\(x^2-4xy+5y^2+6x-10y+10=0\)

\(x^2-2x\left(2y-3\right)+5y^2-10y+10=0\)

\(x^2-2x\left(2y-3\right)+\left(4y^2-12x+9\right)+\left(y^2+2x+1\right)=0\)

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

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

Ta có: \(\hept{\begin{cases}\left(x-2y+3\right)^2\ge0\forall x;y\\\left(y+1\right)^2\ge0\forall y\end{cases}}\)\(\Rightarrow\left(x-2y+3\right)^2+\left(y+1\right)^2\ge0\forall x;y\)

Mà \(\left(x-2y+3\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-2y+3\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2y+3=0\\y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-2y+3=0\\y=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x+2+3=0\\y=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)Vậy \(\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)

Tham khảo nhé~

Sao anh kudo không tách thẳng như vầy luôn cho nhanh?(nhanh hơn đúng 1 dòng ở phần phân tích thôi:v)

\(A=x^2-4xy+5y^2+6x-10y+10=0\)

\(\Leftrightarrow\left(x^2-2.x.2y+4y^2\right)+\left(6x-12y\right)+9+\left(y^2+2y+1\right)=0\)

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

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

Đến đây ez rồi!

Mình không chắc là có đúng không nữa các bạn xem hộ mình với nha!
= (100^2 - 99^2) + (98^2 - 97^2) + ... + (4^2 - 3^2) + (2^2 - 1^2) = 
= (100+99)(100-99) + (98+97)(98-97) + ... + (4+3)(4-3) + (2+1)(2-1) = 
= (100+99).1 + (98+97).1 + ... + (4+3).1 + (2+1).1 = 
= 100 + 99 + 98 + 97 + ... + 4 + 3 + 2 + 1 = 
= (100+1) + (99+2) + (98+3) + ... + (51+50) = 101.50 = 5050 
(50 cặp dấu ngoặc, tổng trong mỗi cặp dấu ngoặc là 101) 

Bài 1 :

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

\(\Rightarrow\left(2x+1\right)\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=5\end{cases}}\)

KL :...

bài 2 :

\(x^2+6x+9-y^2=\left(x+3\right)^2-y^2\)

\(=\left(x+3-y\right)\left(x+3+y\right)\)

Bài 3 :

\(P=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\frac{3}{x-2}\)

\(\frac{x^3-2x^2+4}{x-2}\inℤ\Leftrightarrow x^3-2x^2+4⋮x-2\)

\(\Leftrightarrow x^3-2x^2-\left(x^3-2x^2\right)+4⋮x-2\Leftrightarrow4⋮x-2\)

\(\Leftrightarrow x-2\in\left\{-1;2;-2;1;-4;4\right\}\Leftrightarrow x\in\left\{1;4;0;3;-2;6\right\}\)

b, \(\frac{x^3-x^2+2}{x-1}\inℤ\Leftrightarrow x^3-x^2+2⋮x-1\)

\(\Leftrightarrow x^3-x^2-\left(x^3-x^2\right)+2⋮x-1\)

\(\Leftrightarrow2⋮x-1\Leftrightarrow x-1\in\left\{-1;1;-2;2\right\}\)

\(\Leftrightarrow x\in\left\{0;2;-1;3\right\}\)

Giúp mình nha

\(a,ĐKXĐ:x\ne\pm2\)

\(b,P=\left(\frac{x+2}{2x-4}+\frac{x-2}{2x+4}+\frac{-8}{x^2-4}\right):\frac{4}{x-2}\)

\(=\left(\frac{x+2}{2\left(x-2\right)}+\frac{x-2}{2\left(x+2\right)}+\frac{-8}{\left(x-2\right)\left(x+2\right)}\right).\frac{x-2}{4}\)

\(=\left(\frac{\left(x+2\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{\left(-8\right).2}{2\left(x-2\right)\left(x+2\right)}\right)\)\(.\frac{x-2}{4}\)

\(=\left(\frac{x^2+4x+4+x^2-4x+4-16}{2\left(x-2\right)\left(x+2\right)}\right).\frac{x-2}{4}\)

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

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

chịu rồi bạn ạ

\(Taco:\)

\(A=2\left(3x+1\right)\left(x-1\right)-3\left(2x-3\right)\left(x-4\right)\)

\(A=\left(6x+2\right)\left(x-1\right)-\left(6x-9\right)\left(x-4\right)\)

\(A=\left(6x^2-4x-2\right)-\left(6x^2-24x-9x-36\right)\)

\(A=6x^2-4x-2-6x^2+33x+36=29x+34\)

\(b,x=2\Rightarrow A=58+34=92\)

\(A=-20\Leftrightarrow29x=-20-34=-54\Leftrightarrow x=\frac{-54}{29}\)

\(x^2\ge0.\Rightarrow A+x^2=x\left(x+29\right)+34\ge-176,25\)

Dấu "=" xảy ra khi: x(x+29) đạtGTNN

<=> x=-14,5

Ta có: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

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

\(\Leftrightarrow x^2+2xy+y^2=z^2\)

\(\Leftrightarrow x^2+y^2-z^2=-2xy\)

Chứng minh tương tự ta có:

\(x^2+z^2-y^2=-2xz\)

\(y^2+z^2-x^2=-2yz\)

\(\frac{xy}{x^2+y^2-z^2}+\frac{xz}{x^2+z^2-y^2}+\frac{yz}{y^2+z^2-x^2}\)

\(=\frac{xy}{-2xy}+\frac{xz}{-2xz}+\frac{yz}{-2yz}\)

\(=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\)

\(=-\frac{3}{2}\)

Vậy giá trị biểu thức là \(-\frac{3}{2}\)