\(A=\dfrac{4\left(x^2-4x+4\right)+\left(x^2-8x+16\right)}{x^2-4x+4}=4+\left(\dfrac{x-4}{x-2}\right)^2\ge4\)

\(A_{min}=4\) khi \(x=4\) (A max ko tồn tại)

\(B=\dfrac{6\left(x^2+2x+1\right)+\left(4x^2+12x+9\right)}{x^2+2x+1}=6+\left(\dfrac{2x+3}{x+1}\right)^2\ge6\)

\(B_{min}=6\) khi \(x=-\dfrac{3}{2}\) 

B max ko tồn tại

gửi bạn nhé chúc bn lm bài tốt ^^

Ta có: A = (x+z)(y+t) = xy+zy+xt+zt

Áp dụng BĐT Cô-si, có:

x^2 + y^2 >= 2xy

y^2 + z^2 >= 2yz

z^2 + t^2 >= 2zt

t^2 + x^2 >= 2yt

=> 2(xy+yz+zt+tx) <= 2(x^2+y^2+z^2+t^2)

=>xy+yz+zt+tx <= x^2+y^2+z^2+t^2 = 1

Vậy max A = 1 khi x^2=y^2=z^2=t^2=1/4

đk: x≥1; y≥4; z≥9

Ta có: \(A=\dfrac{yz\sqrt{x-1}+xz\sqrt{y-4}+xy\sqrt{z-9}}{xyz}\)

\(=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-4}}{y}+\dfrac{\sqrt{z-9}}{z}\)

A/dụng bđt côsi cho 2 số không âm \(\sqrt{x-1}\) và 1 có:

\(\sqrt{x-1}=\sqrt{1\left(x-1\right)}\le\dfrac{1+x-1}{2}=\dfrac{x}{2}\)

\(\Rightarrow\dfrac{\sqrt{x-1}}{x}\le\dfrac{x}{2x}=\dfrac{1}{2}\)

Tương tự: \(\dfrac{\sqrt{y-4}}{y}\le\dfrac{1}{4};\dfrac{\sqrt{z-9}}{z}\le\dfrac{1}{6}\)

Cộng theo vế các Bđt trên ta có:

\(\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-4}}{y}+\dfrac{\sqrt{z-9}}{z}\le\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}\)

hay \(A\le\dfrac{11}{12}\)

Dấu ''='' xảy ra khi x = 2; y = 8; z = 18

vậy max_A = \(\dfrac{11}{12}\)

Tìm GTLN 

a) Ta có: A = 15 - 3 | x - 7 | 

Để A đạt GTLN khi 3 | x - 7 |  đạt GTNN

 \(\Rightarrow3\left|x-7\right|=0\Rightarrow\left|x-7\right|=0\Rightarrow x-7=0\Rightarrow x=7\)

Vậy để biểu thức đạt GTLN khi A = 15 và x = 7 

Ừa, nhầm 1 xíu

\(B=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\ge1+\frac{2}{0+1}=3\)

Lý giải:

\(\sqrt{x}\ge0;\forall x\Rightarrow\sqrt{x}+1\ge0+1\)

\(\Rightarrow\frac{2}{\sqrt{x}+1}\le\frac{2}{0+1}\Rightarrow1+\frac{2}{\sqrt{x}+1}\ge1+\frac{2}{0+1}\)

Anh Mai

\(B=2-\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+3}{\sqrt{x}+1}\)

\(B=1+\frac{2}{\sqrt{x}+1}\le1+\frac{2}{0+1}=3\)

\(B_{max}=3\) khi \(x=0\)

\(C=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+7\right)}\)

\(\Rightarrow\frac{1}{C}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+7\right)}{\sqrt{x}-1}=\frac{x+8\sqrt{x}+7}{\sqrt{x}-1}=\sqrt{x}+9+\frac{16}{\sqrt{x}-1}\)

\(\frac{1}{C}=\sqrt{x}-1+\frac{16}{\sqrt{x}-1}+10\ge2\sqrt{\frac{16\left(\sqrt{x}-1\right)}{\sqrt{x}-1}}+10=18\)

\(\Rightarrow\frac{1}{C}\ge18\Rightarrow C\le\frac{1}{18}\)

\(\Rightarrow C_{max}=\frac{1}{18}\) khi \(\sqrt{x}-1=4\Leftrightarrow x=25\)

\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)

\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)

 \(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)

         y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)

         y = \(\dfrac{245}{64}\)

2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)

\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)

\(\dfrac{12}{5}\): y        = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)

 \(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)

        y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)

        y = \(\dfrac{15}{13}\)

các bạn giúp m với =(((((