Câu hỏi đâu em ơi

\(B=9x^4-\left(2x+1\right)^2-\left(9x^4+6x^2+1\right)\\ =9x^4-4x^2-4x-1-9x^4-6x^2-1\\ =-10x^2-4x-2\)

\(B=\left(3x^2+1-2x\right)\left(3x^2+1+2x\right)-\left(3x^2+1\right)^2\\ B=\left(3x^2+1\right)^2-4x^2-\left(3x^2+1\right)^2=-4x^2\)

ĐKXĐ: \(x\ge0\)
\(\dfrac{2\sqrt{x}+x^2+1}{x+2}=\dfrac{\left(\sqrt{x}+1\right)^2}{x+2}\)

\(\dfrac{S}{\dfrac{1}{v1}+\dfrac{1}{v2}}\)= \(\dfrac{S.v1.v2}{v1+v2}\)

\(\dfrac{2sin8a-sin16a}{2sin8a+sin16a}=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(1-2sin^24a\right)}=\dfrac{2sin^24a}{2-2sin^24a}=\dfrac{sin^24a}{1-sin^24a}=\dfrac{sin^24a}{cot^24a}=tan^24a\)

\(=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}\)

\(=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(2cos^24a-1\right)}=\dfrac{2sin^24a}{2cos^24a}=tan^24a\)

\(c,=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x-2}\\ d,=\dfrac{\left(2-x-3\right)\left(2+x+3\right)}{\left(x+5\right)^2}=\dfrac{\left(x+5\right)\left(-x-1\right)}{\left(x+5\right)^2}=\dfrac{-x-1}{x+5}\)

\(A=\left(\dfrac{2-x}{2+x}-\dfrac{16}{4-x^2}-\dfrac{2+x}{2-x}\right)\)

\(\Rightarrow A=\left(\dfrac{\left(2-x\right)^2}{\left(2+x\right)\left(2-x\right)}-\dfrac{16}{\left(2+x\right)\left(2-x\right)}-\dfrac{\left(2+x\right)^2}{\left(2+x\right)\left(2-x\right)}\right)\)\(\Rightarrow A=\left(\dfrac{4-4x+x^2}{\left(2+x\right)\left(2-x\right)}-\dfrac{16}{\left(2+x\right)\left(2-x\right)}-\dfrac{4+4x+x^2}{\left(2+x\right)\left(2-x\right)}\right)\)

\(\Rightarrow A=\dfrac{4-4x+x^2-16-4-4x-x^2}{\left(2+x\right)\left(2-x\right)}\)

\(\Rightarrow A=\dfrac{-8x-16}{\left(2+x\right)\left(2-x\right)}\)

\(\Rightarrow A=\dfrac{-8\left(x+2\right)}{\left(2+x\right)\left(2-x\right)}\)

\(\Rightarrow A=\dfrac{-8}{2-x}\)

\(\Rightarrow A=\dfrac{8}{x-2}\)