\(\left(5n-7\right)^2-9\)

\(=\left(5n-7\right)^2-3^2\)

\(=\left(5n-7-3\right)\left(5n-7+3\right)\)

\(=\left(5n-10\right)\left(5n-4\right)\)

\(5\left(n-2\right)\left(5n-4\right)⋮5\)với mọi số nguyên n \(\left(đpcm\right)\)

a)\(=5c\left(a^2-9b^2\right)+5c^2.\left(3b-a\right)\))\

\(=5c\left(a-3b\right)\left(a+3b\right)-5c^2.\left(a-3b\right)\)

= \(5c.\left(a-3b\right)\left(a+3b-c\right)\)

b)\(=xy.\left(3x-y\right)-xyz.\left(3x-y\right)\)

\(=xy.\left(3x-y\right)\left(1-z\right)\)

c) \(=a^2m^2+2ambp+b^2p^2-a^2p^2-2ambp-b^2m^2\)

\(=a^2m^2-a^2p^2+b^2p^2-b^2m^2\)

\(=a^2.\left(m^2-p^2\right)+b^2\left(p^2-m^2\right)\)

\(=a^2.\left(m^2-p^2\right)-b^2\left(m^2-p^2\right)\)

\(=\left(a^2-b^2\right)\left(m^2-p^2\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(m-p\right)\left(m+p\right)\)

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

\(=\frac{x^6}{y^9}-3.\frac{x^4}{y^6}.\frac{y^3}{z^2}+3\frac{x^2}{y^3}.\frac{y^9}{z^4}-\frac{y^9}{z^6}=\frac{x^6}{y^9}-\frac{3x^4}{y^3.z^2}+\frac{3x^2y^6}{z^4}-\frac{y^9}{z^6}\)

\(\left(\frac{x^2}{y^3}-\frac{y^3}{z^2}\right)^3\)

\(=\left(\frac{x^2}{y^3}\right)^3-3.\left(\frac{x^2}{y^3}\right)^2.\frac{y^3}{z^2}+3.\frac{x^2}{y^3}.\left(\frac{y^3}{z^2}\right)^2-\left(\frac{y^3}{z^2}\right)^3\)

\(=\frac{x^6}{y^9}-3.\frac{x^4}{y^6}.\frac{y^3}{z^2}+3.\frac{x^2}{y^3}.\frac{y^6}{z^4}-\frac{y^9}{z^6}\)

\(=\frac{x^6}{y^9}-\frac{3x^4}{y^3z^2}+\frac{3x^2y^3}{z^4}-\frac{y^9}{z^6}\)

\(=\frac{x^6z^6-3x^4y^6z^4+3x^2y^{12}z^2-y^{18}}{y^9z^6}\)

\(a)\)

\(\left(3-x\right)^2-x\left(x-4\right)=2x-5\)

\(\Leftrightarrow\left(9-6x+x^2\right)-\left(x^2-4x\right)=2x-5\)

\(\Leftrightarrow9-2x=2x-5\)

\(\Leftrightarrow4x=14\)

\(\Leftrightarrow x=\frac{7}{2}\)

\(b)\)

\(x^2-2x+1=25x^2\)

\(\Leftrightarrow x^2-2x+1-25x^2=0\)

\(\Leftrightarrow-24x^2-2x+1=0\)

\(\Leftrightarrow-\left(24x^2+2x-1\right)=0\)

\(\Leftrightarrow24x^2+2x-1=0\)

\(\Leftrightarrow x=\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-1}{4}\end{cases}}\)

\(c)\)

\(4x^2-4x=24\)

\(\Leftrightarrow x^2-x=6\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow x=\orbr{\begin{cases}x=\left(-2\right)\\x=3\end{cases}}\)

a)\(\left(3-x\right)^2-x\left(x-4\right)=2x-5\)

\(< =>\left(9-6x+x^2\right)-\left(x^2-4x\right)=2x-5\)

\(< =>9-6x+x^2-x^2+4x=2x-5\)

\(< =>-6x+4x-2x=-5-9\)

\(< =>-4x=-14\)

\(< =>x=\frac{7}{2}\)

b)\(x^2-2x+1=25x^2\)

\(< =>\left(x-1\right)^2-25x^2=0\)

\(< =>\left(x-1\right)^2-\left(5x\right)^2=0\)

\(< =>\left(x-1-5x\right)\left(x-1+5x\right)=0\)

\(< =>\orbr{\begin{cases}-4x-1=0\\6x-1=0\end{cases}}\)

\(< =>\orbr{\begin{cases}-4x=1\\6x=1\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{6}\end{cases}}\)

c)\(4x^2-4x=24\)

câu này mk ko biết làm @@ sorry nha :))

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

\(\Leftrightarrow x^2+6x+9-x^2+1=6x+10=2\)

\(\Leftrightarrow6x=2-10=-8\Leftrightarrow x=-\frac{8}{6}=-\frac{4}{3}\)

Vậy ......

a.\(\left(x+2\right)\left(2-x\right)-\left(2x-1\right)\left(x+3\right)=4-x^2-\left(2x^2+5x-3\right)=7-5x-x^2\)

b.\(\left(x+1\right)^2-2\left(x^2-1\right)+\left(x-1\right)^2=x^2+2x+1-2x^2+2+x^2-2x+1=4\)

c.\(\left(x+y\right)^3-\left(x-y\right)^3-6x^2y=x^3+3x^2y+3xy^2+y^3-\left(x^3-3x^2y+3xy^2-y^3\right)-6x^2y\)

\(=2y^3\)

2.\(\left(x^2-16y^2\right)-3x+12y=\left(x-4y\right)\left(x+4y\right)-3\left(x-4y\right)=\left(x-4y\right)\left(x+4y-3\right)\)

4.\(x^3+6x^2+12x+8=\left(x+2\right)^3\)

bạn ơi  bạn viết sai đề có đề nào viết  thế này đâu?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

Uchiha Itachi

a. \(A=4x-x^2+3=7-\left(x^2-4x\right)+4=7-\left(x-2\right)^2\le7\)

b.\(B=x-x^2=\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)

c.\(C=2x-2x^2-5=-\frac{9}{2}-2\left(x^2-x+\frac{1}{4}\right)=-\frac{9}{2}-2\left(x-\frac{1}{2}\right)^2\le-\frac{9}{2}\)