5.
\(A=\dfrac{x}{x+\sqrt{x+yz}}+\dfrac{y}{y+\sqrt{y+zx}}+\dfrac{z}{z+\sqrt{z+xy}}\)
\(=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}+\dfrac{y}{y+\sqrt{y\left(x+y+z\right)+zx}}+\dfrac{z}{z+\sqrt{z\left(x+y+z\right)+xy}}\)
\(=\dfrac{x}{x+\sqrt{x^2+xy+yz+zx}}+\dfrac{y}{y+\sqrt{y^2+xy+yz+zx}}+\dfrac{z}{z+\sqrt{z^2+xy+yz+zx}}\)
\(=\dfrac{x\left(\sqrt{x^2+xy+yz+zx}-x\right)}{xy+yz+zx}+\dfrac{y\left(\sqrt{y^2+xy+yz+zx}-y\right)}{xy+yz+zx}+\dfrac{z\left(\sqrt{z^2+xy+yz+zx}-z\right)}{xy+yz+zx}\)
\(=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}-x^2}{xy+yz+zx}+\dfrac{y\sqrt{\left(x+y\right)\left(y+z\right)}-y^2}{xy+yz+zx}+\dfrac{z\sqrt{\left(z+x\right)\left(y+z\right)}-z^2}{xy+yz+zx}\)
Áp dụng BĐT \(ab\le\dfrac{a^2+b^2}{2}\) và BĐT \(a^2+b^2+c^2\ge ab+bc+ca\)
\(A=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}-x^2}{xy+yz+zx}+\dfrac{y\sqrt{\left(x+y\right)\left(y+z\right)}-y^2}{xy+yz+zx}+\dfrac{z\sqrt{\left(z+x\right)\left(y+z\right)}-z^2}{xy+yz+zx}\)
\(=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}+y\sqrt{\left(x+y\right)\left(y+z\right)}+z\sqrt{\left(z+x\right)\left(y+z\right)}-\left(x^2+y^2+z^2\right)}{xy+yz+zx}\)
\(\le\dfrac{x.\dfrac{2x+y+z}{2}+y.\dfrac{x+2y+z}{2}+z.\dfrac{x+y+2z}{2}-\left(x^2+y^2+z^2\right)}{xy+yz+zx}\)
\(=\dfrac{xy+yz+zx}{xy+yz+zx}=1\)
\(maxA=1\Leftrightarrow x=y=z=\dfrac{1}{3}\)
1.
a, \(A=(\dfrac{1}{2};2];B=[\dfrac{2}{3};+\infty)\)
b, \(A\cap B=\left[\dfrac{2}{3};2\right];A\cup B=\left(\dfrac{1}{2};+\infty\right)\)
2.
ĐK: \(x\ne2;x\ne-3\)
\(1+\dfrac{2}{x-2}=\dfrac{10}{x+3}-\dfrac{50}{\left(2-x\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{2}{2-x}+\dfrac{10}{x+3}-\dfrac{50}{\left(2-x\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\dfrac{2\left(x+3\right)}{\left(2-x\right)\left(x+3\right)}+\dfrac{10\left(2-x\right)}{\left(2-x\right)\left(x+3\right)}-\dfrac{50}{\left(2-x\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\dfrac{-8x-24}{-x^2-x+6}=1\)
\(\Leftrightarrow-8x-24=-x^2-x+6\)
\(\Leftrightarrow x^2-7x-30=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=-3\left(l\right)\end{matrix}\right.\)
3.
\(A\left(0;3\right)\in\left(P\right)\Rightarrow c=3\)
\(I\left(2;-1\right)\) là đỉnh của \(\left(P\right)\Rightarrow\left\{{}\begin{matrix}-\dfrac{b}{2a}=2\\-\dfrac{\Delta}{4a}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-4a\\b^2-4ac=4a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-4a\\b^2=16a\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=b=0\left(l\right)\\a=1;b=-4\end{matrix}\right.\)
Vậy \(a=1;b=-4;c=3\)
b, \(y=f\left(x\right)=x^2-4x+3\)
Bảng biến thiên:
Đồ thị hàm số:
c, Yêu cầu bài toán thỏa mãn khi phương trình \(x^2-6x+3-m=0\) có nghiệm duy nhất
\(\Leftrightarrow\Delta'=m+6=0\Leftrightarrow m=-6\)
4.
Hình vẽ:
a, Gọi D là trung điểm AB, khi đó:
\(\overrightarrow{MA}+\overrightarrow{MB}-\overrightarrow{MC}=\overrightarrow{0}\)
\(\Leftrightarrow2\overrightarrow{MD}=\overrightarrow{MC}\)
\(\Rightarrow\) M đối xứng với C qua D
\(\Rightarrow ACBM\) là hình bình hành
b, Gọi E là trung điểm AC
\(AC=\sqrt{AB^2+BC^2}=\sqrt{25a^2}=5a\)
\(MC=2CD=2\sqrt{\dfrac{1}{4}AB^2+BC^2}=2\sqrt{\dfrac{73}{4}a^2}=a\sqrt{73}\)
\(\Rightarrow ME^2=\dfrac{MA^2+MC^2}{2}-\dfrac{AC^2}{4}=\dfrac{153}{4}a^2\Rightarrow ME=\dfrac{\sqrt{153}}{2}a\)
\(\left|2\overrightarrow{MA}+\overrightarrow{MB}\right|=\left|\overrightarrow{MA}+2\overrightarrow{MD}\right|=\left|\overrightarrow{MA}+\overrightarrow{MC}\right|=2\left|\overrightarrow{ME}\right|=2ME=a\sqrt{153}\)
c, \(\overrightarrow{MB}.\overrightarrow{MC}=MB.MC.cosBMC=...\)
