a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

c) \(C=4x-10-x^2=-\left(x^2-4x+10\right)\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2+6\right]\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2\right]-6\le-6< 0\forall x\)

x^2-xy+y^2

=x^2-2*x*1/2y+1/4y^2+3/4y^2

=(x-1/2y)^2+3/4y^2>0 với mọi x,y thỏa mãn \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

\(B=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\)

Áp dụng bđt AM-GM ta có

\(x^2-xy+y^2\ge x^2+y^2-\frac{x^2+y^2}{2}=\frac{x^2+y^2}{2}\)

\(\Rightarrow\frac{x+y}{x^2-xy+y^2}\le\frac{2\left(x+y\right)}{x^2+y^2}\le\frac{2\sqrt{2\left(x^2+y^2\right)}}{x^2+y^2}=\frac{2\sqrt{2}}{\sqrt{x^2+y^2}}\)

Dấu "=" xảy ra khi x=y=1

a: Ta có: \(a^2+b^2+c^2=ab+bc+ac\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

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

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

=>a=b=c

b: ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

Ta có : x² + 2x + 2

= x² + 2x + 1 + 1

= (x + 1)² + 1 \(\ge1\forall x\)

Vậy  x² + 2x + 2 \(>0\forall x\)

Ta có : x² + 2x + 2

=> x² + 2x + 1 + 1

=> ( x + 1)² + 1  >  1\(\forall x\)

Vậy x² + 2x + 2   > \(0\forall x\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

rtvb5tb4tyn 4h

\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)

\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)

ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)

Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)

T i c k cho mình 1 cái nha mới bị trừ 50 đ