\(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(B=x^3-9x^2+27x-27-\left(x^3-3x^2+9x+3x^2-9x+27\right)+\left(9x^2-1\right)\)

\(B=x^3-9x^2+27x-27-\left(x^3+27\right)+9x^2-1\)

\(B=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(B=27x-55\)

giải cho tui cả phần c nx đi

`Q=(3x-1)(9x^2-3x+1)-(1-3x)(1+3x+9x^2)`

`=(3x-1)(9x^2-3x+1)+(3x-1)(9x^2+3x+1)`

`=(3x-1)(9x^2-3x+1+9x^2+3x+1)`

`=(3x-1)(18x^2+2)`

Thay `x=10` vào biểu thức: `Q=(3.10-1)(18 .10^2+2)=52258`

a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)

\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)

\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)

c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)

\(=\dfrac{-x+7x-4}{3x-2}\)

\(=\dfrac{6x-4}{3x-2}=2\)

ĐKXĐ: \(x\ne\left\{-\dfrac{1}{3};\dfrac{1}{3};0;-\dfrac{4}{3}\right\}\)

\(M=\left(\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right):\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

\(=\left(\dfrac{x\left(3x+5\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)

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

\(A=27x^3+1-\left(1-27x^3\right)\\ A=27x^3+1-1+27x^3=54x^3\\ A=54\cdot10^3=54000\)

A = (3x + 1)(9x² - 3x + 1) - (1 - 3x)(1 + 3x + 9x²) 

A = (27x³ + 1) - (1 - 27x³)

A = 27x³ + 1 - 1 + 27x³

A = 54x³

Thay x = 10 vào, ta được:

A = 54 . 10³

A = 54 . 1000

A = 54000

A = (3x + 1)[(3x)² - 3x + 1²] - (1 - 3x)[1² + 3x + (3x)²]

= (3x)³ + 1³ - [1³ - (3x)³]

= 27x³ + 1 - 1 + 27x³ = 54x³

Tại x = 10, giá trị của A là:

54 . 10³ = 54000

1: cot x=-6 nên cosx/sinx=-6

=>cosx=-6*sinx

\(F=\dfrac{sinx-3\cdot cosx}{cosx+2\cdot sinx}=\dfrac{sinx+18\cdot sinx}{-6\cdot sinx+2\cdot sinx}=\dfrac{20}{-4}=-5\)

2: cotx=1

=>cosx/sinx=1

=>cosx=sinx

\(I=\dfrac{sin^3x-4\cdot sin^3x}{sinx+3sinx}=\dfrac{5\cdot sin^3x}{4\cdot sinx}=\dfrac{5}{4}\cdot sin^2x\)

\(1+cot^2x=\dfrac{1}{sin^2x}\)

=>\(\dfrac{1}{sin^2x}=1+1=2\)

=>sin^2=1/2

=>\(I=\dfrac{5}{4}\cdot\dfrac{1}{2}=\dfrac{5}{8}\)

3: cotx=3

=>cosx/sinx=3

=>cosx=3*sinx

1+cot^2x=1/sin^2x

=>\(\dfrac{1}{sin^2x}=1+9=10\)

=>\(sin^2x=\dfrac{1}{10}\)

\(I=\dfrac{2\cdot sin^3x+cos^3x}{4\cdot sinx-6\cdot cosx}\)

\(=\dfrac{2\cdot sin^3x+\left(3\cdot sinx\right)^3}{4\cdot sinx-6\cdot\left(3\cdot sinx\right)}=\dfrac{2\cdot sin^3x+27\cdot sin^3x}{4\cdot sinx-18\cdot sinx}\)

\(=\dfrac{29}{-14}\cdot sin^2x=\dfrac{-29}{14}\cdot\dfrac{1}{10}=-\dfrac{29}{140}\)

\(\Leftrightarrow3^{x-1}\left(1+3+3^2\right)=39\\ \Leftrightarrow3^{x-1}\cdot13=39\\ \Leftrightarrow3^{x-1}=3=3^1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\)

\(\Leftrightarrow3^x\cdot\dfrac{13}{3}=39\)

\(\Leftrightarrow x=2\)

\(3^{x-1}+3^x+3^{x+1}=39\\ \Rightarrow3^x:3+3^x+3^x.3=39\\ \Rightarrow3^x.\dfrac{1}{3}+3^x+3^x.3=39\\ \Rightarrow3^x\left(\dfrac{1}{3}+1+3\right)=39\\ \Rightarrow3^x.\dfrac{13}{3}=39\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2\)

a) Rút gọn Q = 54 x 3 , thay x = 10 vào tính được Q = 54000;

b) Gợi ý x 4 + y 2 = x + 2 y 4 = 8 4 = 2 .  Kết quả P = 2.