giúp mình với mình cần gấp

a)A=(2x+y)²-(y-2x)²

\(A=(2x+y-y+2x)(2x+y+y-2x)\)

\(A=4x.2y=8xy\)

b)B=x²-y²+(x-y)²

\(B=x^2-y^2+x^2-2xy+y^2\)

\(B=2x^2-2xy\)

Lần sau bạn đăng vào môn Toán nhé!!!

\(a,A=7\sqrt{5}+6\sqrt{5}-5\sqrt{5}-6\sqrt{5}=2\sqrt{5}\\ b,B=12-5\cdot2=2\\ c,C=\left[2-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\right]\left[2+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}+1}\right]\\ C=\left(2-\sqrt{7}\right)\left(2+\sqrt{7}\right)=4-7=-3\)

\(A=\dfrac{3}{2\left(2x-1\right)}\cdot x^2\left|2x-1\right|\cdot2\sqrt{2}\)

\(=\pm3\sqrt{2}x^2\)

\(B=\dfrac{a-b}{b^2}\cdot\dfrac{b^2\cdot\left|a\right|}{\left|a-b\right|}\)

\(=\pm\left|a\right|\)

a) \(\sqrt{4a^2}=2\left|a\right|=-2a\) ( do a<0)

b) \(\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=\left|2x-3\right|=3-2x\)(do \(x< \dfrac{3}{2}\Leftrightarrow2x-3< 0\))

a: M=A:B

\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{1}=\dfrac{x+7}{\sqrt{x}+3}\)

b: \(M=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)

=>\(M=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)

Dấu = xảy ra khi (căn x+3)^2=16

=>căn x+3=4

=>x=1

\(b.\)

\(=\sqrt{\left(3a\right)^2\cdot\left(b-2\right)^2}\)

\(=\left|3a\right|\cdot\left|b-2\right|\)

Với : \(a=2,b=-\sqrt{3}\)

\(2\cdot3\cdot\left(-\sqrt{3}-2\right)=6\cdot\left(-\sqrt{3}-2\right)\)

\(a.\)

\(=\sqrt{4\cdot\left(3x+1\right)^2}=2\cdot\left|3x+1\right|\)

Với : \(x=-\sqrt{2}\)

\(2\cdot\left|3\cdot-\sqrt{2}+1\right|=2\cdot\left|1-\sqrt{6}\right|\)

a) Ta có:\(\sqrt{4\left(9x^2+6x+1\right)^2}\)

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

\(=2\cdot\left(-3\cdot\sqrt{2}+1\right)^2\)

\(=2\left(19-6\sqrt{2}\right)\)

\(=38-12\sqrt{2}\)

b) Ta có: \(\sqrt{9a^2\left(b^2-4b+4\right)}\)

\(=3\left|a\right|\left|b-2\right|\)

\(=3\cdot\left|2\right|\cdot\left|-\sqrt{3}-2\right|\)

\(=6\left(2+\sqrt{3}\right)=12+6\sqrt{3}\)