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

\(x^6-2x^3y-x^4+y^2+7=0\)

\(\Leftrightarrow\left(x^6-2x^3y+y^2\right)-x^4+7=0\)

\(\Leftrightarrow\left(x^3-y\right)^2-\left(x^2\right)^2=-7\)

\(\Leftrightarrow\left(x^3-y+x^2\right)\left(x^3-y-x^2\right)=-7\)

Liệt kê ước 7 ra rồi lm đc

Ta có:

\(x^2-y^2+2x-4y-10=0\)

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

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

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

Vì \(x,y\) nguyên dương 

Nên \(x+y+3>x-y-1>0\)

\(\Rightarrow\hept{\begin{cases}x+y+3=7\\x-y-1=1\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=1\end{cases}}}\)

Vậy phương trình có nghiệm nguyên dương duy nhất \(\left(x,y\right)=\left(3;1\right)\)

a) \(x^3-2x^2-5x+6=0\)

\(x^3-x^2-x^2+x-6x+6=0\)

\(x^2\left(x-1\right)-x\left(x-1\right)-6\left(x-1\right)=0\)

\(\left(x-1\right)\left(x^2-x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2-x-6=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x^2-2x+3x-6=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\left\{2;-3\right\}\end{cases}}\)

\(a,x^3-2x^2-5x+6=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(x^2-x\right)-\left(6x-6\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)-x\left(x-1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^2-3x\right)+\left(2x-6\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow x-1=0\left(h\right)x+2=0\left(h\right)x-3=0\)

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

Vậy \(x\in\left\{-2;1;3\right\}\)

P/S: (h) là hoặc nhé