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

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.

làm ơn giúp em với

\(a,=x^2+6x+9+2x^2+5xy^2=3x^2+6x+5xy^2+9\\ b,=9x^2-12x+4-9x^2+1=-12x+5\)

b: \(=9x^2-12x+4-9x^2+1=-12x+5\)

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

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

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

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

\(\left(x+2y\right)^2-\left(x-2y\right)^2\\ =\left[\left(x+2y\right)-\left(x-2y\right)\right]\left[\left(x+2y\right)+\left(x-2y\right)\right]\\ =\left(x+2y-x+2y\right)\left(x+2y+x-2y\right)\\ =4y.\left(2x\right)\\ =8xy\)

\(\left(3x+y\right)^2+\left(x-y\right)^2\\ =\left[\left(3x\right)^2+2.3x.y+y^2\right]+\left(x^2-2xy+y^2\right)\\ =6x^2+6xy+y^2+x^2-2xy-y^2\\ =7x^2+4xy\)

\(-\left(x+5\right)^2-\left(x-3\right)^2\\ =-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\\ =-x^2-10x-25-x^2+6x-9\\ =-2x^2-4x-34\)

\(\left(3x-2\right)^2-\left(3x-1\right)^2\\ =\left[\left(3x-2\right)-\left(3x-1\right)\right]\left[\left(3x-2\right)+\left(3x-1\right)\right]\\ =\left(3x-2-3x+1\right)\left(3x-2+3x-1\right)\\ =-1.\left(6x-3\right)\\ =-6x+3\)

Sửa: \(A=\left(3x+2\right)^2+\left(2x-7\right)^2-2\left(3x+2\right)\left(2x-7\right)\)

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

\(A=\left[\left(3x+2\right)-\left(2x-7\right)\right]^2\)

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

\(A=\left(x+9\right)^2\)

Thay \(x=-19\) vào A ta có:

\(A=\left(-19+9\right)^2=\left(-10\right)^2=100\)

Vậy: ...

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

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

\(=4x^2\)

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

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

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

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

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

(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