bạn giải đi rùi mình tick cho

a, \(x^2-7xy+12y^2\)

\(=x^2-3xy-4xy+12y^2\)

\(=x\left(x-3y\right)-4y\left(x-3y\right)\)

\(=(x-4y)(x-3y)\)

b, \(x^3-3x+2\)

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

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

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

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

\(=\left[x\left(x+2\right)-\left(x+2\right)\right]\left(x-1\right)\)

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

\(x+y=3\Rightarrow x^2+2xy+y^2=9\)

\(\Rightarrow2xy=4\Rightarrow xy=2\)

Ta có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.3=9\)

Vậy...

\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(\Rightarrow a^2-ab-ac+ad+ab-b^2-bc+bd+ca-bc-c^2+cd+da-db-dc+d^2=a^2+ab-ac-ad-ba-b^2+bc+bd+ca+bc-c^2-cd-da-bd+cd+d^2\) \(\Rightarrow a^2-b^2-c^2+d^2+2ad-2bc=a^2-b^2-c^2+d^2-2ad+2bc\)

\(\Rightarrow2ad-2bc=2bc-2ad\)

\(\Rightarrow ad-bc=bc-ad\)

\(\Rightarrow2ad=2bc\)

\(\Rightarrow ad=bc\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Rightarrowđpcm\)

Bài 1: xem lại đề

Bài 2:

\(\left(10^{12}+25\right)^2-\left(10^{12}-25\right)^2=10^n\)

\(\Rightarrow\left(10^{12}+25-10^{12}+25\right)\left(10^{12}+25+10^{12}-25\right)=10^n\)

\(\Leftrightarrow50.2.10^{12}=10^n\)

\(\Leftrightarrow10^{14}=10^n\Rightarrow n=14\)

Vậy n = 14

Ko có câu hỏi à? Làm sao giải được!

\(x-2\sqrt{x}=0\)

\(\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(x-2\sqrt{x}=0\)

\(\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\)

\(d,3x^2+6x-3y^2+3\)

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

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

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

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

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

c, \(4xy-x^2-4y^2+16\)

\(=-\left(x^2-4xy+4y^2-16\right)\)

\(=-\left[\left(x-2y\right)^2-16\right]\)

\(=16-\left(x-2y\right)^2\)

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