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Rút gọn
\(\left(2x+1\right)\left(4x^2-3x+1\right)+\left(2x-1\right)\left(4x^2+3x+1\right)\)
\(=8x^3-12x^2+2x+4x^2-3x+1+8x^3+12x^2+2x-4x^2-3x-1\)
\(=16x^3-2x\)
Phân tích đa thức thnahf nhân tử
\(4y^2+16y-x^2-8x\)
\(=\left(4y^2-x^2\right)+\left(16y-8x\right)\)
\(=\left(2y-x\right)\left(2y+x\right)+8\left(2y-x\right)\)
\(=\left(2y-x\right)\left(2y+x+8\right)\)
Chứng minh .............
Có: \(x^2+x+1=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì: \(\left(x+\frac{1}{2}\right)^2\ge0\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Kết luận......
\(a,VT=\left(a+b+c\right)\left(a-b+c\right)\)
\(=\left(a+c+b\right)\left(a+c-b\right)\)
\(=\left(a+c\right)^2-b^2\)
\(=a^2+2ac+c^2-b^2=VP\)
\(b,VT=\left(3x+2y\right)\left(3x-2y\right)-\left(4x-2y\right)\left(4x+2y\right)\)
\(=9x^2-4y^2-16x^2+4y^2=-7x^2=VP\)
\(c,VT=x^3-1-x^3-1=-2=VP\)
\(d,VT=8x^3+1-8x^3+1=2=VP\)
\(e,VT=\left(x^2+2xy+4y^2\right)\left(x-2y-2x+1\right)\)
\(=\left(x^2+2xy+4y^2\right)\left(-x-2y+1\right)\)
\(=-x^3-2x^2y+x^2-2x^2y-4xy^2+2xy-4xy^2-8y^3+4y^2\)
( bn kiểm tra lại đề nhé)
a) \(=\left(x-2y\right)\left(x^2+5x\right)\)
b) \(=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c) \(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(3-x+3\right)\)
\(=\left(x+3\right)\left(6-x\right)\)
e) \(=\left(x^2-\frac{1}{3}x\right)\left(x^2+\frac{1}{3}x\right)\)
f) \(=2x\left(x-y\right)-16\left(x-y\right)\)
\(=2\left(x-y\right)\left(x-8\right)\)
2: \(=3\left(x-2y\right)+y\left(x-2y\right)=\left(x-2y\right)\left(y+3\right)\)
3: \(=x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
4: \(x^2+2x+1-16y^2\)
\(=\left(x+1\right)^2-16y^2\)
\(=\left(x+1+4y\right)\left(x+1-4y\right)\)
5: \(x^2-y^2+5x+5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+5\right)\)
6: \(=25-\left(x^2-2xy+y^2\right)\)
\(=25-\left(x-y\right)^2\)
\(=\left(5-x+y\right)\left(5+x-y\right)\)
a) \(x^2-6x+9=x^2-2\cdot x\cdot3+3^2=\left(x-3\right)^2\)
b) \(4x^2-12xy+9y^2=\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2=\left(2x-3y\right)^2\)
c) \(4x^2-2x+1=\left(2x-1\right)^2\)
d) \(x^2+8xy+16y^2=\left(x+4y\right)^2\)
Ta có công thức :
\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Rightarrow m^2-n^2=\left(m-n\right)\left(m+n\right)\)
a) Ta có: \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
b) Ta có: \(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
c) Ta có: \(2x^2-5xy+3y^2\)
\(=2x^2-2xy-3xy+3y^2\)
\(=2x\left(x-y\right)-3y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-3y\right)\)
d) Ta có: \(16y^3-2x^3-6x\left(x+1\right)-2\)
\(=16y^3-2x^3-6x^2-6x-2\)
\(=2\left[8y^3-x^3-3x^2-3x-1\right]\)
\(=2\left[\left(2y\right)^3-\left(x^3+3x^2+3x+1\right)\right]\)
\(=2\left[\left(2y\right)^3-\left(x+1\right)^3\right]\)
\(=2\left(2y-x-1\right)\left[\left(2y\right)^2+2y\left(x+1\right)+\left(x+1\right)^2\right]\)
\(=2\left(2y-x-1\right)\left(4y^2+2xy+2y+x^2+2x+1\right)\)
\(x^2+2x+1-16y^2\)
\(=\left(x+1\right)^2-\left(4y\right)^2\)
\(=\left(x+1-4y\right)\left(x+1+4y\right)\)