a) hpt \(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=11\\\left(x+y\right)^2-2xy-\left(x+y\right)=8\end{matrix}\right.\)

Đặt S=x+y; P =xy, ta có hệ :

\(\left\{{}\begin{matrix}S+P=11\\S^2-S-2P=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\S^2-S-2\left(11-S\right)=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\S^2+S-30=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\\left[{}\begin{matrix}S=5\\S=-6\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=11-\left(x+y\right)\\\left[{}\begin{matrix}x+y=5\\x+y=-6\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\curlyvee\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\\\text{hệ vô nghiệm}\end{matrix}\right.\)

Vậy...

b)ĐK : \(\left\{{}\begin{matrix}x\le0\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{2x^2-4x}-4=\frac{2x^2-2x+12}{x+8}-4\)

\(\Leftrightarrow\frac{2x^2-4x-16}{\sqrt{2x^2-4x}+4}=\frac{2x^2-6x-20}{x+8}\)

\(\Leftrightarrow\frac{\left(x-4\right)\left(x+2\right)}{\sqrt{2x^2-4x}+4}=\frac{\left(x-5\right)\left(x+2\right)}{x+8}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\frac{x-4}{\sqrt{2x^2-4x}+4}=\frac{x+5}{x+8}\left(1\right)\end{matrix}\right.\)

Giải tiếp pt 1 và kết hợp vs đk t tìm được nghiệm

thay m=2 vào HPT ta có
\(\left\{{}\begin{matrix}x+2y=2+1\\2x+y=2.2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=3\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+4y=6\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y=2\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
vậy ..........
 

Chắc đề phải là tìm a nguyên chứ bạn?

\(\left\{{}\begin{matrix}\left(a+1\right)x-ay=5\\x+ay=a^2+4a\end{matrix}\right.\)

\(\Rightarrow\left(a+2\right)x=a^2+4a+5\)

- Với \(a=-2\) hệ vô nghiệm

- Với \(a\ne-2\Rightarrow x=\dfrac{a^2+4a+5}{a+2}=a+2+\dfrac{1}{a+2}\)

\(x\in Z\Rightarrow\dfrac{1}{a+2}\in Z\Rightarrow a+2=\left\{-1;1\right\}\)

\(\Rightarrow a=\left\{-3;-1\right\}\)

a, Khi \(m=-1\)ta có HPT : \(\hept{\begin{cases}-x+y=-2\\x-y=0\end{cases}}\)

=> HPT vô nghiệm

b, \(\hept{\begin{cases}mx+y=2m\\x+my=m+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2m-mx\\x+m\left(2m-mx\right)=m+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2m-mx\\\left(1-m^2\right)x=-2m^2+m+1\end{cases}}\)( * )

HPT vô nghiệm

<=> ( * ) vô nghiệm

\(\Leftrightarrow\hept{\begin{cases}1-m^2=0\\-2m^2+m+1\end{cases}}\ne0\)

<=> m = 1 hoặc m = -1 mà m khác 1 và -1/2 

<=> m = -1

a. Khi m = 2, ta được:

\(\left\{{}\begin{matrix}3x-y=2.2+3\\x+2y=3.2+2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-y=7\\x+2y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2.\left(3x-7\right)=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+6x-14=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\7x=22\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x=\dfrac{22}{7}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{22}{7}\\y=\dfrac{17}{7}\end{matrix}\right.\)

@Nhã Doanh giúp mk với

sos

(x-4)(y-4)=416

=>xy-4x-4y+16=416

=>600-4(x+y)+16=416

=>4(x+y)=200

=>x+y=50

mà xy=600

nên x,y là hai nghiệm của pt:

a^2-50a+600=0

=>a=20 hoặc a=30

=>(x,y)=(20;30) hoặc (x,y)=(30;20)

Bài 1:

\(x^4+2x^3+10x-25=0\)

\(\Leftrightarrow x^4+2x^3-5x^2+5x^2+10x-25=0\)

\(\Leftrightarrow x^2\left(x^2+2x-5\right)+5\left(x^2+2x-5\right)=0\)

\(\Leftrightarrow\left(x^2+5\right)\left(x^2+2x-5\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+5=0\\x^2+2x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+5>0\forall x\rightarrow Vn\\\Delta_{x^2+2x-5}=2^2-\left[-4\left(1.5\right)\right]=24\end{array}\right.\)

\(\Leftrightarrow x_{1,2}=\frac{-2\pm\sqrt{24}}{2}\)

Bài 2:

Đặt \(\begin{cases}\sqrt{x-1}=a\left(a\ge1\right)\\\sqrt{y}=b\left(b\ge0\right)\end{cases}\)(*) hệ đầu thành:

\(\begin{cases}3a+2b=13\left(1\right)\\2a-b=4\left(2\right)\end{cases}\).Từ \(\left(2\right)\Rightarrow b=2a-4\) thay vào (1) ta có:

\(\left(1\right)\Rightarrow3a+2\left(2a-4\right)=13\)

\(\Rightarrow3a+4a-8=13\Rightarrow7a=21\Rightarrow a=3\) (thỏa mãn)

\(a=3\Rightarrow b=2a-4=2\cdot3-4=2\) (thỏa mãn)

Thay \(\begin{cases}a=3\\b=2\end{cases}\) vào (*) ta có:

(*)\(\Leftrightarrow\begin{cases}\sqrt{x-1}=3\\\sqrt{y}=2\end{cases}\)\(\Leftrightarrow\begin{cases}x-1=9\\y=4\end{cases}\)\(\Leftrightarrow\begin{cases}x=10\\y=4\end{cases}\)