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

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

\(X=1\)

( 2x + 1 )² = ( x + 2 )²

<=> ( 2x + 1 )² - ( x + 2 )² = 0

<=> [ ( 2x + 1 ) - ( x + 2 ) ][ ( 2x + 1 ) + ( x + 2 ) ] = 0

<=> ( 2x + 1 - x - 2 )( 2x + 1 + x + 2 ) = 0

<=> ( x - 1 )( 3x + 3 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\3x+3=0\end{cases}}\Leftrightarrow x=\pm1\)

\(\frac{1}{9}x^2-\frac{1}{16}=\left(\frac{1}{3}x\right)^2-\left(\frac{1}{4}\right)^2=\left(\frac{1}{3}x-\frac{1}{4}\right)\left(\frac{1}{3}x+\frac{1}{4}\right)\)

\(2-x^2=\left(\sqrt{2}\right)^2-x^2=\left(\sqrt{2}-x\right)\left(\sqrt{2}+x\right)\)

\(x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

\(5-x^2=\left(\sqrt{5}\right)^2-x^2=\left(\sqrt{5}-x\right)\left(\sqrt{5}+x\right)\)

\(\frac{1}{4}x^4-9\)

\(=\left(\frac{1}{2}x^2\right)^2-3^2\)

\(=\left(\frac{1}{2}x^2-3\right)\left(\frac{1}{2}x^2+3\right)\)

HAHA :) 4 năm vẫn chưa có ai trả lời, nhục !

a) \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)

\(A=y\left(x^4-y^4\right)-y\left(y^4-y^4\right)=0\)

=> đpcm

b) \(B=\left(\frac{1}{3}+2x\right)\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\) (đã sửa đề)

\(B=\left(\frac{1}{27}+8x^3\right)-\left(8x^3-\frac{1}{27}\right)\)

\(B=\frac{2}{27}\)

=> đpcm

c) \(C=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\) (đã sửa đề)

\(C=x^3-3x^2+3x-1-x^3+1+3x^2-3x\)

\(C=0\)

=> đpcm

\(M=4x^2+9y^2-12xy\)

\(M=\left(4x^2+12xy+9y^2\right)-24xy\)

\(M=\left(2x+3y\right)^2-24xy\)

\(M=2^2-288=-284\)

Ta có: \(x-y=7\Rightarrow x=y+7\)

Thay vào: \(y\left(y+7\right)=60\)

\(\Leftrightarrow y^2+7y-60=0\)

\(\Leftrightarrow\left(y-5\right)\left(y+12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=5\\y=-12\left(ktm\right)\end{cases}}\Rightarrow y=5\Rightarrow x=12\)

Từ đó:

\(N=5^4+12^4=625+20736=21361\)