\(F=2x^2+2xy+5y^2-8x-22y\)

<=> \(2F=4x^2+4xy+10y^2-16x-44\)

\(=\left(4x^2+4xy+y^2-16x+16-8y\right)+9y^2-36y-16\)

\(=\left(2x+y-4\right)^2+\left(3y-6\right)^2-52\ge-52\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x+y-4=0\\3y-6=0\end{cases}}\Leftrightarrow y=2;x=1\)

=> min 2F = -52

=> min F = -26. 

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-4\left(a^3+b^3+c^3\right)-12abc\)

\(=-3\left(a^3+b^3+c^3\right)+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-12abc\)

\(=-3\left(\left(a^3+b^3+c^3\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)+4abc\right)\)

XONG NHAAAAA :3333333

\(I=1^2+2^2+3^2+4^2+...+2017^2+2018^2\)

\(I=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2018^2-2017^2\right)\)

\(I=\left(1+2\right)\left(2-1\right)+\left(3+4\right)\left(4-3\right)+...+\left(2017+2018\right)\left(2018-2017\right)\)

\(I=1+2+3+4+...+2017+2018\)

\(I=\frac{\left(2018+1\right).2018}{2}=2037171\)

I = 1² + 2² + 3² + ... + 2017²

= 1.1 + 2.2 + 3.3 + ... + 2017.2017

= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 2017.(2018 - 1)

= 1.2 + 2.3 + 3.4 + ... + 2017.2018 - (1 + 2 + 3 + 4 + ... + 2017)

= 1.2 + 2.3 + 3.4 + ... + 2017.2018 - 2035153

Đặt K = 1.2 + 2.3 + 3.4 + ... + 2017.2018

=> 3K = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2017.2018.3

=> 3K = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.2018.(2019 - 2016)

=> 3K = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 2017.2018.2019 - 2016.2017.2018

=> 3K = 2017.2018.2019

=> K = 2017.2018.2019 : 3 

=> K = 2739315938

Lại có I = K - 2035153

= 2739315938 - 2035153

= 2 737 280 785

pt => \(2x+5=1-x\)

<=> \(3x=4\)

<=> \(x=\frac{4}{3}\)

VẬY    \(x=\frac{4}{3}\)là no duy nhất

Hermit :) làm nhầm rồi em.

\(\sqrt{2x+5}=\sqrt{1-x}\Leftrightarrow\hept{\begin{cases}1-x\ge0\\2x+5=1-x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\3x=-4\end{cases}}\Leftrightarrow x=-\frac{4}{3}\)

Vậy x = -4/3

lớp 9 thì mình dùng cách lớp 9 

\(\sqrt{x+2\sqrt{x}-1}=2\left(đk:x\ge1\right)\)

\(< =>x+2\sqrt{x}-1=4\)(bình phương 2 vế)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)(*)

\(< =>t^2+2t-5=0\)

\(\Delta=2^2-4.\left(-5\right)=4+20=24\)

\(\orbr{\begin{cases}t_1=\frac{-2+2\sqrt{6}}{2}=-1+\sqrt{6}\left(tm\right)\\t_2=\frac{-2-2\sqrt{6}}{2}=-1-\sqrt{6}\left(ktm\right)\end{cases}}\)

Khi đó thế vào * ta được :

 \(\sqrt{x}=\sqrt{6}-1< =>x=7-2\sqrt{6}\left(tmđk\right)\)

Vậy nghiệm của phương trình trên là \(7-2\sqrt{6}\)

ĐK: \(x\ge1\)

\(\sqrt{x+2\sqrt{x-1}}=2\)

<=> \(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=2\)

<=> \(\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)

<=> \(\sqrt{x-1}+1=2\)

<=> \(\sqrt{x-1}=1\)

<=> x - 1 = 1 

<=> x = 2 thỏa mãn

M = \(54-\frac{1}{2}\left(1+2\right)-\frac{1}{3}\left(1+2+3\right)-...-\frac{1}{12}\left(1+2+3+...+12\right)\)

\(54-\left(\frac{3}{2}+\frac{4}{2}+...+\frac{13}{2}\right)=54-\frac{1}{2}\left(3+4+5+...+13\right)=54-\frac{1}{2}.176=54-88\)

= -24

\(I=1\left(2-1\right)+2\left(3-1\right)+...+2017\left(2018-1\right)\)

\(I=\left(1.2+2.3+...+2017.2018\right)-\left(1+2+...+2017\right)\)

\(3I=\left(1.2.3+2.3\left(4-1\right)+...+2017.2018\left(2019-2016\right)\right)-\frac{3.2017.2018}{2}\)

=> \(I=\frac{2017.2018.2019}{3}-2017.1009\)

=> \(I=.....\)

dài :vv

a) \(\left|2x-5\right|=4\Leftrightarrow\hept{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Leftrightarrow\hept{\begin{cases}2x=9\\2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)

b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)

\(\Leftrightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)

Bài 1 :

a) \(|2x-5|=4\)

\(\Rightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Rightarrow}\orbr{\begin{cases}2x=9\\2x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)

b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)

\(\Rightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)

c) \(\left|\frac{-2}{3}\right|+\left|x-\frac{1}{3}\right|=\left|-1\right|-\left|\frac{-1}{3}\right|\)

\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=1-\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=\frac{2}{3}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=0\Rightarrow x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)

d) \(\left|-\frac{1}{2}\right|-\left|x+\frac{1}{4}\right|=\left|-\frac{3}{4}\right|\)

\(\Rightarrow\frac{1}{2}-\left|x+\frac{1}{4}\right|=\frac{3}{4}\)

\(\Rightarrow\left|x+\frac{1}{4}\right|=-\frac{1}{4}\)

Vì \(\left|x\right|\ge0\Rightarrow\)ko có gtri nào của x thỏa mãn đề bài

Bài 2 :

a) \(\left|x-1\right|=3x+2\)

\(\Rightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}x-3x=2+1\\x+3x=-2+1\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}-2x=3\\4x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{-1}{4}\end{cases}}\)

b|) \(\left|9+x\right|=2x\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-2x=-9\\x+2x=-9\end{cases}\Rightarrow\orbr{\begin{cases}-x=-9\\3x=-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=9\\x=-3\end{cases}}}\)

c) \(\left|x+6\right|-9=2x\Rightarrow\left|x+6\right|=2x+9\)

\(\Rightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=-2x-9\end{cases}\Rightarrow}\orbr{\begin{cases}x-2x=9-6\\x+2x=-9-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=3\\3x=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}}\)

Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^