2.

\(x^2-2x-11-y^2\)

\(\Leftrightarrow\left(x-2\right)x-y^2-11\)

\(\Leftrightarrow\left(x-1\right)^2-y^2-12\)

\(\Leftrightarrow y=-\sqrt{x^2-2x-11}\)

\(\Leftrightarrow y=\sqrt{x^2-2x-11}\)

\(x=4\left(y^2+12\right)\)

\(\Rightarrow x=-3;y=⊥2\)

\(x=5;y=⊥2\)\(\)

phê phết chú ạ

1. Ta có : \(-x^2+4x+4=-\left(x^2-4x-4\right)=-\left(x^2-2\cdot x\cdot2+2^2\right)+8\)

\(=-\left(x-2\right)^2+8\)

Vì \(\left(x-2\right)^2\ge0\forall x\)

=> \(-\left(x-2\right)^2\le0\forall x\)

=> \(-\left(x-2\right)^2+8\le8\forall x\)

Dấu " = " xảy ra khi và chỉ khi -(x - 2)² = 0 => x = 2

Vậy GTLN là 8 khi x = 2

2. \(4-16x^2-8x=16x^2-8x-4\)

\(=\left[\left(4x\right)^2-2\cdot4x\cdot1+1^2\right]-5\)

\(=\left(4x-1\right)^2-5\)

Vì \(\left(4x-1\right)^2\ge0\forall x\)

=> \(\left(4x-1\right)^2-5\le-5\forall x\)

Dấu " = " xảy ra khi và chỉ khi (4x - 1)² = 0 => x = 1/4

Vậy GTLN là -5 khi x = 1/4

2. Ta có : \(x^2+2x+y^2-6y+10=0\)

=> \(\left(x^2+2x+1\right)+\left(y^2-6y+9\right)=0\)

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

Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x+1\right)^2+\left(y-3\right)^2\ge0\forall x,y\)

Dấu " = " xảy ra khi và chỉ khi

+) (x + 1)² = 0 => x = -1

+) (y - 3)² = 0 => y = 3

Vậy GTNN bằng 0 khi x = -1,y = 3

Bài 3 làm nốt nhé

P/S : K chắc :<

Giải thích các bước giải:CÂU 3 

3a = (4-1) (4+1) (4^2+1) (4^4+1) (4^8+1) (4^16+1)

=(4^2-1) (4^2+1) (4^8+1) (4616+1)

=(4^8-1) (4^8+1 ) (4^16+1)

=(4^16-1)(4^16+1)

=4^32-1 =b ( dpcm)

câu 2: (x+1)^2 +(y-3)^2=0 nếu x=-1 và ngược lại

a) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)

\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)

b) \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

c) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-bc^2\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(a^2+bc-ab-ca\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Nhầm đoạn cuối là \(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)