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2) Để sau đi (em chưa nghĩ ra)
3) \(A=\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)^2\left(x-y\right)+\left(y+z\right)^2\left(y-z\right)+\left(z+x\right)^2\left(z-x\right)\)
Đặt x - y = a; y - z = b => z - x = -(a+b)
\(A=\left(x+y\right)^2a+\left(y+z\right)^2b-\left(z+x\right)^2a-\left(z+x\right)^2b\)
\(=a\left[\left(x+y\right)^2-\left(z+x\right)^2\right]+b\left[\left(y+z\right)^2-\left(z+x\right)^2\right]\)
\(=\left(x-y\right)\left(x+y-z-x\right)\left(x+y+z+x\right)+\left(y-z\right)\left(y+z-z-x\right)\left(y+z+z+x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(2x+y+z\right)-\left(y-z\right)\left(x-y\right)\left(2z+x+y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Em tính sai sót chỗ nào thì thông cảm cho em ạ :>
1) a4+b4+c4−2a2b2−2a2c2−2b2c2
=2(a4+b4+c4-4a2b2-4a2c2-4b2c2)
=2a4+2b4+2c4-4a2b2-4a2c2-4b2c2
=(a4-2a2b2+b4)+(a4-2a2c2+c4)+(b4-2b2c2+c4
a)\(x^3+y^3+z^3-3xyz\\ \left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-\left[3xyz+3xy\left(x+y\right)\right]\\=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right] \\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+x^2-xy-xz-yz\right)\)
Bạn tự tách hđt nhé! Gõ mỏi tay :v~
\(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2=\left(y+z-2x\right)^2+\left(z+x-2y\right)^2+\left(y+z-2z\right)^2\)
⇔ \(y^2-2yz+z^2+z^2-2xz+x^2+x^2-2xy+y^2=\)\(6(z^2-yz-xz+y^2-xy+x^2)\)
⇔ \(2\left(x^2+y^2+z^2-yz-xz-xy\right)\)=\(6(z^2-yz-xz+y^2-xy+x^2)\)
⇔ \(x^2+y^2+z^2-yz-xz-xy\) = \(3(z^2-yz-xz+y^2-xy+x^2)\)
⇔ \(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
⇔ \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z\)
Do đó \(\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)
⇒ \(x=y=z\)
j lắm thế :)))
Bài 2 : ~ bài 1 ngán quá =)))
a, Có
\(5x^2+10y^2-6xy-4x-2y+3\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\forall x;y\)
Do đó không tồn tại x , y tm \(5x^2+10y^2-6xy-4x-2y+3=0\)
b, \(x^2+4y^2+z^2-2x-6x+6y+15=0\)
Câu này đề sai :v bài ngta không cho 2 lần x vậy đâu bạn :)))
Lời giải:
a)
$yz(y+z)+xz(z-x)-xy(x+y)=yz(y+z)+xz^2-x^2z-x^2y-xy^2$
$=yz(y+z)+x(z^2-y^2)-x^2(z+y)$
$=yz(y+z)+x(z-y)(z+y)-x^2(z+y)$
$=(y+z)(yz+xz-xy-x^2)$
$=(y+z)[z(x+y)-x(x+y)]=(y+z)(x+y)(z-x)$
b)
$2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc$
$=(2a^2b+4ab^2)-(a^2c+2abc)+(ac^2+2bc^2)-(4b^2c+2abc)$
$=2ab(a+2b)-ac(a+2b)+c^2(a+2b)-2bc(a+2b)$
$=(a+2b)(2ab-ac+c^2-2bc)$
$=(a+2b)[2b(a-c)-c(a-c)]$
$=(a+2b)(2b-c)(a-c)$
c)
$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$
$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$
Lời giải:
a)
$yz(y+z)+xz(z-x)-xy(x+y)=yz(y+z)+xz^2-x^2z-x^2y-xy^2$
$=yz(y+z)+x(z^2-y^2)-x^2(z+y)$
$=yz(y+z)+x(z-y)(z+y)-x^2(z+y)$
$=(y+z)(yz+xz-xy-x^2)$
$=(y+z)[z(x+y)-x(x+y)]=(y+z)(x+y)(z-x)$
b)
$2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc$
$=(2a^2b+4ab^2)-(a^2c+2abc)+(ac^2+2bc^2)-(4b^2c+2abc)$
$=2ab(a+2b)-ac(a+2b)+c^2(a+2b)-2bc(a+2b)$
$=(a+2b)(2ab-ac+c^2-2bc)$
$=(a+2b)[2b(a-c)-c(a-c)]$
$=(a+2b)(2b-c)(a-c)$
c)
$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$
$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$