=(y+1)² +(2^x-1)² =0

cặp nghiệm là: y=-1; x= 0

Ai giúp em với ạ

1. Ta có: \(x^2-2xy-x+y+3=0\)

<=> \(x^2-2xy-2.x.\frac{1}{2}+2.y.\frac{1}{2}+\frac{1}{4}+y^2-y^2-\frac{1}{4}+3=0\)

<=> \(\left(x-y-\frac{1}{2}\right)^2-y^2=-\frac{11}{4}\)

<=> \(\left(x-2y-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=-\frac{11}{4}\)

<=> \(\left(2x-4y-1\right)\left(2x-1\right)=-11\)

Th1: \(\hept{\begin{cases}2x-4y-1=11\\2x-1=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-3\end{cases}}\)

Th2: \(\hept{\begin{cases}2x-4y-1=-11\\2x-1=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)

Th3: \(\hept{\begin{cases}2x-4y-1=1\\2x-1=-11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)

Th4: \(\hept{\begin{cases}2x-4y-1=-1\\2x-1=11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=3\end{cases}}\)

Kết luận:...

2. \(y^2+1\ge1>0;2x^2+x+1>0\) với mọi x; y 

=> x + 5 > 0 

=>  \(y^2+1=\frac{x+5}{2x^2+x+1}\ge1\)

<=> \(x+5\ge2x^2+x+1\)

<=> \(x^2\le2\)

Vì x nguyên => x = 0 ; x = 1; x = -1 

Với x = 0 ta có: \(y^2+1=5\Leftrightarrow y=\pm2\)

Với x = 1 ta có: \(y^2+1=\frac{3}{2}\)loại vì y nguyên 

Với x = -1 ta có: \(y^2+1=2\Leftrightarrow y=\pm1\)

Vậy Phương trình có 4 nghiệm:...

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-4\right)\\x_1x_2=-m^2+4\end{matrix}\right.\)

\(\dfrac{x_1+x_2}{x_1x_2}+\dfrac{4}{x_1x_2}=1\)

Thay vào ta được : \(\dfrac{2\left(m-4\right)+4}{-m^2+4}=1\Leftrightarrow\dfrac{2m-4}{\left(2-m\right)\left(m+2\right)}=1\Leftrightarrow\dfrac{-2}{m+2}=1\Rightarrow-2=m+2\Leftrightarrow m=-4\)

a, x + 2 chia hết cho x^2 - 7

=> (x + 2)(x - 2) chia hết cho x^2 - 7

=> x^2 - 4 chia hết cho x^2 - 7

=> x^2 - 7 + 3 chia hết cho x^2 - 7

=> 3 chia hết cho x^2 - 7

=> x^2 - 7 thuộc Ư(3)

=> x^2 - 7 thuộc {-1; 1; -3; 3}

=> x^2 thuộc {6; 8; 4; 10}

mà x là số nguyên

=> x = 2 hoặc x = -2

\(A=\left(\frac{2X-1}{x^2-4}+\frac{x+2}{x^2-x-2}\right):\frac{x-2}{x^2+3x+2}ĐK:x\ne\left\{2,-2,-1\right\}\)

a)  \(A=\left[\frac{\left(2x-1\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x-2\right)}\right]:\frac{x-2}{\left(x+2\right)\left(x+1\right)}\)

\(A=\left[\frac{\left(2x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\frac{\left(x+2\right)\left(x+2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\right].\frac{\left(x+2\right)\left(x+1\right)}{x-2}\)

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

\(A=\frac{3x^2+5x+3}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)

\(A=\frac{3x^2+5x+3}{\left(x-2\right)^2}\)

Ta có :\(3x^2+5x+3\)

\(=3\left(x^2+\frac{5}{3}x+1\right)\)

\(=3\left[x^2+2.\frac{5}{6}x+\frac{25}{36}+\frac{9}{36}\right]\)

\(=3\left[\left(x+\frac{5}{6}\right)^2+\frac{9}{36}\right]>0\)

Mà \(\left(x-2\right)^2>0\)

\(\Rightarrow A>0\left(dpcm\right)\)

\(b,A=11\Leftrightarrow\frac{3x^2+5x+3}{\left(x-2\right)^2}=11\)

\(\Rightarrow3x^2+5x+3=11.\left(x-2\right)^2\)

\(\Rightarrow3x^2+5x+3=11.\left(x^2-4x+4\right)\)

\(\Rightarrow8x^2-49x+41=0\)

\(\Rightarrow8x^2-8x-41x+41=0\)

\(\Rightarrow8x\left(x-1\right)-41\left(x-1\right)=0\)

\(\Rightarrow\left(8x-41\right)\left(x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}8x-41=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{41}{8}\\x=1\end{cases}}}\)(Thỏa mãn)

\(\Delta=\left(m-1\right)^2-4\left(m+3\right)=m^2-6m-11>0\) (1)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m-1\\x_1x_2=m+3\end{matrix}\right.\)

Ta có:

\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=\left(m-1\right)^2-2\left(m+3\right)=m^2-4m-5\)

Biểu thức này ko tồn tại cả min lẫn max với điều kiện m từ (1)