P=3xy^2+xy^2-6xy+8xz-10xz

=4xy^2-6xy-2xz

Khi x=-3 và y=-1/2 và z=3 thì P=4*(-3)*1/4-6*(-3)(-1/2)-2*(-3)*3

=-3+18/-2+6*3

=15-9

=6

a) \(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\)

\(A=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]+1\)

\(A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)

Đặt \(a=x^2-5x+5\)

\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)+1\)

\(\Leftrightarrow A=a^2-1^2+1\)

\(\Leftrightarrow A=a^2\)

Thay \(a=x^2-5x+5\)vào A ta có :

\(A=\left(x^2-5x+5\right)^2\)

b) \(B=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\)

\(B=\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)+1\)

\(B=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]+1\)

\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

Làm tương tự câu a)

c) \(12x^2-3xy-8xz+2yz\)

\(=3x\left(4x-y\right)-2z\left(4x-y\right)\)

\(=\left(4x-y\right)\left(3x-2z\right)\)

a) Ta có: \(27m\left(m+n\right)-m-n\)

\(=27m\left(m+n\right)-\left(m+n\right)\)

\(=\left(m+n\right)\left(27m-1\right)\)

b) Ta có: \(15x\left(x-y\right)-25x+25y\)

\(=15x\left(x-y\right)-25\left(x-y\right)\)

\(=5\left(x-y\right)\left(3x-5\right)\)

c) Ta có: \(12x^2-3xy+8xz-2yz\)

\(=3x\left(4x-y\right)+2z\left(4x-y\right)\)

\(=\left(4x-y\right)\left(3x+2z\right)\)

d) Ta có: \(x^3+x^2y-x^2z-xyz\)

\(=x^2\left(x+y\right)-xz\left(x+y\right)\)

\(=x\left(x+y\right)\left(x-z\right)\)

Bài b nhé bạn!

\(\hept{\begin{cases}\frac{xyz}{x+y}=2\\\frac{xyz}{y+z}=\frac{6}{5}\\\frac{xyz}{x+z}=\frac{3}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{x+y}{xyz}=\frac{1}{2}\\\frac{y+z}{xyz}=\frac{5}{6}\\\frac{x+z}{xyz}=\frac{2}{3}\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{1}{yz}+\frac{1}{xz}=\frac{1}{2}\\\frac{1}{xz}+\frac{1}{xy}=\frac{5}{6}\\\frac{1}{xy}+\frac{1}{yz}=\frac{2}{3}\end{cases}}\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=\frac{\frac{1}{2}+\frac{5}{6}+\frac{2}{3}}{2}=1\)

Trừ lại từng phương trình trong hệ:

\(\hept{\begin{cases}\frac{1}{xy}=\frac{1}{2}\\\frac{1}{yz}=\frac{1}{6}\\\frac{1}{xz}=\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}xy=2\\yz=6\\xz=3\end{cases}\Rightarrow xyz=\sqrt{2.6.3}=6}\)

Chia lại từng phương trình trong hệ mới, được:

\(\hept{\begin{cases}z=3\\x=1\\y=2\end{cases}}\)

Vậy \(\left(x;y;z\right)=\left(1;2;3\right)\)

Xong rồi đó!!!

1) 2x²-8xy-5x+20y

=2x(x-4y)-5(x-4y)

=(2x-5)(x-4y)

2) x³-x²y-xy+y²

=x²(x-y)-y(x-y)

=(x²-y)(x-y)

3) x²-2xy-4z²+y²

=(x-y)²-(2z)²

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

4) a³+a²b-a²c-abc

=a²(a+b)-ac(a+b)

=(a²-ac)(a+b)

=a(a-c)(a+b)

5) x³+y³+3x²y+3xy²-x-y

=(x+y)(x²-xy+y²)+3xy(x+y)-(x+y)

=(x+y)(x²-xy+y²+3xy-1)

=(x+y)[(x+y)²-1)]

=(x+y)(x+y+1)(x+y-1)

6) x³+x²y-x²z-xyz

=x²(x+y)-xz(x+y)

=(x²-xz)(x+y)

=x(x-z)(x+y)

7) =[x(y+z)²-2xyz]+[y(z+x)²-2xyz]+z(x+y)²

=x(y²+z²)+y(z²+x²)+z(x+y)²

=xy(x+y)+z²(x+y)+z(x+y)²

=(x+y)(xy+z²+zx+zy)

=(x+y)(x+z)(y+z)

8) x³(z-y)+y³(x-z)+z³(y-x)

Tách x-z= -[z-y+y-x]