Câu hỏi đâu em ơi

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

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

mà

a: \(\sqrt{\frac47}-10\sqrt{\frac{7}{25}}-6\cdot\sqrt{\frac{1}{28}}\)

\(=\frac{2}{\sqrt7}-10\cdot\frac{\sqrt7}{5}-6\cdot\frac{1}{2\sqrt7}\)

\(=\frac27\sqrt7-2\sqrt7-\frac{3\sqrt7}{7}=-\frac{15}{7}\sqrt7\)

b: \(\left(\sqrt{10}+\sqrt2\right)\cdot\sqrt{3-\sqrt5}\)

\(=\left(\sqrt5+1\right)\cdot\sqrt{6-2\sqrt5}\)

\(=\left(\sqrt5+1\right)\cdot\sqrt{\left(\sqrt5-1\right)^2}=\left(\sqrt5+1\right)\left(\sqrt5-1\right)\)

=5-1

=4

c: \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt6+\sqrt2}\)

\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{14-2\sqrt{33}}}{2\left(\sqrt3+1\right)}\)

\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-\sqrt3\right)^2}}{2\left(\sqrt3+1\right)}=\frac{\sqrt{11}+1-\sqrt{11}+\sqrt3}{2\left(1+\sqrt3\right)}\)

\(=\frac{1+\sqrt3}{2\left(1+\sqrt3\right)}=\frac12\)

d: \(\frac{5\sqrt3-3\sqrt5}{\sqrt5-\sqrt3}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt5+3\sqrt3}{\sqrt5+\sqrt3}\)

\(=\frac{\sqrt{15}\left(\sqrt5-\sqrt3\right)}{\sqrt5-\sqrt3}+\frac{2\left(4-\sqrt{15}\right)}{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}-\frac{\left(\sqrt5+\sqrt3\right)\left(8-\sqrt{15}\right)}{\sqrt5+\sqrt3}\)

\(=\sqrt{15}+2\left(4-\sqrt{15}\right)-\left(8-\sqrt{15}\right)\)

\(=2\sqrt{15}-8+8-2\sqrt{15}\)

=0

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

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

b: \(\sqrt{1\frac{9}{16}}+\frac{\sqrt{15}-\sqrt{12}}{\sqrt5-2}-\frac{3}{\sqrt3}\)

\(=\sqrt{\frac{25}{16}}+\frac{\sqrt3\left(\sqrt5-2\right)}{\sqrt5-2}-\sqrt3\)

\(=\frac54+\sqrt3-\sqrt3=\frac54\)

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

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

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

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

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

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

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

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

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

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

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