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

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

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

a, mình nghĩ đề là cm đẳng thức nhé 

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

Vậy ta có đpcm 

b, \(VT=\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)=\frac{5xy^2}{-xy}+\frac{9xy}{-xy}-\frac{x^2y^2}{-xy}\)

\(=-5y-9+xy=VP\)

Vậy ta có đpcm 

c, \(VT=\left(x^3y^3-x^2y^3-x^3y^2\right):x^2y^2=\frac{x^3y^3}{x^2y^2}-\frac{x^2y^3}{x^2y^2}-\frac{x^3y^2}{x^2y^2}=xy-y-x=VP\)

Vậy ta có đpcm 

bạn đặt nhân tử chung là xong bài rồi

1. \(x^3-x^2+x-1=(x^3-x^2)+(x-1)\)

\(=x^2(x-1)+(x-1)=(x^2+1)(x-1)\)

2. \(6x^2y-2xy^2+3x-y=2xy(3x-y)+(3x-y)\)

\(=(3x-y)(2xy+1)\)

3. \(4x^2+1\) thì còn cái gì để phân tích hả bạn? Hay ý bạn là \(4x^4+1\)?

\(4x^4+1=(2x^2)^2+1=(2x^2)^2+1+4x^2-4x^2\)

\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)

4. \(x^2-9x+8=(x^2-x)-(8x-8)\)

\(=x(x-1)-8(x-1)=(x-1)(x-8)\)

5. \(x^3-2x^2y+3xy^2=x(x^2-2xy+3y^2)\)

6. \(x^2-6x+y-y^2\) (sai đề)

7. \(x^2-xy-2x+2y=(x^2-xy)-(2x-2y)\)

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

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

b. \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}=\frac{2\left(x^2-y^2\right)+2y^2}{x}=\frac{2x^2-2y^2+2y^2}{x}=2x\)

c. \(\frac{3x+1}{x+y}-\frac{2x-3}{x+y}=\frac{3x+1-2x+3}{x+y}=\frac{x+4}{x+y}\)

d. \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}=\frac{xy}{2x-y}-\frac{1-x^2}{2x-y}=\frac{xy-1+x^2}{2x-y}\)

e. \(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}=\frac{4x-1-7x+1}{3x^2y}=\frac{-3x}{3x^2y}=\frac{-1}{xy}\)

P = 3x² - 2x + 3y² - 2y + 6xy +2018

P = 3(x² + y² + 2xy) - 2(x + y) + 2018

P = 3[(x + y)² - 2xy + 2xy] -2.5 + 2018

P = 3[ 5² +0] - 10 + 2018

P = 3.25 + 2008

P = 75 + 2008

P = 2083