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1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
b/
\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)
\(=16+8+20=44\)
\(\Rightarrow B\ge11\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
\(F=a^2\left(a+1\right)-b^2\left(b-1\right)+ab-3ab\left(1-1\right)\)(vì a-b=1)
\(F=a^2\left(a+1\right)-b^2\left(b-1\right)+ab\)
\(F=a^3+a^2-b^3+b^2+ab\)
\(F=\left(a^3-b^3\right)+a^2+b^2+ab\)
\(F=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a^2+ab+b^2\right)\)
\(F=\left(a^2+ab+b^2\right)+\left(a^2+ab+b^2\right)\)(vì a-b=1)
\(F=2\left(a^2+ab+b^2\right)\)
\(F=2\left(a^2-2ab+b^2+3ab\right)\)
\(F=2\left(\left(a-b\right)^2+3ab\right)\)
\(F=2\left(1+3ab\right)\)
\(F=2+6ab\)
ta có x+y+z=0
=> \(\left(x+y+z\right)^2=0\)
\(< =>x^2+y^2+z^2+2xy+2xz+2yx=0\)
\(< =>x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(< =>x^2+y^2+z^2+2.0=0\)(vì xy+xz+yz=0)
\(< =>x^2+y^2+z^2=0\)
\(< =>\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}< =>x=y=z=0}\)
thay x=y=z=0 vào
\(K=\left(x-1\right)^{2014}+y^{2015}+\left(z+1\right)^{2016}\)
\(K=\left(0-1\right)^{2014}+0^{2015}+\left(0+1\right)^{2016}\)
\(K=1+0+1=2\)
\(\)
Bài 1 :
\(e,x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y-1\right)^2\)
Bài 2:
\(b,2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\left(x^2+1>0\right)\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{5x+2}{4-x^2}\left(x\ne\pm2\right)\)
\(=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x-8+3x+6-5x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x-2\right)\left(x+2\right)}\)
f) \(x^2+1-\frac{x^4-3x^2+2}{x^2-1}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x^2-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\left(x^2-2\right)\)
\(=x^2+1-x^2+2\)
\(=3\)
1.a) Chịu, nếu thay -4x (hoặc 4y) thành +4x (hoặc -4y) hoặc có thêm gì đó thì tui làm được
b) \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow\left(x+y+z\right)^2-2\left(xy+yz+zx\right)=0\)
\(\Rightarrow x^2+y^2+z^2=0\)\(\Rightarrow x=y=z=0\)
Vậy \(B=-1+0+1=0\)
2.a) \(9x^2+5x+1=\left(3x\right)^2+2.3x.\frac{5}{6}+\frac{25}{36}+\frac{11}{36}\)
\(=\left(3x+\frac{5}{6}\right)^2+\frac{11}{36}\ge\frac{11}{36}\). "=" xảy ra khi \(x=-\frac{5}{18}\). Vậy ....
b) \(1+6x-4x^2=-\left(4x^2-2.2x\frac{3}{2}+\frac{9}{4}\right)+\frac{9}{4}+1\)
\(=-\left(2x-\frac{3}{4}\right)^2+\frac{13}{4}\le\frac{13}{4}\). "=" xảy ra khi \(x=\frac{3}{4}\)