nhìn cái đề con hơi bị ''sốc'' , thế này ạ ??? 

Sửa đề \(4+\frac{1}{3}x\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}x\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(4+\frac{1}{3}x\left(-\frac{1}{3}\right)\le x\le\frac{2}{3}x\left(-\frac{11}{12}\right)\)

\(4-\frac{1}{9}x\le x\le-\frac{11}{18}x\)

b: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)

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

\(\Leftrightarrow12x=12\)

hay x=2

d: Ta có: \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)

\(\Leftrightarrow3x^2-6x+3-3x^2+15x=1\)

\(\Leftrightarrow9x=-2\)

hay \(x=-\dfrac{2}{9}\)

Ta có

( 3 x   –   1 ) 2   +   2 ( x   +   3 ) 2   +   11 ( 1   +   x ) ( 1   –   x )   =   6     ⇔   ( 3 x ) 2   –   2 . 3 x . 1   +   1 2   +   2 ( x 2   +   6 x   +   9 )   +   11 ( 1   –   x 2 )   =   6     ⇔   9 x 2   –   6 x   +   1   +   2 x 2   +   12 x   +   18   +   11   –   11 x 2   =   6     ⇔   ( 9 x 2   +   2 x 2   –   11 x 2 )   +   ( - 6 x   +   12 x )   =   6   –   1   –   11   –   18

ó 6x = -24 ó x = -4

Vậy x = -4

Đáp án cần chọn là: A

x^3 -9x^2 +27x -27 -(x^3 -27) +6(x^2 +2x+1) +3x^2 =-33

x^3 -9x^2 +27x -27 -x^3 +27 +6x^2 + 12x+ 6 +3x^2 =-33

39x+6=-33

39x=-39

x=-1

Vậy x=-1

Đặt bt trên là A nha

Đổi |x-1|=|1-x|

Suy ra A=|1-x|+x-2|+|x-3|

Áp dụng BĐTGTTĐ ta có

A=|1-x|+x-2|+|x-3|\(\ge\)|1-x+x-3|=2

Dấu = xảy ra khi   \(\hept{\begin{cases}x-2=0\\1< x< 3\end{cases}}\)đồng thời xảy ra

Vậy x =2

b,

\(\left|3x+\frac{1}{2}\right|\ge0\)

\(\left|3x+\frac{1}{6}\right|\ge0\)

..........

\(\left|3x+380\right|\ge0\)

Suy ra đề bài \(\ge\)0

 suy ra 58x \(\ge\)0

Suy ra \(3x+\frac{1}{2}+3x+\frac{1}{6}+......+3x+380=58x\)

Tự tính nhé hok tốt

a) \(\left(x+2\right)\left(x+3\right)-\left(x+1\right)\left(x+7\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-8x-7=6\)

\(\Leftrightarrow-3x=7\)

\(\Leftrightarrow x=-\frac{7}{3}\)

b) \(\left(8x-3\right)\left(3x+2\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

\(\Leftrightarrow\left(8x-3\right)\left(9x^2+12x+4\right)-4x^2-23x-28=10x^2+3x-1-33\)

\(\Leftrightarrow72x^3+69x^2-4x-12-14x^2-26x+6=0\)

\(\Leftrightarrow72x^3+55x^2-30x-6=0\)

Nghiệm vô tỉ: \(x_1=-1,078...\) ; \(x_2=0,476...\) ; \(x_3=-0,162...\)

a) (x + 2)(x + 3) - (x + 1)(x + 7) = 6

=> x(x + 3) + 2(x + 3) - x(x + 7) - 1(x + 7) = 6

=> x² + 3x + 2x + 6 - x² - 7x - x - 7 = 6

=> x² + 5x + 6 - x² - 7x - x - 7 = 6

=> (x² - x²) + (5x - 7x - x) + (6 - 7) = 6

=> -3x - 1 = 6

=> -3x = 7

=> x = -7/3

b) (8x - 3)(3x + 2)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33

=> (8x - 3)(9x² + 12x + 4) - [4x(x + 4) + 7(x + 4)] = 2x(5x - 1) + 1(5x - 1) - 33

=> 8x(9x² + 12x + 4) - 3(9x² + 12x + 4) - (4x² + 16x + 7x + 28) = 10x² - 2x + 5x - 1 - 33

=> 72x³ + 96x² + 32x - 27x² - 36x - 12  - 4x² - 16x - 7x - 28 - 10x² + 2x - 5x + 1 + 33 = 0

=> 72x³ + (96x² - 27x² - 10x² - 4x²) + (32x - 36x  - 16x -  7x + 2x - 5x)  + (-12  - 28 + 1 +  33) = 0

=> 72x³ + 55x² - 30x - 6 = 0

=> x vô nghiệm