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Mk sửa r đó. H giúp mk vs. Cảm ơn

\(pt\Leftrightarrow x\left(y-2\right)=-3y-1\)

\(\Leftrightarrow x=\frac{-3y-1}{y-2}=\frac{\left(-3y+6\right)-7}{y-2}=-3-\frac{7}{y-2}\)

Để \(x\inℤ\)thì \(\frac{7}{y-2}\inℤ\)

\(\Leftrightarrow y-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

lần lượt thay các giá trị của y-2 ta tìm dc các cặp nghiệm (x;y) là:

(-2; -5); (4; 1); (-10; 3); (-4; 9)

ábfaksjbdbczdcbsabv

Ta có :

  \(\frac{2,7}{m_{dd}}\times100=10\left(\%\right)\)

\(\Rightarrow m_{dd}=\frac{2,7\times100}{10}\)

\(\Rightarrow m_{dd}=27\left(g\right)\)

\(\Rightarrow A=m_{dd}-m_{Al}=27-2,7=24,3\left(g\right)\)

Vậy ...

ai giúp mình với mai mình kiểm tra 1 tiết rồi

1

\(4x^4+81\)

\(=\left(2x^2\right)^2+9^2+2.2x^2.9-2.2x^2.9\)

\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)

\(=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)

2

\(4a^4+b^4\)

\(=\left(2x^2\right)^2+\left(b^2\right)^2+2.2a^2.b^2-2.2a^2.b^2\)

\(=\left(2a^2+b^2\right)^2-\left(2ab\right)^2\)

\(=\left(2a^2+b^2-2ab\right)\left(2a^2+b^2+2ab\right)\)

3

\(x^7+x^5+1\)

\(=\left(x^7-x\right)+\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

\(=x\left(x^6-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

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

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

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

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

4

\(x^5+x^4+1\)

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

\(=x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)

\(=\left(x^3+1-x\right)\left(x^2+x+1\right)\)