\(m_{Mg}+m_{O_2}\rightarrow m_{MgO}\Leftrightarrow4,8g+m_{O_2}\rightarrow8\Leftrightarrow m_{O_2}=3,2g\)

MIK ĐANG CẦN GẤP GIÚP MIK VỚI

b, lấy cái dễ nhất làm cho nhanh :PP làm tương tự thôi nhé !!!

\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)

\(=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x^2+2\right)\left(x-1\right)}=\frac{\left(4-x\right)\left(x-1\right)}{x\left(x^2+2\right)\left(x-1\right)}-\frac{x\left(x+5\right)}{x\left(x^2+2\right)\left(x-1\right)}\)

\(=\frac{4x-4-x^2+x-x^2-5x}{MTC}=\frac{-4-2x^2}{MTC}\)

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

\(\frac{x-1}{x+2}-\frac{x+1}{2-x}-\frac{x^2-2x+4}{x^2-4}\)

= \(\frac{x-1}{x+2}+\frac{x+1}{x-2}-\frac{x^2-2x+4}{x^2-4}\)

= \(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\)

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

im,em moi lop 3m :)

\(\frac{x^2+2}{2xy^3}-\frac{2x+2}{2xy^3}=\frac{x^2+2-2x-2}{2xy^3}=\frac{x^2-2x}{2xy^3}=\frac{x\left(x-2\right)}{2xy^3}=\frac{x-2}{2y^3}\)

\(\frac{4}{x-5}-\frac{1}{x+5}+\frac{13x-x^2}{25-x^2}=\frac{4}{x-5}-\frac{1}{x+5}+\frac{x^2-13x}{x^2-25}\)

\(=\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-13x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{4x+20-x+5+x^2-13x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)

a) x(x - 3) + 5x = x² - 8

=> x² - 3x + 5x - x² + 8 = 0

=> 2x + 8 = 0

=> 2x = -8

=> x = -4

b) 3(x + 4) - x² - 4x = 0

=> 3(x + 4) - x(x + 4) = 0

=> (3 - x)(x + 4) = 0

=> \(\orbr{\begin{cases}3-x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

Vậy \(x\in\left\{3;-4\right\}\)là giá trị cần tìm

c) 7x³ + 12x² - 4x = 0

=> x(7x² + 12x - 4) = 0

=> x(7x² + 14x - 2x - 4) = 0

=> x[7x(x + 2) - 2(x + 2)] = 0

=> x(x + 2)(7x - 2) = 0

=> x = 0 hoặc x + 2 = 0 hoặc 7x - 2 = 0

=> x = 0 hoặc x = -2 hoặc x = 2/7

Vậy \(x\in\left\{0;-2;\frac{2}{7}\right\}\)là giá trị cần tìm

x( x - 3 ) + 5x = x² - 8

⇔ x² - 3x + 5x - x² + 8 = 0

⇔ 2x + 8 = 0

⇔ 2x = -8

⇔ x = -4

3( x + 4 ) - x² - 4x = 0

⇔ 3( x + 4 ) - x( x + 4 ) = 0

⇔ ( x + 4 )( 3 - x ) = 0

⇔ x = -4 hoặc x = 3

7x³ + 12x² - 4x = 0

⇔ x( 7x² + 12x - 4 ) = 0

⇔ x( 7x² + 14x² - 2x - 4 ) = 0

⇔ x[ 7x( x + 2 ) - 2( x + 2 ) ] = 0

⇔ x( x + 2 )( 7x - 2 ) = 0

⇔ x = 0 hoặc x = -2 hoặc x=  2/7

a, \(5x^2y+10xy=5xy\left(x+2\right)\)

b, \(x^2-2xy+y^2-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)

c, \(x^3-8+2x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\)

\(=\left(x-2\right)\left[\left(x^2+2x+4\right)+2x\right]=\left(x-2\right)\left(x+2\right)^2\)

d, \(x^4+x^2y^2+y^4\):< 

a, 5xy(  x + 2)

b, ( x - y -5 )(x-y+5)

c,( x-2)( x+ 2)²