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

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

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

\(=\left(-4\right)^2=16\)

(2x + 1)2 + (3x – 1)2 + 2(2x + 1)(3x – 1)

= (2x + 1)2 + 2.(2x + 1)(3x – 1) + (3x – 1)2

= [(2x + 1) + (3x – 1)]2

= (2x + 1 + 3x – 1)2

= (5x)2

= 25x2

\(=\left(4x-5+3x+5\right)^2=\left(7x\right)^2=49x^2\)

A = (2x + 3)² + (3x - 2)² + 2(2x + 3)(3x - 2) = (2x + 3 + 3x - 2)² = (5x + 1)²

( 3 x   +   1 ) 2   –   2 ( 3 x   +   1 ) ( 3 x   +   5 )   +   ( 3 x   +   5 ) 2 =   ( ( 3 x   +   1 )   –   ( 3 x   +   5 ) ) 2 =   ( 3 x   +   1   –   3 x   –   5 ) 2     =   ( - 4 ) 2   =   16

Đáp án cần chọn là: B

\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)

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

b: \(=\dfrac{\left(x-1\right)\cdot\left(x+1\right)}{\left(x-1\right)^2}=\dfrac{x+1}{x-1}\)

c: \(=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\)

Bạn lưu ý viết đề bằng công thức toán để được hỗ trợ tốt hơn. Viết thế này nhìn khá khó đọc.

Để viết công thức toán bạn nhấn biểu tượng $\sum$ góc trái khung soạn thảo.

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

\(=9x^2-12x+4-4x^2+9\)

\(=5x^2-12x+13\)

b) \(3x\left(5x-2\right)-\left(2x^2-1\right)\left(2-x\right)\)

\(=15x^2-6x-\left(4x^2-2x^3-2+x\right)\)

\(=15x^2-6x-4x^2+2x^3+2-x\)

\(=11x^2-7x+2x^3+2\)