Rút gọn biểu thức

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

Ta có :

\(a\left(b^2+c^2+bc\right)+b\left(c^2+a^2+ac\right)+c\left(a^2+b^2+ab\right)\)

\(=ab^2+ac^2+abc+bc^2+a^2b+a^2c+b^2c+abc\)

\(=\left(ab^2+a^2b+abc\right)+\left(bc^2+b^2c+abc\right)+\left(ac^2+a^2c+abc\right)\)

\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\)

\(=\left(ab+bc+ac\right)\left(a+b+c\right)\)

Thanks rat nhieu

Phân tích 1000027 thành tích các thừa số được : 1000027 = 7x19x73x103

=> 1000027 là hợp số. 

a) (9+1)³  = 1000

Moi ng lam chi tiet nha!  Thanks trc

\(\frac{2x+2}{\left(x+1\right)\left(x-1\right)}.ĐKXĐ:x\ne\pm1\)

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

~~

Bài 2:

a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)

\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}\)

\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)

b: Để A>0 thì x-3>0

hay x>3

1.

\(x^2\)+\(y^2\)+2y-6x+10=0

=> \(x^2\)-6x+9 +\(y^2\)+2y+1=0

=> (x-3)\(^2\)+(y+1)\(^2\)=0

pt vô nghiệm

4.

=> \(x^2\)+8x+16+(3y)\(^2\)-2.3.2y+4=0

=> (x+4)\(^2\)+(3y-2)\(^2\)=0

pt vô nghiệm

khó lắm

hôm nay bọn mik vừa hok về chưa thấm đâu vô đâu nên kg giúp đc xin lỗi nhe!