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\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)
\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)
\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)
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
\(A=x^2+3xy+6x+5y^2+7y-2\)
\(=\left[x^2+2x\left(3+\dfrac{3}{2}y\right)+\left(3+\dfrac{3}{2}y\right)^2\right]+5y^2+7y-2-\left(3+\dfrac{3}{2}y\right)^2\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+5y^2+7y-2-9-9y-\dfrac{9}{4}y^2\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+\dfrac{11}{4}y^2-2y-11\)
\(=\left(x+3+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\left(y^2-\dfrac{8}{11}y+\dfrac{16}{121}\right)-\dfrac{125}{11}\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+\dfrac{11}{4}\left(x-\dfrac{4}{11}\right)^2-\dfrac{125}{11}\ge\dfrac{-125}{11}\)Vậy \(Min_A=\dfrac{-125}{11}\) khi \(\left[{}\begin{matrix}x+3+\dfrac{3}{2}y=0\\x-\dfrac{4}{11}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{74}{33}\\x=\dfrac{4}{11}\end{matrix}\right.\)
Biết số nhọ nhưng vẫn làm tiếp:)
\(2,x^4+3x^2+2x+2=\left(x^4+2x^2+1\right)+\left(x^2+2x+1\right)=\left(x^2+1\right)^2+\left(x+1\right)^2>0\left(đpcm\right)\)
\(b,x^2+y^2+z^2+xy+yz+zx\ge0\)
\(\Leftrightarrow2\left(x^2+y^2+z^2+xy+yz+zx\right)\ge0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2xz+z^2\right)+\left(y^2+2yz+z^2\right)\ge0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+z\right)^2+\left(y+z\right)^2\ge0\)
Đúng với mọi x , y ,z
c,\(x^2+y^2+xy+x+y+1\ge0\)
\(\Leftrightarrow2\left(x^2+y^2+xy+y+x+1\right)\ge0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)\ge0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2\ge0\)
Đúng với mọi x , y
Có: \(\left(\frac{x+y}{x-y}\right)^2=\frac{x^2+y^2+2xy}{x^2+y^2-2xy}=\frac{3xy+2xy}{3xy-2xy}=5\)
Mà: \(x>y>0\Rightarrow x+y>0;x-y>0\)
\(\Rightarrow\frac{x+y}{x-y}>0\)
Do đó \(\frac{x+y}{x-y}=\sqrt{5}\)
a: \(A=4\cdot15^2-70^2=-4000\)
b: \(B=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+y+1\right)^2\)
\(=100^2=10000\)
c: \(C=b^2-3b+a^2+3a-2ab\)
\(=\left(a-b\right)^2+3\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+3\right)\)
\(=\left(-5\right)\cdot\left(-5+3\right)=\left(-5\right)\cdot\left(-2\right)=10\)
d: \(D=\left(x-y\right)^3+3xy\left(x-y\right)+3xy\)
\(=\left(-1\right)^3-3xy+3xy\)
=-1
\(H=x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)-95\)
\(H=x^3+x^2-y^3+y^2+xy-3x^2y+3xy^2-3xy-95\)
\(\Leftrightarrow H=x^3-3x^2y+3xy^2-y^3+x^2-2xy+y^2-95\)
\(\Leftrightarrow\left(x-y\right)^3+\left(x-y\right)^2-95\)
\(\Leftrightarrow H=7^3+7^2-95=297\)
1) \(\left(x-3\right)\left(x-5\right)+44\)
\(=x^2-3x-5x+15+44\)
\(=x^2-8x+59\)
\(=x^2-2.x.4+4^2+43\)
\(=\left(x-4\right)^2+43\ge43>0\)
\(\rightarrowĐPCM.\)
2) \(x^2+y^2-8x+4y+31\)
\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)
\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)
\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)
\(\rightarrowĐPCM.\)
3)\(16x^2+6x+25\)
\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)
\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)
\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)
\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)
-> ĐPCM.
4) Tương tự câu 3)
5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)
\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)
-> ĐPCM.
6) Tương tự câu 5)
7) 8) 9) Tương tự câu 3).
Sai đề. Ví dụ: x = y = 1 => x2 - 3xy + y2 = 12 - 3.1.1 + 12 = -1