=6.\left(\frac{1}{2}\right)^3-9.\left(\frac{1}{2}\right)^2-5.\frac{1}{2}+4\)

\(A=\frac{3}{4}-\frac{9}{4}-\frac{5}{2}+4\)

\(\Rightarrow A=0\)

 \(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)

\(=\frac{1+1+1+1+1}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)

\(=\frac{5}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)

\(=\frac{5}{\left(x+1\right)\left(x+11\right)\left(x+3\right)\left(x+9\right)\left(x+5\right)\left(x+7\right)}\)

\(=\frac{5}{\left(x^2+11x+x+11\right)\left(x^2+9x+3x+27\right)\left(x^2+7x+5x+35\right)}\)

\(=\frac{5}{\left(x^2+12x+11\right)\left(x^2+12x+27\right)\left(x^2+12x+35\right)}\)

A=\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+11}\)

Rút gọn hết đi ta có \(\frac{1}{x+1}-\frac{1}{x+11}\)=\(\frac{x+11}{\left(x+1\right).\left(x+11\right)}-\frac{x+1}{\left(x+1\right).\left(x+11\right)}\)

A=\(\frac{x+11-x-1}{\left(x+1\right).\left(x+11\right)}\)

A=\(\frac{10}{x^2+12x+11}\)

\(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)

\(A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+...+\frac{1}{x+9}-\frac{1}{x+11}\)

\(A=\frac{1}{x+1}-\frac{1}{x+11}\Leftrightarrow A=\frac{x+11}{\left(x+1\right)\left(x+11\right)}+\frac{x+1}{\left(x+1\right)\left(x+11\right)}\)

\(\Leftrightarrow A=\frac{2x+12}{\left(x+1\right)\left(x+11\right)}\)

1. \(\left(x+5\right)\left(x^2-5x+25\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3+125-\left(x^3-8\right)=x^3+125-x^3+8=133\)

1,

\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x-2\right)\left(x^2+2x+4\right)\\ =\left(x^3+5^3\right)-\left(x^3-2^3\right)\\ =x^3+125-x^3+8\\ =\left(x^3-x^3\right)+\left(125+8\right)\\ =133\)

b,

\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+1\right)^3\\ =\left[\left(2x\right)^3-3^3\right]-\left[\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x+1+1\right]\\ =\left(8x^3-27\right)-\left(8x^3+12x^2+6x+1\right)\\ =8x^3-27-8x^3-12x^2-6x-1\\ =\left(8x^3-8x^3\right)-\left(12x^2+6x\right)-\left(27+1\right)\\ =-6x\left(2x+1\right)-28\\ =\left(-2\right)\left[3x\left(2x+1\right)+14\right]\)

\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)

\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)

\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)

\(M=343-8x^3-64+8x^3\)

\(M=279\)

Vậy M có giá trị 279 với mọi x

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

\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)

\(P=16x^3-8x^2+4x-2\)

Thay \(x=10\) vào P ta có:

\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)

Vậy P có giá trị 15238 tại x=10

a: M=343-8x^3-64+8x^3=279

b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3

=16x^3-8x^2+4x-2

=16*10^3-8*10^2+4*10-2=15238

1:  \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)

\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)

\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)

=56

2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)

\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)

\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)

\(=6\)