PTĐTTNT`3x-3y+2x^2-2y^2`
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\(3x^2y-12x^3y^2=3x^2y\left(1-4xy\right)\)
\(2x^2\left(x-y\right)-4xy\left(x-y\right)\)
\(=2x\left(x-y\right)\left(x-2y\right)\)
1) \(x^2-2x+1+x^2y-xy=\left(x-1\right)^2+xy\left(x-1\right)=\left(x-1\right)\left(x+xy-1\right)\)
2) \(x^2+6x+9+x^2y+3xy\)
\(=\left(x+3\right)^2+xy\left(x+3\right)\)
\(=\left(x+3\right)\left(x+xy+3\right)\)
\(\left(2x+3y\right)^2+\left(3x-2y\right)^2-2.\left(2x+3y\right).\left(3x-2y\right)\)
\(=\left(2x+3y\right)^2-2.\left(2x+3y\right).\left(3x-2y\right)+\left(3x-2y\right)^2\)
\(=[\left(2x+3y\right)-\left(3x-2y\right)]^2\)
\(=\left(2x+3y-3x+2y\right)^2\)
\(=\left(5y-x\right)^2\)
em ko lam dc anh ehh
em chua gap bai nao nhu the nay. noi dung hon la chua den lop lam haha
Tìm x:
\(8x^2-\left(2x+5\right)\left(4x-2\right)-9=0\)
\(\Leftrightarrow8x^2-\left(8x^2-4x+20x-10\right)-9=0\)
\(\Leftrightarrow8x^2-8x^2+4x-20x+10-9=0\)
\(\Leftrightarrow-16x+1=0\)
\(\Leftrightarrow-16x=-1\)
\(\Leftrightarrow x=\dfrac{-1}{-16}=\dfrac{1}{16}\)
Vậy \(x=\dfrac{1}{16}\)
Bài 1:
\(a,8x^2-\left(2x+5\right)\left(4x-2\right)-9=0\)
\(\Rightarrow8x^2-\left(8x^2+16x-10\right)-9=0\)
\(\Rightarrow8x^2-8x^2-16x+10-9=0\)
\(\Rightarrow-16x+1=0\)
\(\Rightarrow x=\dfrac{1}{16}\)
\(=3\left(x-y\right)+2\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(3+2x+2y\right)\)