b: \(B=\left(x+2\right)^2-\left(2x-1\right)^2\)

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

\(=-3x^2+8x+3\)

*Đa thức \(B=-4x^3-2x^2-2+2x\left(3+x\right)-9x+2x^3\)

Ta có: \(B=-4x^3-2x^2-2+2x\left(3+x\right)-9x+2x^3\)

\(=-2x^3-2x^2-2+6x+2x^2-9x\)

\(=-2x^3-3x-2\)

*Đa thức \(C=x^3-2x\left(3x-1\right)+4\)

Ta có: \(C=x^3-2x\left(3x-1\right)+4\)

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

\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)

\(a,\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)

\(b,\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)

\(c,\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)

\(d,\left(\dfrac{2x+2}{3x}\right)^2=\dfrac{\left(2x+2\right)^2}{9x^2}=\dfrac{4x^2+8x+4}{9x^2}\)

\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)

\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)

a. \(\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)

b. \(\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)

c. \(\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)

d. \(\left(2x+\dfrac{2}{3x}\right)^2=4x^2+\dfrac{8}{3}+\dfrac{4}{9x^2}\)

e. (a - 1)(a + 1) = a² - 1

f. (5x² - 2)(5x² + 2) = 25x⁴ - 4 

g. (2a - 3)(2a + 3) = 4a² - 9

\(A=\left(2x-3y\right)\left(5x-2y\right)=10x^2-4xy-15xy+6y^2=10x^2-19xy+6y^2\)

a) Do \(\left|1+2x\right|\ge0\Rightarrow\dfrac{-1}{4}\left|1+2x\right|\le0\)

\(\Rightarrow A=2,25-\dfrac{1}{4}\left|1+2x\right|\le2,25\)

\(maxA=2,25\Leftrightarrow x=-\dfrac{1}{2}\)

b) Do \(\left|2x-3\right|\ge0\Rightarrow3+\dfrac{1}{2}\left|2x-3\right|\ge3\)

\(\Rightarrow B=\dfrac{1}{3+\dfrac{1}{2}\left|2x-3\right|}\le\dfrac{1}{3}\)

\(maxB=\dfrac{1}{3}\Leftrightarrow x=\dfrac{3}{2}\)

mình ghi nhầm đề bài là Tìm giá trị lớn nhất nhé

\(\left(3x^2-2x+1\right)\left(x^2+2x+3\right)-4x\left(x^2-1\right)-3x^2\left(x^2+2\right)=3x^4+6x^3+9x^2-3x^3-4x^2-6x-4x^3+4x-3x^4-6x^2=0\)