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, \(A\cap B=\left[\dfrac{2}{3};2\right];A\cup B=\left(\dfrac{1}{2};+\infty\right)\)

Cả lớp mình đi làm lễ 20/11 nà^^!

ukm

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simp!

câu 1:

a²+b²+c²+42 = 2a+8b+10c

<=> a²-2a+1+b² -8b+16+c²-10c+25=0

<=> (a-1)²+(b-4)²+(c-5)²=0

<=>a=1 và b=4 và c=5

=> a+b+c = 10

ta có 2(a²+b²)=5ab

<=> 2a²+2b²-5ab=0

<=> 2a²-4ab-ab+2b²=0

<=> 2a(a-2b)-b(a-2b)=0

<=> (a-2b)(2a-b)=0

<=> a=2b(thỏa mãn)

hoặc b=2a( loại vì a>b)

với a=2b =>P=5b/5b=1

\(cos\alpha=\frac{1}{2}\Leftrightarrow\alpha=\frac{-\pi}{3}\)(vì \(\frac{-\pi}{2}< \alpha< 0\))

\(cot\left(\frac{\pi}{3}-\alpha\right)=cot\left(\frac{2\pi}{3}\right)=\frac{-\sqrt{3}}{3}\)