\(\text{Thực hiện phép tính}\)

   \(3\text{ × }\left(\frac{1}{7}+\frac{1}{3}-\frac{3}{14}\right) : \frac{11}{4}\)

\(=3\text{ x }\left(\frac{3}{21}+\frac{7}{21}-\frac{3}{14}\right)\text{ : }\frac{11}{4}\)

\(=3\text{ x }\left(\frac{10}{21}-\frac{3}{14}\right)\text{ : }\frac{11}{4}\)

\(=3\text{ x }\left(\frac{140}{294}-\frac{63}{294}\right)\text{ : }\frac{11}{4}\)

\(=3\text{ x }\frac{77}{294}\text{ : }\frac{11}{4}\)

\(=\frac{231}{294}:\frac{11}{4}\)

\(=\frac{231}{294}\text{ x }\frac{4}{11}\)

\(=\frac{924}{3234}=\text{Phân số này có thể rút gọn đó bạn ! Bạn tự rút gọn nha !}\)

   \(\left(\frac{21}{8}+\frac{1}{2}\right)\text{ : }\frac{5}{16}\)

\(=\left(\frac{21}{8}+\frac{4}{8}\right)\text{ : }\frac{5}{16}\)

\(=\frac{25}{8}\text{ : }\frac{5}{16}\)

\(=\frac{25}{8}\text{ x }\frac{16}{5}\)

\(=\frac{400}{40}=10\)

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

chưa biết

1a) -3x²(2x³ - 2x + 1/3) = -6x⁵ + 6x³ - x²

b) (x⁴ + 2x³ - 2/3).(-3x⁴) = -3x⁸ - 6x⁷ + 2x⁴

c) (x + 3)(x - 4) = x² - 4x + 3x - 12 = x² - x - 12

d)(x - 4)(x² + 4x + 16) = (x - 4)(x² + 4x + 4²) = x³ - 64

e) 4(x - 1/2)(x + 1/2)(4x² + 1) =4(x² - 1/4)(4x²  + 1) = 4(4x⁴ + x² - x² - 1/4) = 4(4x⁴ - 1/4) = 16x⁴ - 1

B2. a) (2 - x)(x² + 2x + 4) + x(x - 3)(x + 4) - x² + 24 = 0

=> 8 - x³ + x(x² + 4x - 3x - 12) - x² + 24 = 0

=> 8 - x³ + x³ + x² - 12x - x² + 24 = 0

=> -12x + 32 = 0

=> -12x = -32

=> x = -32 : (-12) = 8/3

b) (x/2 + 3)(5 - 6x) + (12x - 2)(x/4 + 3) = 0

=> 5x/2 - 3x² + 15 - 18x + 3x² + 36x - x/2 - 6 = 0

=> 20x + 9 = 0

=> 20x = -9

=> x = -9/20

Giúp mk câu hỏi này vs

\(2x\left(3x^2-7x+2\right)\)

\(=6x^3-14x^2+4x\)

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

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

\(\left(3x^2y^2+6x^2y^3-12xy\right)\div3xy\)

\(=xy+2xy-4\)

x^2/x-2+4-4x/x-2 ???

1) 2x(3x^2-7x+2)

<=> \(6x^3-14x^2+4x\)

2) (x-2)(3x^2+2x+4)

<=> \(3x^3-4x^2-8\)

3) (3x^2y^2+6x^2y^3-12xy)÷3xy

<=> xy + 2xy² -4 

4) X^2/x-2+4-4x/x-2

<=> \(\dfrac{x^2-4x+4}{x-2}=\dfrac{\left(x-2\right)^2}{x-2}=x-2\)