`24)`

\(\left\{{}\begin{matrix}2\left(x+y\right)-\left(x-y\right)=1\\\left(x+y\right)-3\left(x-y\right)=-5\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}2x+2y-x+y=1\\x+y-3x+3y=-5\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x+3y=1\\-2x+4y=-5\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=1-3y\\-2\left(1-3y\right)+4y=-5\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}x=1-3y\\-2+6y+4y=-5\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=1-3y\\10y=-3\end{matrix}\right.\)  `<=>` \(\left\{{}\begin{matrix}x=1-3.\left(-\dfrac{3}{10}\right)\\y=-\dfrac{3}{10}\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=\dfrac{19}{10}\\y=-\dfrac{3}{10}\end{matrix}\right.\)

Vậy hpt có nghiệm `(x;y)=((19)/(10) ; (-3)/(10))`

`25)`

\(\left\{{}\begin{matrix}2x+1=x+2y\\x-y=2x+y+1\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}2x-x-2y=-1\\x-y-2x-y=1\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}x-2y=-1\\-x-2y=1\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}2x=-2\\-x-2y=1\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}x=-1\\-\left(-1\right)-2y=1\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)

Vậy hpt có nghiệm `(x;y)=(-1;0)`

`26)`

\(\left\{{}\begin{matrix}2\left(x-2y\right)+3\left(x+2y\right)=4\\\left(x-y\right)+2\left(x+y\right)=1\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}2x-4y+3x+6y=4\\x-y+2x+2y=1\end{matrix}\right.\)   `<=>` \(\left\{{}\begin{matrix}5x+2y=4\\3x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+2y=4\\6x+2y=2\end{matrix}\right.\)  `<=>` \(\left\{{}\begin{matrix}-x=2\\6x+2y=2\end{matrix}\right.\)

`<=> {(x=-2),(6.(-2) + 2y=2):}`

`<=>{(x=-2),(y=7):}`

Vậy hpt có nghiệm `(x;y)=(-2;7)`

`27)`

\(\left\{{}\begin{matrix}\left(x+4\right)\left(y+4\right)=xy+216\\\left(x+2\right)\left(y-5\right)=xy-50\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}xy+4x+4y+16=xy+216\\xy-5x+2y-10=xy-50\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}xy+4x+4y-xy=216-16\\xy-5x+2y-xy=-50+10\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}4x+4y=200\\-5x+2y=-40\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x+y=50\\-5x+2y=-40\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}x=50-y\\-5\left(50-y\right)+2y=-40\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=50-y\\-250+5y+2y=-40\end{matrix}\right.\) 

`<=>` \(\left\{{}\begin{matrix}x=50-y\\7y=210\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=50-30\\y=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=30\end{matrix}\right.\) 

Vậy hpt có nghiệm `(x;y)=(20;30)`

`4x^2 - 4x+1 = 16`

`<=>4x^2 - 4x + 1 - 16=0`

`<=> (2x)^2 - 2.2x.1+1^2 - 16=0`

`<=> (2x-1)^2 - 16=0`

`<=>(2x-1)^2 - 4^2 =0`

`<=>(2x-1-4)(2x-1+4)=0`

`<=>(2x-5)(2x+3)=0`

`<=>` `2x-5=0` hoặc `2x+3=0`

`<=> 2x=5` hoặc `2x=-3`

`<=>x=5/2` hoặc `x=-3/2`

Vậy `x=5/2` hoặc `x=-3/2`

Đổi `200ml=0,2l`

Số mol `H_2SO_4` là:

`n_(H_2SO_4)=0,2.2=0,4(mol)`

PTHH: `Fe+H_2SO_4->FeSO_4+H_2`

Theo pthh: +`n_(Fe)=n_(H_2SO_4)=0,4(mol)`

+ `n_(H_2)=n_(H_2SO_4)=0,4(mol)`

+ `n_(FeSO_4)=n_(H_2SO_4)=0,4(mol)`

`a)`Khối lượng của sắt `(Fe)` là:

`m_(Fe)=0,4.56=22,4(g)`

`b)`  Thể tích khí `H_2` là:

`V_(H_2)=0,4.22,4=8,96(l)`

`c)` Khối lượng `FeSO_4` sau phản ứng:

`m_(FeSO_4)=0,4.152=60,8(g)`

\(\left\{{}\begin{matrix}2x+3y=2\\x+2y=3\end{matrix}\right.\)

`<=>`  \(\left\{{}\begin{matrix}2x+3y=2\\x=3-2y\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}2\left(3-2y\right)+3y=2\\x=3-2y\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}6-4y+3y=2\\x=3-2y\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}-y=-4\\x=3-2y\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}y=4\\x=3-2.4\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}y=4\\x=-5\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm `(x;y)=(-5;4)`

:)) xóa đi