vào tìm kiems có câu tương tự nhé

\(M=9x^2-6x+1+x+\frac{1}{9x}+2019\)

\(M=\left(3x-1\right)^2+x+\frac{1}{9x}+2019\ge\left(3x-1\right)^2+\frac{2}{3}+2019\left(AM-GM\right)\)

\(MinM=\frac{6059}{3}\)

Đẳng thức xảy ra khi x=1/3

My Nguyễn ơi,bạn truy cập vào đường link này để tìm câu hỏi tương tự của câu a/Bài 1 nhé

https://vn.answers.yahoo.com/question/index?qid=20110206184834AAokV5m&sort=N

Ko biết đợi đứa khác đê

\(\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4\sqrt{x}}+\frac{4x\sqrt{x}+4\sqrt{x}}{4x^2+9x+18\sqrt{x}+9}-2=\frac{\left(-4x\sqrt{x}+4x^2+9x+22\sqrt{x}+9\right)^2}{\left(4x^2+9x+18\sqrt{x}+9\right)\left(4x\sqrt{x}+4\sqrt{x}\right)}\ge0\)

Đặt \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}\left(x>0\right)\Rightarrow M>0\)

Đặt \(y=\sqrt{x}>0\)ta có \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}=\frac{4y^4+9y^2+18y+9}{4y^3+4y^2}\)\(=\frac{3\left(4y^3+4y^2\right)+\left(4y^2-12y^3-3y^2+18y+9\right)}{4y^3+4y^2}=3+\frac{\left(2y^2-3y-3\right)^2}{4y^3+4y^2}\ge3\)

\(y>0\Rightarrow\hept{\begin{cases}4y^3+4y^2>0\\\left(2y^2-3y-3\right)^2\ge0\end{cases}\Rightarrow\frac{\left(2y-3y-3\right)^2}{4y^3+4y^2}\ge0}\)

Đẳng thức xảy ra \(\Leftrightarrow2y^2-3y-3=0\Leftrightarrow y=\frac{3+\sqrt{33}}{4}\left(y>0\right)\)

\(\Rightarrow x=\left(\frac{3+\sqrt{33}}{4}\right)^2=\frac{21+3\sqrt{33}}{8}\)

Khi đó \(A=M+\frac{1}{M}=\frac{8M}{9}+\left(\frac{M}{9}+\frac{1}{M}\right)\ge\frac{8\cdot3}{9}+2\sqrt{\frac{M}{9}\cdot\frac{1}{M}}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}M=3\\\frac{M}{9}=\frac{1}{M}\end{cases}\Leftrightarrow M=3\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}}\)

Vậy \(A_{min}=\frac{10}{3}\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}\)

ta có: \(4x^2+9x+18\sqrt{x}+9=4x^2+9\left(\sqrt{x}+1\right)^2\),\(4x\sqrt{x}+4x=4x\left(\sqrt{x}+1\right)\)
Đặt \(a=x,b=\sqrt{x}+1\)ta có:
\(A=\frac{4a^2+9b^2}{4ab}+\frac{4ab}{4a^2+9b^2}=t+\frac{1}{t},t=\frac{4a^2+9b^2}{4ab}\)
có \(\frac{4a^2+9b^2}{4ab}=t\Rightarrow4a^2-t.4ab+9b^2=0\Leftrightarrow4.\left(\frac{a}{b}\right)^2-4t.\frac{a}{b}+9=0,\)do a khác 0.
Đặt \(\frac{a}{b}=y\Rightarrow4y^2-t.4y+9=0\), \(\Delta=16t^2-36\ge0\Leftrightarrow t\ge\frac{3}{2}\left(t>0\right)\)
xét \(f\left(t\right)=t+\frac{1}{t}\left(t\ge\frac{3}{2}\right)\)
lấy \(\frac{3}{2}< t_1< t_2\)
\(\Rightarrow f\left(t_1\right)-f\left(t_2\right)=\left(t_1-t_2\right)\left(\frac{t_1.t_2-1}{t_1.t_2}\right)< 0\)
suy ra với t càng tăng thì f(t) càng lớn vậy min \(f\left(t\right)=\frac{3}{2}+\frac{2}{3}=\frac{13}{6}\)
các em tự tìm x nhé.

bài này bạn áp dụng BĐT cô si cko 2 số dương là đc.

đáp án: Min A=  2

\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

dùng côsi ra = 1 chắc v

ê tuấn nếu cô-si thì mk nghĩ phải =2 chứ sao =1 được 

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

\(A=\sqrt{1^2-2\cdot3x\cdot1+\left(3x\right)^2}+\sqrt{\left(3x\right)^2-2\cdot2\cdot3x+2^2}\)

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

\(A=\left|1-3x\right|+\left|3x-2\right|\)

\(A=\left|1-3x+3x-2\right|\)

\(A=\left|-1\right|=1\)

Dấu "=" xảy ra \(\left(1-3x\right)\left(3x-2\right)\ge0\)

\(\Rightarrow\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

Vậy: \(A_{min}=1\) khi \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

\(a,ĐK:9x^2-1\ne0\Leftrightarrow x^2\ne\frac{1}{9}\Leftrightarrow x\ne\pm\frac{1}{3}\)

\(b,M=\frac{\sqrt{9x^2-6x+1}}{9x^2-1}=\frac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\frac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}\)

với \(3x-1>0\) ta có \(M=\frac{3x-1}{\left(3x-1\right)\left(3x+1\right)}=\frac{1}{3x+1}\)

với \(3x-1< 0\) ta có \(M=\frac{-\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}=-\frac{1}{3x+1}\)

\(c,\) th1 : \(M=\frac{1}{3x+1}\)  khi \(x>\frac{1}{3}\) mà \(M=\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{3x+1}=\frac{1}{4}\Leftrightarrow x=1\left(thoaman\right)\) 

th2 : \(M=-\frac{1}{3x+1}\) khi \(x< \frac{1}{3}\) mà \(M=\frac{1}{4}\)

\(\Leftrightarrow\frac{-1}{3x+1}=\frac{1}{4}\Leftrightarrow3x+1=-4\Leftrightarrow x=-\frac{5}{3}\left(thoaman\right)\)

\(d,M=\frac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}< 0\) có \(\left|3x-1\right|>0\)

\(\Rightarrow\left(3x-1\right)\left(3x+1\right)< 0\)

th1 : \(\hept{\begin{cases}3x-1>0\\3x+1< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{1}{3}\\x< -\frac{1}{3}\end{cases}\left(voli\right)}}\)

th2 : \(\hept{\begin{cases}3x-1< 0\\3x+1>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{1}{3}\\x>-\frac{1}{3}\end{cases}\Leftrightarrow-\frac{1}{3}< x< \frac{1}{3}}\)