\(\left(x+2\right)^2-2\left(x+2\right)\left(2x-3\right)+\left(2x-3\right)^2=25\\ < =>\left[\left(x+2\right)-\left(2x-3\right)\right]^2=25\\ < =>\left(x+2-2x+3\right)^2-25=0\\ < =>\left(-x+5\right)^2-5^2=0\\ < =>\left(-x+5-5\right)\left(-x+5+5\right)=0\\ < =>-x\left(-x+10\right)=0\\ < =>x\left(x-10\right)=0\\ < =>\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

Vậy: ... 

\(\left(x+2\right)^2-2\left(x+2\right)\left(2x-3\right)+\left(2x-3\right)^2=25\\ \Leftrightarrow\left(x+2-2x+3\right)^2=5^2\\\Leftrightarrow\left(-x+5\right)^2=5^2\\ \Leftrightarrow\left[{}\begin{matrix}-x+5=5\\-x+5=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy...

Ta có:

\(M=\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\\ =\left(y^2-5y+8y-40\right)-\left(y^2+4y-y-4\right)\\ =y^2+3y-40-y^2-3y+4\\ =-36\)

=> Giá trị của bt không phụ thuộc vào biến y

\(M=\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\)

\(=y^2+8y-5y-40-\left(y^2-y+4y-4\right)\)

\(=y^2+3y-40-y^2-3y+4\)

=-36

=>M không phụ thuộc vào biến

`2x^3-3x^2-32x-15`

`=2x^3-10x^2+7x^2-35x+3x-15`

`=2x^2(x-5)+7x(x-5)+3(x-5)`

`=(x-5)(2x^2+7x+3)`

`=(x-5)(2x^2+x+6x+3)`

`=(x-5)[x(2x+1)+3(2x+1)]`

`=(x-5)(2x+1)(x+3)`

\(P=2x^4+3x^2y^2+y^4+y^2\)

\(=2x^4+2x^2y^2+x^2y^2+y^4+y^2\)

\(=2x^2\left(x^2+y^2\right)+y^2\left(x^2+y^2\right)+y^2\)

\(=2x^2+y^2+y^2=2\left(x^2+y^2\right)=2\)

Ta có :

\(P\left(x\right)=2x^4+3x^2y^2+y^4+y^2\)

\(\Rightarrow P\left(x\right)=x^4+2x^2y^2+y^4+x^4+x^2y^2+y^2\)

\(\Rightarrow P\left(x\right)=\left(x^2+y^2\right)^2+x^2\left(x^2+y^2\right)+y^2\)

\(\Rightarrow P\left(x\right)=1^2+x^2.1+y^2\) Vì \(\left(x^2+y^2=1\right)\)

\(\Rightarrow P\left(x\right)=1^2+x^2+y^2=1+1=2\)

Vậy \(P\left(x\right)=2\)

\(0< a< 2\Rightarrow a\left(a-2\right)< 0\Rightarrow a^2< 2a\)

Tương tự: \(\left\{{}\begin{matrix}b\left(b-2\right)< 0\\c\left(c-2\right)< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b^2< 2b\\c^2< 2c\end{matrix}\right.\)

\(\Rightarrow a^2+b^2+c^2< 2\left(a+b+c\right)\)

\(\Rightarrow a^2+b^2+c^2< 2.3=6\)

\(x^2+4y^2-2xy+2x-14y+9=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+2\left(x-y\right)+3y^2-12y+12-3=0\)

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

\(\Leftrightarrow\left(x-y+1\right)^2+3\left(y-2\right)^2=4\) (1)

Do \(\left(x-y+1\right)^2\ge0;\forall x;y\)

\(\Rightarrow3\left(y-2\right)^2\le4\)

\(\Rightarrow\left(y-2\right)^2\le\dfrac{4}{3}\Rightarrow\left[{}\begin{matrix}\left(y-2\right)^2=0\\\left(y-2\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=2\\y=3\\y=1\end{matrix}\right.\)

Thế vào (1): 

Với \(y=1\) \(\Rightarrow x^2=1\Rightarrow x=\pm1\)

Với \(y=2\Rightarrow\left(x-1\right)^2=4\Rightarrow x=\left\{3;-1\right\}\)

Với \(y=3\Rightarrow\left(x-2\right)^2=1\Rightarrow x=\left\{3;1\right\}\)

Vậy \(\left(x;y\right)=\left(-1;1\right);\left(1;1\right);\left(-1;2\right);\left(3;2\right);\left(1;3\right);\left(3;3\right)\)

x=24 nên x+1=25

Sửa đề: \(f\left(x\right)=x^{50}-25x^{49}+25x^{48}-...+25x^2-25x+18\)

\(=x^{50}-x^{49}\left(x+1\right)+x^{48}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+18\)

\(=x^{50}-x^{50}-x^{49}+x^{49}+...+x^3+x^2-x^2-x+18\)

=-x+18=-24+18=-6

Đoạn cuối là \(+25x^2+25x+18\) hay \(+25x^2-25x+18\) em?

\(3x^2+6xy+3y^2-12\left(x+y\right)+4y^2-8y+4-4=0\)

\(\Leftrightarrow3\left(x+y\right)^2-12\left(x+y\right)+12+4\left(y-1\right)^2=16\)

\(\Leftrightarrow3\left(x+y-2\right)^2+4\left(y-1\right)^2=16\) (1)

Do \(3\left(x+y-2\right)^2\ge0;\forall x;y\)

\(\Rightarrow4\left(y-1\right)^2\le16\Rightarrow\left(y-1\right)^2\le4\)

\(\Rightarrow\left(y-1\right)^2=\left\{0;1;4\right\}\)

\(\Rightarrow y=\left\{1;0;2;3;-1\right\}\)

Thế vào (1)

Với \(y=-1\Rightarrow3\left(x-3\right)^2=0\Rightarrow x=3\)

Với \(y=0\Rightarrow3\left(x-2\right)^2=12\Rightarrow x=\left\{0;4\right\}\)

Với \(y=1\Rightarrow...\) các trường hợp còn lại em tự giải tương tự

Đề bài yêu cầu gì em?

\(6k+5\)Do \(p;q>5\Rightarrow p;q\) đều là số lẻ ko chia hết cho 3

\(\Rightarrow p;q\) có dạng \(6k+1\) hoặc \(6k+5\)

Mặt khác \(p< q< p+6\Rightarrow0< q-p< 6\)

\(\Rightarrow q-p\) không chia hết cho 6

\(\Rightarrow q;p\) không thể có cùng dạng \(6k+1\) hoặc cùng dạng \(6k+5\)

\(\Rightarrow\) 1 số có dạng \(6k+1\) và 1 số có dạng \(6k+5\)

Hay 1 số chia 6 dư 1, một số chia 6 dư 5

\(\Rightarrow p+q\) chia 6 dư 0

\(\Rightarrow p+q⋮6\)

\(4x^2+12xy+9y^2-8\left(2x+3y\right)-y^2+2y-1+15=0\)

\(\Leftrightarrow\left(2x+3y\right)^2-8\left(2x+3y\right)+16-\left(y-1\right)^2=1\)

\(\Leftrightarrow\left(2x+3y-4\right)^2-\left(y-1\right)^2=1\)

\(\Leftrightarrow\left(2x+4y-5\right)\left(2x+2y-3\right)=1\)

TH1: \(\left\{{}\begin{matrix}2x+4y-5=1\\2x+2y-3=1\end{matrix}\right.\) \(\Rightarrow x;y=...\)

TH2: \(\left\{{}\begin{matrix}2x+4y-5=-1\\2x+2y-3=-1\end{matrix}\right.\) \(\Rightarrow x;y=...\)