\(\left(\frac{1}{2}-\frac{1}{3}\right).6x+6x+2=67+64\)

\(\frac{\Rightarrow1}{6}.6x+6x+2=131\)

\(\Rightarrow x+6x=131-2\)

\(\Rightarrow7x=129\)

\(\Rightarrow x=\frac{129}{7}\)

Ta có: \(x^3+12x^2+48x+64=8x^3-12x^2+6x-1\)

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

\(\Leftrightarrow\left(x+2\right)^3-\left(2x-1\right)^3=0\)

\(\Leftrightarrow\left[\left(x+2\right)-\left(2x-1\right)\right]\left[\left(x+2\right)^2+\left(x+2\right)\left(2x-1\right)+\left(2x-1\right)^2\right]=0\)

\(\Leftrightarrow\left(x+2-2x+1\right)\left(x^2+4x+4+2x^2+3x-2+4x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(7x^2+3x-6\right)=0\)

\(\Leftrightarrow7\left(3-x\right)\cdot\left(x^2+\frac{3}{7}x-\frac{6}{7}\right)=0\)

mà 7>0

nên \(\left[{}\begin{matrix}3-x=0\\x^2+\frac{3}{7}x-\frac{6}{7}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+2\cdot x\cdot\frac{3}{14}+\frac{9}{196}-\frac{177}{196}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x+\frac{3}{14}\right)^2=\frac{177}{196}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+\frac{3}{14}=\frac{\sqrt{177}}{14}\\x+\frac{3}{14}=-\frac{\sqrt{177}}{14}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3+\sqrt{177}}{14}\\x=\frac{-3-\sqrt{177}}{14}\end{matrix}\right.\)

Vậy: \(S=\left\{3;\frac{-3+\sqrt{177}}{14};\frac{-3-\sqrt{177}}{14}\right\}\)

x³ + 12x² + 48x + 64 = 8x³ - 12x² + 6x - 1

\(\Leftrightarrow\) x³ + 12x² + 48x + 64 - 8x³ + 12x² - 6x + 1 = 0

\(\Leftrightarrow\) -7x³ + 24x² + 42x + 65 = 0

Bn cho đề thế này ai mà giải được :vvv

a)6x²-12x=6x(x-2)

b)x²-12x+36=(x-6)²

c)8xy-16x²-y²=-(4x-y)²

d)125x³+1/64=(5x+1/4)(25x²-5/4x+1/16)

e)1/64x²+1/64=1/64(x²+1)

a, 6x^2 - 12x = 6x ( x-2)

b, x^2 - 12x + 36 = x^2 - 2.x.6 + 36 = ( x -6)^2

c, 8xy - 16x^2 - y^2 = - [ 16x^2 - 8xy + y^2) = - (4x - y)^2

d, 125y^3 + 1/64 = ( 5y + 1/4)(25y^2 - 5/4y + 1/16)

e, 1/64 x^2 + 1/64 = 1/64( x^2+1)

1: \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)

2: \(=8x^2+12x-10x-15-4\left(2x^2-x+4x-2\right)+10x+7\)

\(=8x^2+12x-8-8x^2+4x-16x+8\)

\(=0\)

3: \(=\left(6x+1-6x+1\right)^2=4\)

5: \(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)

\(=3x^3+6x-3x^3+3x=9x\)

Bài 4:

a, \(x^3+12x^2+48x+64=x^3+4x^2+8x^2+32x+16x+64\)

\(=x^2.\left(x+4\right)+8x.\left(x+4\right)+16.\left(x+4\right)\)

\(=\left(x+4\right).\left(x^2+8x+16\right)=\left(x+4\right).\left(x^2+4x+4x+16\right)\)

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

Thay \(x=6\) vào (1) ta được:

\(\left(6+4\right)^3=10^3=1000\)

Vậy...........

b, \(x^3-6x^2+12x-8=x^3-2x^2-4x^2+8x+4x-8\)

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

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

\(=\left(x-2\right).\left(x-2\right)^2=\left(x-2\right)^3\)(2)

Thay \(x=22\) vào (2) ta được:

\(\left(22-2\right)^3=20^3=8000\)

Vậy.............

Chúc bạn học tốt!!!

Bài 2:

a, \(\left(x+9\right)^3=27=3^3\)

\(\Rightarrow x+9=3\Rightarrow x=-6\)

Vậy.........

b, \(8-12x-x^3+6x^2=-64\)

\(\Rightarrow-\left(x^3-6x^2+12x-8\right)=-64\)

\(\Rightarrow x^3-2x^2-4x^2+8x+4x-8=64\)

\(\Rightarrow x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-4x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-2x-2x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x-2\right)^2=64\)

\(\Rightarrow\left(x-2\right)^3=4^3\Rightarrow x-2=4\Rightarrow x=6\)

Vậy............

Chúc bạn học tốt!!!

4. Tính giá trị biểu thức

a) x³ + 12x² + 48x + 64 khi x = 6

Ta có:

x³ + 12x² + 48x + 64 =

= (x³ + 64) + (12x² + 48x)

= (x³ + 4³) + 12x(x + 4)

= (x + 4)(x² - 4x + 4²) + 12x(x + 4)

= (x + 4)(x² - 4x + 16 +12x)

= (x + 4)(x² + 8x + 16)

= (x + 4)(x + 4)²

= (x + 4)³

Thế x = 6 vào biểu thức vừa tìm, ta được:

(x + 4)³ = (6 + 4)³ = 10³ = 1000

Vậy 1000 là giá trị của biểu thức x³ + 12x² + 48x + 64 khi x = 6.

b) x³ - 6x² + 12x - 8 khi x = 22

Ta có:

x³ - 6x² + 12x - 8 =

= (x³ - 8) - (6x² - 12x)

= (x³ - 2³) - 6x(x - 2)

= (x - 2)(x² + 2x + 2²) - 6x(x - 2)

= (x - 2)(x² + 2x + 4 - 6x)

= (x - 2)(x² - 4x + 4)

= (x - 2)(x - 2)²

= (x - 2)³

Thế x = 22 vào biểu thức vừa tìm, ta được:

(x - 2)³ = (22 - 2)³ = 20³ = 8000

Vậy 8000 là giá trị của biểu thức x³ - 6x² + 12x - 8 khi x = 22.

5. Tìm x a) (x + 9)³ = 27 \(\Leftrightarrow\) (x + 9)³ = 3³ \(\Leftrightarrow\) x + 9 = 3 \(\Leftrightarrow\) x = - 6 Vậy x = -6 b) 8 - 12x - x³ + 6x² = -64 \(\Leftrightarrow\) (8 - x³) - (12x - 6x²) = -64 \(\Leftrightarrow\) (2³ - x³) - 6x(2 - x) = -64 \(\Leftrightarrow\) (2 - x)(2² + 2x + x²) - 6x(2 - x) = -64 \(\Leftrightarrow\) (2 - x)(4 + 2x + x² - 6x) = -64 \(\Leftrightarrow\) (2 - x)(x² - 4x + 4) = -64 \(\Leftrightarrow\) -(x - 2)(x - 2)² = -64 \(\Leftrightarrow\) -(x - 2)³ = -4³ \(\Leftrightarrow\) x - 2 = 4 \(\Leftrightarrow\) x = 6 Vậy x = 6

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

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

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

 

[(6x² - 12) : 3] . 16 = 64

(6x² - 12) : 3 = 64 : 16

(6x² - 12) : 3 = 4

6x² - 12 = 4 . 3

6x² - 12 = 12

6x² = 12 + 12

6x² = 24

x² = 24 : 6

x² = 4

x = 2 hoặc x = -2

\(\left[\left(6x^2-12\right):3\right]\times16=64\\ \left(6x^2-12\right):3=4\\ 6x^2-12=12\\ 6x^2=24\\ x^2=4\\ x^2=2^2\\ x=2.\)

Đặt \(A=\sqrt{x^2-6x+36}+\sqrt{x^2-6x+64}=18\)

\(B=\sqrt{x^2-6x+64}-\sqrt{x^2-6x+36}\)

\(\Rightarrow A.B=\left(x^2-6x+64\right)-\left(x^2-6x+36\right)=28\)

mà \(A=18\Rightarrow B=\frac{28}{18}=\frac{14}{9}\)