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

\(=\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)+\left(y^2-6y+9\right)-7\)

\(=\left(x-y\right)^2+\left(x+2\right)^2+\left(y-3\right)^2-7\)

Đề hình như có gì đó không đúng

Ta có: \(A=x^2+2y^2-2xy+4x-6y+6=\left(x^2-2xy+y^2\right)\)          \(+4\left(x-y\right)+4+y^2-2y+1+1=\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]\)\(+\left(y-1\right)^2+1=\left(x-y+2\right)^2+\left(y-1\right)^2+1\)

Ta có: \(\left(x-y+2\right)^2\ge0\forall x,y\); \(\left(y-1\right)^2\ge0\forall y\)nên \(\left(x-y+2\right)^2+\left(y-1\right)^2+1>0\forall x,y\)

Vậy \(A=x^2+2y^2-2xy+4x-6y+6>0\forall x,y\)(đpcm)

phần a sai đề nha bạn 

b,Ta có

      \(2\equiv2\left(mod13\right)\)

\(\Rightarrow2^{12}\equiv1\left(mod13\right)\)

\(\Rightarrow2^{12.5}.2^{10}\equiv1.2^{10}\left(mod13\right)\)

\(\Rightarrow2^{60}.2^{10}\equiv1024\left(mod13\right)\)

\(\Rightarrow2^{70}\equiv10\left(mod13\right)\)\(\left(1\right)\)

Lại có:

\(3\equiv3\left(mod13\right)\)

\(\Rightarrow3^6\equiv1\left(mod13\right)\)

\(\Rightarrow3^{6.11}.3^4\equiv1.3^4\left(mod13\right)\)

\(\Rightarrow3^{66}.3^4\equiv81\left(mod13\right)\)

\(\Rightarrow3^{70}\equiv3\left(mod13\right)\)\(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow2^{70}+3^{70}\equiv13\equiv0\left(mod13\right)\)

c, Ta có

\(17\equiv-1\left(mod18\right)\)

\(\Rightarrow17^{19}\equiv-1\left(mod18\right)\)\(\left(1\right)\)

Lại có

\(19\equiv1\left(mod18\right)\)

\(\Rightarrow19^{17}\equiv1\left(mod18\right)\)\(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow17^{19}+19^{17}\equiv0\left(mod18\right)\)

\(\Rightarrow17^{19}+19^{17}⋮18\)

Bài 1

a) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x-1\right)\left(x+1\right)\)

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

\(=3x^3+6x-3x^3+3x=9x\)

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

\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2bc-2ca+4a^2-4ab+b^2\)

\(=6a^2+3b^2+2c^2+4ab-4ab=6a^2+3b^2+2c^2\)

Bài 2 

a) \(x^2-20x+101=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)

Dấu = xảy ra \(< =>\left(x-10\right)^2=0< =>x-10=0< =>x=10\)

b) \(4a^2+4a+2=4\left(a^2+a+\frac{1}{4}\right)+1=4\left(a+\frac{1}{2}\right)^2+1\ge1\)

Dấu = xảy ra \(< =>4\left(a+\frac{1}{2}\right)^2=0< =>a+\frac{1}{2}=0< =>a=-\frac{1}{2}\)

c) \(x^2-4xy+5y^2+10x-22y+28=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+y^2-2y+1+27\)

\(=\left(x-2y\right)^2+2.5.\left(x-2y\right)+25+\left(y-1\right)^2+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu = xảy ra \(< =>\hept{\begin{cases}y-1=0\\x-2y+5=0\end{cases}< =>\hept{\begin{cases}y=1\\x=-3\end{cases}}}\)

Bài 3 

a) \(4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Dấu = xảy ra \(< =>\left(x-2\right)^2=0< =>x-2=0< =>x=2\)

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

Dấu = xảy ra \(< =>\left(x-\frac{1}{2}\right)^2=0< =>x-\frac{1}{2}=0< =>x=\frac{1}{2}\)

(x + 3)(x² - 3x + 5) = x² + 3x

=> x(x² - 3x + 5) + 3(x² - 3x + 5) = x² + 3x

=> x³ - 3x² + 5x + 3x² - 9x + 15 = x² + 3x

=> x³ - 3x² + 5x + 3x² - 9x + 15 - x² - 3x = 0

=> x³ + (-3x² + 3x² - x²) + (5x - 9x - 3x) + 15 = 0

=> x³ - x² - 7x + 15 = 0

=> \(\left(x+3\right)\left(x^2-4x+5\right)=0\)

=> x = -3 ( vì x² - 4x + 5 = (x - 2)² + 1 \(\ge\)1\(\forall\)x)

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

\(\Leftrightarrow x^3-3x^2+5x+3x^2-9x+15-x^2-3x=0\)

\(\Leftrightarrow x^3-x^2-7x+15=0\)( vô nghiệm ) 

a) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(Vt\ge0\left(\forall a,b,c\right)\) nên dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow a=b=c\)

Ta có : a² + b² + c² = ab + bc + ca

=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

= (a² - 2ab + b²) +  (b² - 2bc + c²) + (c² - 2ca + a²) = 0

=> (a - b)² + (b - c)² + (c - a)² = 0

=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)

b) Ta có :  2(x² + t²) + (y + t)(y - t) = 2x(y + t)

=> 2x² + 2t² + y² - t² = 2xy + 2t

=> 2x² + t² + y² = 2xt + 2xy

=> 2x² + t² + y² - 2xt - 2xy = 0

=> (x² - 2xy + y²) + (x² + t² - 2xt)  = 0

=> (x - y)² + (x - t)² = 0

=> \(\hept{\begin{cases}x-y=0\\x-t=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y\\x=t\end{cases}}\Rightarrow x=y=t\left(\text{đpcm}\right)\)

c) Ta có a + b + c = 0 

=> (a + b + c)² = 0

=> a² + b² + c² + 2ab + 2bc + 2ca = 0

=> a² + b² + c² + 2(ab + bc + ca) = 0

=> a² + b² + c² = 0

=> a = b = c = 0

Khi đó A = (0 - 1)²⁰⁰³ + 0²⁰⁰⁴ + (0 + 1)²⁰⁰⁵

= - 1 + 0 + 1 = 0

Vậy A = 0

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

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

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

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

\(=VP\left(đpcm\right)\)         

Ta có: \(a^3-b^3+ab\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)\)

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

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