Pha tich da thuc thanh nhan tu
\(a,6x^4-11x^2+3\)
\(b,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
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Câu a :
\(\left(x-5\right)^2+\left(x-5\right)\left(x+5\right)-\left(5-x\right)\left(2x+1\right)\)
\(=x^2-10x+25+x^2-25-10x-5+2x^2+x\)
\(=4x^2-19x-5\)
Câu b :
\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=12x^2-9x-8x+6-2x+2+3x^2-3x-6x^2-6x+4x+4\)
\(=9x^2-24x+2\)
(*)\(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(x-3\right)\left(3x-2\right)\)
(*)\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-5\right)\left(x-1\right)\)
(*)\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+1+x\right)\left(x^2+1-x\right)\)
(*)\(x^4-4x^2+3=x^4-x^2-3x^2+3=x^2\left(x^2-1\right)-3\left(x^2-1\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
(*)\(6x^2+7xy+2y^2=6x^2+4xy+3xy+2y^2=2x\left(3x+2y\right)+y\left(3x+2y\right)=\left(2x+y\right)\left(3x+2y\right)\)
a, \(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(3x-2\right)\left(x-3\right)\)
b, \(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
c, \(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
d, \(x^4-4x^2+3=x^4-4x^2+4-1=\left(x^2-2\right)^2-1=\left(x^2-1\right)\left(x^2-3\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
e, \(6x^2+7xy+2y^2=6x^2+3xy+4xy+2y^2=3x\left(2x+y\right)+2y\left(2x+y\right)=\left(2x+y\right)\left(3x+2y\right)\)
\(9\left(x+1\right)^2-\left(3x-2\right)^2\)
\(=9\left(x^2+2x+1\right)-\left(9x^2-12x+4\right)\)
\(=9x^2+18x+9-9x^2+12x-4\)
\(=30x+5\)
\(=5\left(6x+1\right)\)
\(9\left(x+1\right)^2-\left(3x-2\right)^2\)
\(=\left[3\left(x+1\right)+3x-2\right]\left[3\left(x+1\right)-3x+2\right]\)
\(=\left(3x+3+3x-2\right)\left(3x+3-3x+2\right)\)
\(=5\left(6x+1\right)\)
Đặt \(x^2+3x+1=t\)
\(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
\(=t\left(t-4\right)-5\)
\(=t^2-4t-5\)
tự làm nốt ý này nhé.
những ý kia lát nx mình làm.
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
\(\Leftrightarrow\left(3x-1\right)^2-4^2=0\)
\(\Leftrightarrow\left(3x-1-4\right)\left(3x-1+4\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\Leftrightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
Chúc bạn học tốt!!!
\(6x^4-11x^2+3=6x^4-9x^2-2x^2+3\)
\(=3x^2\left(2x^2-3\right)-\left(2x^2-3\right)=\left(2x^2-3\right)\left(3x^2-1\right)\)