Phân tích đa thức thành nhân tử : (x + y + z)2 + (x + y – z)2 – 4z2
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Những câu hỏi liên quan
TT
0
DN
4
TG
31 tháng 8 2021
\(A=-x-z\left(x-y\right)+y=-x-xz+zy+y=-x\left(1+z\right)+y\left(1+z\right)=\left(1+z\right)\left(y-x\right)\)
DN
2
31 tháng 8 2021
\(4\left(x^2y^2+z^2t^2+2xyzt\right)-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left[2\left(xy+zt\right)\right]^2-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy+2zt\right)^2-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy+2zt-x^2-y^2+z^2+t^2\right)\left(2xy+2zt+x^2+y^2-z^2-t^2\right)^2\)
31 tháng 8 2021
Ta có: \(4\left(x^2y^2+2xyzt+z^2t^2\right)-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy+2tz\right)^2-\left(x^2+y^2-z^2-t^2\right)^2\)
\(=\left(2xy+2tz-x^2-y^2+z^2+t^2\right)\left(2xy+2tz+x^2+y^2-z^2-t^2\right)\)
\(=\left[-\left(x^2-2xy+y^2\right)+\left(z^2+2tz+t^2\right)\right]\left[\left(x^2+2xy+y^2\right)-\left(t^2-2tz+z^2\right)\right]\)
\(=\left(z+t-x+y\right)\left(z+t+x-y\right)\left(x+y-t+z\right)\left(x+y+t-z\right)\)
12 tháng 9 2021
\(4(x^2y^2+z^2t^2+2xyzt)-(x^2+y^2-z^2-t^2)^2\)
\(=[2(xy+zt]^2-(x^2+y^2-z^2-t^2)^2\)
\(=(2xy+2zt)^2-(x^2+y^2-z^2-t^2)^2\)
\(=(2xy+2zt-x^2-y^2+z^2+t^2)(2xy+2zt+x^2+y^2-z^2-t^2)^2\)
TT
0
DN
2
4 tháng 9 2021
\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)
\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)
4 tháng 9 2021
\(\left(x+y\right)^2+3\left(x+y\right)-10\)
\(=\left(x+y\right)^2+5\left(x+y\right)-2\left(x+y\right)-10\)
\(=\left(x+y+5\right)\left(x+y-2\right)\)
DN
2
2 tháng 9 2021
\(\left(xy+1\right)^2-\left(x+y\right)^2=\left(xy+1-x-y\right)\left(xy+1+x+y\right)=\left[x\left(y-1\right)-\left(y-1\right)\right]\left[x\left(y+1\right)+\left(y+1\right)\right]=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
2 tháng 9 2021
\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\left(xy-x-y+1\right)\left(xy+1+x+y\right)\)
\(=\left(y-1\right)\left(x-1\right)\left(y+1\right)\left(x+1\right)\)
DN
1
12 tháng 9 2021
\(=x\left[x^2\left(x-y\right)^2-36y^2\right]\\ =x\left[x\left(x-y\right)-6y\right]\left[x\left(x-y\right)+6y\right]\\ =x\left(x^2-xy-6y\right)\left(x^2-xy+6y\right)\)
\(\left(x+y+z\right)^2+\left(x+y-z\right)^2-4z^2=\left(x+y+z\right)^2+\left(x+y-z-2z\right)\left(x+y-z+2z\right)=\left(x+y+z\right)^2+\left(x+y-3z\right)\left(x+y+z\right)=\left(x+y+z\right)\left(x+y+z+x+y-3z\right)=\left(x+y+z\right)\left(2x+2y-2z\right)=2\left(x+y+z\right)\left(x+y-z\right)\)
Ta có:
(x + y + z)2 + (x + y – z)2 – 4z2
\(=\left(x+y-z\right)^2+\left(x+y-z\right)\left(x+y+3z\right)\)
\(=\left(x+y-z\right)\left(x+y+3z+x+y-z\right)\)
\(=2\left(x+y-z\right)\left(x+y+z\right)\)