x5+y3-(x+y)2=3*x2
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6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
1. Ta có: hằng đẳng thức: \(x^3+y^3+z^3=3xyz\) nếu x+y+z=0
đặt b-c=x, c-a=y, a-b=z⇒x+y+z=0
\(\Rightarrow\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3=3\left(a-b\right)\left(c-a\right)\left(b-c\right)\)
2. \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
3. Tham khảo: https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-y-5-x-5-y-5-thanh-nhan-tu-faq447273.html
\(5,=x^3+2x^2y-7x^2y-14xy^2\\ =x^2\left(x+2y\right)-7xy\left(x+2y\right)\\ =x\left(x-7y\right)\left(x+2y\right)\)
\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^3-x^3y^2\)
\(=\left(x^2+y^2\right)\left(x^3+y^3\right)-\left(xy\right)^2\left(x+y\right)\)
\(=10.26-\left(-3\right)^2.2=...\)
(x+y)5=32
⇔ x5+5x4y+10x3y2+10x2y3+5xy4+y5 = 32
⇔ x5+y5 = 32-5xy(x3+y3)-10x2y2(x+y)
= 32-5.(-3).26-10.(-3)2.2
= 242
2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
\(\text{a) x^2 + y^2 = (x+y)^2 - 2xy = a^2 - 2b}\)
\(\text{b) x^3 + y^3 = (x+y)^3 - 3xy(x+y) = a^3 - 3ab}\)
\(\text{c) x^4 + y^4 = (x^2+y^2)^2 - 2x^2y^2 = (a^2-2b)^2 - 2b^2 = a^4 - 4a^2b + 2b^2}\)
\(\text{d) x^5 + y^5 = (x^3+y^3)(x^2+y^2) - x^2y^2(x+y) = a^5 - 5a^3b + 5ab^2}\)
mình quên , đề là : M =(x+y).x2-y3.(x+y) +(x2- y3) +3 biết x+y+1=0