Câu 4:

5x + 7y = 112

5(x+ y) = 112 - 2y

5(x + y) = 2(56 - y)

\(\begin{cases}x+y=2\\ 56-y=5\end{cases}\)

\(\begin{cases}x+y=2\\ y=56-5\end{cases}\)

\(\begin{cases}x=2-y\\ y=51\end{cases}\)

\(\begin{cases}x=2-51\\ y=51\end{cases}\)

\(\begin{cases}x=-49\\ y=51\end{cases}\)

Vậy (x ; y) = (-49; 51)

1)<=> \(\left(x^2+2x+1\right)-\left(y^2+4x+4\right)-7=0\)

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

\(\Rightarrow\left(x+1-y-2\right)\left(x+1+y+2\right)=7\)

\(\left(x-y-1\right)\left(x+y+3\right)=7\)

vì x;y là số nguyên dương=> \(x;y\ge1\Rightarrow x+y+3\ge1+1+3=5\)

=> \(x+y+3=7\Rightarrow x+y=4\)

\(x-y-1=1\Rightarrow x-y=2\)

cộng hai phương trình ta có:

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

=> \(2x=6\)

=> x=3

=> y=4-3=1

2)<=> \(\left(x^2+xy+2x\right)+\left(2xy+2y^2+4y\right)+\left(x+y+2\right)=17\)

=> \(x\left(x+y+2\right)+2y\left(x+y+2\right)+1\left(x+y+2\right)=17\)

=> \(\left(x+y+2\right)\left(x+2y+1\right)=17\)

tớ không biết

cj lậy chú

nhây vừa thoi

a/ \(x^3+2x^2+3x+2=y^3\)

Với \(\orbr{\begin{cases}x>1\\x< -1\end{cases}}\)thì

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

Nên không tồn tại số nguyên x, y thỏa mãn đề bài.

Từ đây ta suy ra \(-1\le x\le1\)

Với \(x=-1\Rightarrow y=0\)

\(x=0\Rightarrow y=\sqrt[3]{2}\left(l\right)\)

\(x=1\Rightarrow y=2\)

b/ \(y^2+2\left(x^2+1\right)=2y\left(x+1\right)\)

\(\Leftrightarrow2y^2+4\left(x^2+1\right)=4y\left(x+1\right)\)

\(\Leftrightarrow\left(y^2-4xy+4x^2\right)+\left(y^2-4y+4\right)=0\)

\(\Leftrightarrow\left(y-2x\right)^2+\left(y-2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}y=2x\\y=2\end{cases}}\)

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

1/ Ta có

 \(x^2+9x+20=x^2+4x+5x+20=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)

Tương tự

\(x^2+11x+30=\left(x+5\right)\left(x+6\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

Đk: x khác 4, 5, 6, 7

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\frac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}+\frac{\left(x+7\right)-\left(x+6\right)}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\) EM tự làm tiếp nhé

em cần đoạn tiếp mak

a) \(x^3-5x^2+8x-4\)

\(=x^3-2x^2-3x^2+6x+2x-4\)

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

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

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

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

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

b) \(A=10x^2-15x+8x-12+7\)

\(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\)

\(A=\left(2x-3\right)\left(5x+4\right)+7\)

Dễ thấy \(\left(2x-3\right)\left(5x+4\right)⋮\left(2x-3\right)=B\)

Vậy để \(A⋮B\)thì \(7⋮\left(2x-3\right)\)

\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x\in\left\{2;1;5;-2\right\}\)

Vậy.......