Cách hack điểm hỏi đáp trên OLM: https://www.youtube.com/watch?v=sMvl8_N_N54

chung minh tu giac abek noi tiep duoc mot duong tron

Pika...........................chịu!

>-<

Giải

a) Ta có : 2.x² -2.x = 5.x 

<=> 2.x² -3.x-5=0 : a = 2 ; b = 3 ; c = -5 

b) Ta có : x² +2.x = m. x + m 

<=> x² + ( 2-m ) .x - m = 0 : a = 1 ; b=2-m ; c=-m

c) Ta có : 2.x² \(+\sqrt{2}.\left(3.x-1\right)=1+\sqrt{2}\)

<=>  2.x²  + 3.\(\sqrt{2}.x-2.\sqrt{2}-1=0\): a = 2 ; b= 3\(\sqrt{2};c=-2\sqrt{2}-1\)

a) \(2x^2-2x=5+x\)

\(\Leftrightarrow2x^2-x-5=0\)với \(\hept{\begin{cases}a=2\\b=-3\\c=-5\end{cases}}\)

b) \(x^2+2x=mx+m\)

\(\Leftrightarrow x^2+\left(2-m\right)x-m=0\)với \(\hept{\begin{cases}z=1\\b=3-m\\c=-m\end{cases}}\)

c) \(2x^2+\sqrt{2}\left(3x-1\right)=1+\sqrt{2}\)

\(\Leftrightarrow2x^2+3\sqrt{2}\cdot x-2\sqrt{2}-1=0\)

với \(\hept{\begin{cases}a=2\\b=3\sqrt{2}\\c=-2\sqrt{2}-1\end{cases}}\)

hệ \(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x+2y\right)+\left(x-y\right)=0\\x^2-y^2+x+y=6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x+2y+1\right)=0\left(1\right)\\x^2-y^2+x+y=6\left(2\right)\end{cases}}\)

Th1: x=y

pt 2<=> 2x=6

<=> x=y=3

Th2: x+2y+1=0

<=> x=-1-2y

=> pt (2) <=> \(\left(-1-2y\right)^2-y^2-1-2y+y=6\)

\(\Leftrightarrow4y^2+4y+1-y^2-1-2y+y=6\)

\(\Leftrightarrow3y^2+3y-6=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)

KL:............................

Gọi số hs lớp 9A là x => số hsg của lớp 9A là \(\frac{x.60}{100}\)

Gọi số hs lớp 9B là y => số hsg của lớp 9b là \(\frac{y.75}{100}\)

=> Ta có pt (1) \(\frac{60x}{100}+\frac{75y}{100}=51\Leftrightarrow12x+15y=1020\)

Ta có hệ PT

\(\hept{\begin{cases}x+y=76\\12x+15y=1020\end{cases}}\)

Giải hệ PT trên 

Ta có \(\frac{x^3}{\left(y+z\right)^2}=\frac{x^3}{\left(2018-x\right)^2}\)

Xét \(\frac{x^3}{\left(2018-x\right)^2}\ge x-\frac{1009}{2}\)

<=> \(x^3\ge\left(2018^2-2.2018.x+x^2\right)\left(x-\frac{1009}{2}\right)\)

<=> \(x^3\ge x^3-x^2\left(\frac{1009}{2}+2018.2\right)+x\left(2018.1009+2018^2\right)-\frac{2018^2.1009}{2}\)

<=> \(\frac{9081}{2}x^2-6.1009^2.x+2018.1009^2\ge0\)

<=> \(\frac{9081}{2}\left(x^2-\frac{2.2018}{3}.x+\left(\frac{2018}{3}\right)^2\right)\ge0\)

<=> \(\frac{9081}{2}\left(x-\frac{2018}{3}\right)^2\ge0\)( luôn đúng)

=> \(\frac{x^3}{\left(y+z\right)^2}\ge x-\frac{1009}{2}\)

Khi đó \(VT\ge x-\frac{1009}{2}+y-\frac{1009}{2}+z-\frac{1009}{2}=2018-\frac{3}{2}.1009=\frac{1009}{2}\)(ĐPCM)

Dấu bằng xảy ra khi \(x=y=z=\frac{2018}{3}\)

Ta có : \(\frac{x^3}{\left(y+z\right)^2}=\frac{x^3}{\left(2018-x\right)^2}\)

xét \(\frac{x^3}{\left(2018-x\right)^2}\ge x-\frac{1009}{2}\)

<=> \(x^3\ge\left(x^2-2.2018.x+2018^2\right)\left(x-\frac{1009}{2}\right)\)

<=> \(x^3\ge x^3-x^2\left(\frac{1009}{2}+2.2018\right)+x\left(2018^2+1009.2018\right)-\frac{2018^2.1009}{2}\ge0\)

<=> \(\frac{9081}{2}x^2-6.1009^2.x+2018.1009^2\ge0\)

<=> \(\frac{9081}{2}.\left(x-\frac{2018}{3}\right)^2\ge0\)( luôn đúng)

=> \(\frac{x^3}{\left(y+z\right)^2}\ge x-\frac{1009}{2}\)

Khi đó \(P\ge x+y+z-\frac{3.1009}{2}=\frac{1009}{2}\)(ĐPCM)

Dấu bằng xảy ra khi \(x=y=z=\frac{2018}{3}\)