Bài 1:

a: Sửa đề \(x^3y-2x^2y+xy\)

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

\(=x\cdot y\cdot\left(x^2-2x+1\right)\)

\(=xy\left(x-1\right)^2\)

b: Sửa đề: \(x^2-9-2xy+y^2\)

\(=\left(x^2-2xy+y^2\right)-9\)

\(=\left(x-y\right)^2-9\)

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

Bài 2:

a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)

b: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)

\(=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{x-3}\cdot\dfrac{1}{x+1}\)

\(=\dfrac{-5x-5}{\left(x-3\right)\left(x+1\right)}=-\dfrac{5\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}=-\dfrac{5}{x-3}\)

c: \(x^2-x-2=0\)

=>\(\left(x-2\right)\left(x+1\right)=0\)

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

Thay x=2 vào A, ta được:

\(A=\dfrac{-5}{2-3}=\dfrac{-5}{-1}=5\)

mình không biết làm:)

Bài 1 : 

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

Ta có : \(x+y+z=0\Rightarrow x+y=-z;y+z=-x;x+z=-y\)

hay \(-z.\left(-x\right)\left(-y\right)=-zxy\)

mà \(xyz=2\Rightarrow-xyz=-2\)

hay N nhận giá trị -2 

Bài 2 : 

\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)Đặt \(a=10k;b=3k\)

hay \(\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)

hay biểu thức trên nhận giá trị là 24 

c, Ta có : \(a-b=3\Rightarrow a=3+b\)

hay \(\frac{3+b-8}{b-5}-\frac{4\left(3+b\right)-b}{3\left(3+b\right)+3}=\frac{-5+b}{b-5}-\frac{12+4b-b}{9+3b+3}\)

\(=\frac{-5+b}{b-5}-\frac{12+3b}{6+3b}\)quy đồng lên rút gọn, đơn giản rồi 

1.Ta có:\(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)

\(\Rightarrow N=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\left(-x\right)\left(-y\right)=-2\)

2.Ta có:\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)

Đặt \(\frac{a}{10}=\frac{b}{3}=k\Rightarrow a=10k;b=3k\)

Ta có:\(A=\frac{3a-2b}{a-3b}=\frac{3.10k-2.3k}{10k-3.3k}=\frac{30k-6k}{10k-9k}=\frac{k\left(30-6\right)}{k\left(10-9\right)}=24\)

Vậy....

2.

\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)

ĐKXĐ là :

\(a\ne0;-3;-2\)

Vs a = 1 ta có:

=> P=3

1.

\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)

a) ĐKXĐ: \(x\ne\pm1\)

b) \(A=\dfrac{x^3-1}{x^2-1}\cdot\left(\dfrac{1}{x-1}-\dfrac{x+1}{x^2+x+1}\right)\left(dkxd:x\ne\pm1\right)\)

\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\left[\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)

\(=\dfrac{x^2+x+1}{x+1}\cdot\dfrac{x^2+x+1-\left(x^2-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(=\dfrac{x+2}{x^2-1}\)

c) Có: \(\left|x+3\right|=1\Leftrightarrow\left[{}\begin{matrix}x+3=1\\x+3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\left(tmdk\right)\)

+) Với \(x=-2\), thay vào \(A\), ta được:

\(A=\dfrac{-2+2}{\left(-2\right)^2-1}=0\)

+) Với \(x=-4\), thay vào \(A\), ta được:

\(A=\dfrac{-4+2}{\left(-4\right)^2-1}=-\dfrac{2}{15}\)

\(\text{#}Toru\)

Bài 2: 

3x + 2(5 - x) = 0

<=> 3x + 10 - 2x = 0

<=> x + 10 = 0

<=> x = 0 - 10

<=> x = -10

=> x = -10

Bài 3: 

6(3q + 4q) - 8(5p - q) + (p - q)

= 6.3p + 6.4q - 8.5p - (-8).q + p - q

= 18p + 24q - 40p + 8q + p - q

= (18p - 40p + p) + (24q + 8q - q)

= -21p + 31q

Bài 4: 

5x(4x² - 2x + 1) - 2x(10x² - 5x - 2)

= 5x.4x² + 5x.(-2x) + 5x.1 - 2x.10x² + (-2x).(-5x) + (-2x).(-2)

= 20x³ - 10x² + 5x - 20x³ + 10x² + 4x

= (20x² - 20x²) + (-10x² + 10x²) + (5x + 4x)

= 0 + 0 + 9x

= 9x (1)

Thay x = -15 vào (1), ta có:

9.(-15) = -135

Vậy: Giá trị biểu thức sau khi rút gọn với x = -15 là: -135

Bài 4:

a. $y=kx$. Thay $x=5; y=3$ vào thì:

$3=k.5\Rightarrow k=\frac{3}{5}$

b. Khi $x=10$ thì: $y=\frac{3}{5}x=\frac{3}{5}.10=6$

c. Khi $y=\frac{-3}{4}$ thì: $\frac{-3}{4}=\frac{3}{5}x$

$\Rightarrow x=\frac{-3}{4}: \frac{3}{5}=\frac{-5}{4}$

Bài 3:

a. $y=kx$. Thay $x=6$ và $y=4$ thì:

$4=k.6\Rightarrow k=\frac{4}{6}=\frac{2}{3}$

b. Khi $x=-20$ thì: $y=\frac{2}{3}x=\frac{2}{3}.(-20)=\frac{-40}{3}$

c. Khi $y=\frac{1}{6}$ thì:

$\frac{1}{6}=\frac{2}{3}x\Rightarrow x=\frac{1}{6}: \frac{2}{3}=\frac{1}{4}$