Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz
= xy(X + y + z) + yz(x + y + z) + xz(X + y + z)
= (x + y +z)(xy + yz+ xz)
b) xy(x + y) - yz(y + z) - xz(z - x)
= x2y + xy2 - y2z - yz2 - xz2 + x2z
= x2(y + z) - yz(y + z) + x(y2 - z2)
= x2(y + z) - yz(y + z) + x(y + z)(y - z)
= (y + z)(x2 - yz + xy - xz)
= (y + z)[x(x + y) - z(x + y)]
= (y + z)(x + y)(x - z)
c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)
= x(y - z)(y + z) + yz2 - yx2 + x2z - y2z
= x(y - z)(y + z) - yz(y - z) - x2(y - z)
= (y - z)((xy + xz - yz - x2)
= (y - z)[x(y - x) - z(y - x)]
= (y - z)(x - z)(y -x)
Đẳng thức ban đầu \(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=4x^2+4y^2+4z^2-4xy-4yz-4zx\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
\(\Leftrightarrow x=y=z\)
Câu a :
\(VT=\) \(\left(x-1\right)\left(x^2+x+1\right)=x^3-1^3=VP\)
Câu b :
\(VT=\)\(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-y^4=VP\)
Tương tự bạn khai triển là ra nhé
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
1.
\(3x^2-16x+5\\ =3x^2-x-15x+5\\ =x\left(3x-1\right)-5\left(3x-1\right)\\ =\left(x-5\right)\left(3x-1\right)\)
2.
\(3x^3-14x^2+4x+3\\ =\left(3x^3+x^2\right)-\left(15x^2+5x\right)+\left(9x+3\right)\\ =x^2\left(3x+1\right)-5x\left(3x+1\right)+3\left(3x+1\right)\\ =\left(x^2-5x+3\right)\left(3x+1\right)\)
3. \(x^8+x^7+1\\ =\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\\ =x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x^3-1\right)+x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x\left(x^3+1\right)\left(x+1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)[x^2\left(x^3+1\right)\left(x-1\right)+x\left(x^3+1\right)\left(x-1\right)+1]\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+x^5-x^4+x^2-x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)4.
\(64x^4+y^4\\ =\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\\ =\left(8x^2+y^2\right)^2-16x^2y^2\\ =\left(8x^2+y^2-4xy\right)\left(8x^2+y+4xy\right)\)
5.
\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\\ =\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\\ =\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+4a^2+2a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)+2a^2\left(x^2+5ax+4a^2\right)+a^4\\ =\left(x^2+5ax+5a^2\right)^2\)