bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc

1). x²y²(y-x)+y²z²(z-y)-z²x²(z-x)

2)xyz-(xy+yz+xz)+(x+y+z)-1

3)yz(y+z)+xz(z-x)-xy(x+y)

5)y(x-2z)²+8xyz+x(y-2z)²-2z(x+y)²

6)8x³(y+z)-y³(z+2x)-z³(2x-y)

7) (x2+y2)³+(z2-x2)³-(y2+z2)³

y(x−2z)²+8xyz+x(y−2z)²−2z(x+y)²
=y(x²−4xz+4z²)+8xyz+x(y²−4yz+4z²)−2z(x²+y²+2xy)

=(yx²+xy²)+(4yz²+4xz²)−(2zx²+2zy²+4xyz)

=xy(x+y)+4z²(x+y)−2z(x²+y²+2xy)

=xy(x+y)+4z²(x+y)−2z(x+y)²

=(x+y)(xy+4z²−2xz−2yz)

=(x+y)[y(x−2z)−2z(x−2z)]

=(x+y)(y−2z)(x−2z)

soyeon_Tiểubàng giải, giúp mình với!!!!!

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)

$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$

$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$

1. x⁵ + x + 1

=x⁵-x²+x²+x+1

=(x⁵-x²)+(x²+x+1)

=x²(x³-1)+(x²+x+1)

=x²(x-1)(x²+x+1)+(x²+x+1)

=(x²+x+1)[x²(x-1)+1]

=(x²+x+1)(x³-x2+1)

2. x⁵ + x⁴ +1

=x⁵+x⁴+x³-x³+1

=(x⁵+x⁴+x³)-(x³-1)

=x³(x²+x+1)-(x-1)(x²+x+1)

=(x²+x+1)[x³-(x-1)]

=(x²+x+1)(x³-x+1)

Bài 2:

a, \(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)

Vậy...

b, \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy...

c, \(x^3-\dfrac{1}{4}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy...

Bài 3:

1, Đặt \(A=x^2+\dfrac{1}{2}x+\dfrac{1}{16}=x^2+\dfrac{1}{4}.x.2+\dfrac{1}{16}\)

\(=\left(x+0,25\right)^2\)

Thay x = 49,75 vào A ta có:
\(A=50^2=2500\)

2, tương tự

bài 1 bn ơi