tìm min của
a) \(A=x+\frac{1}{4x}\)
b) \(B=x+\frac{16}{x-2}\)
c) \(C=3+\sqrt{x^2-6x+11}\)
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20 tháng 7 2017
3,
đặt \(\hept{\begin{cases}\sqrt{x^2+y^2}=a\\\sqrt{y^2+z^2}=b\\\sqrt{z^2+x^2}=c\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+y^2=a^2\\y^2+z^2=b^2\\z^2+x^2=c^2\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=\frac{a^2+c^2-b^2}{2}\\y^2=\frac{b^2+a^2-c^2}{2}\\z^2=\frac{b^2+c^2-a^2}{2}\end{cases}}}\)
\(\Leftrightarrow M=\frac{a^2+c^2-b^2}{2\left(y+z\right)}+\frac{b^2+a^2-c^2}{2\left(z+x\right)}+\frac{c^2+b^2-a^2}{2\left(x+y\right)}\)
áp dụng bunhia ta có:
\(\hept{\begin{cases}\left(x^2+y^2\right)\left(1+1\right)\ge\left(x+y\right)^2\\\left(y^2+z^2\right)\left(1+1\right)\ge\left(y+z\right)^2\\\left(z^2+x^2\right)\left(1+1\right)\ge\left(z+x\right)^2\end{cases}\Leftrightarrow\hept{\begin{cases}2a^2\ge\left(x+y\right)^2\\2b^2\ge\left(y+z\right)^2\\2c^2\ge\left(z+x\right)^2\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt{2}a\ge x+y\\\sqrt{2}b\ge y+z\\\sqrt{2}c\ge z+x\end{cases}}}\)
\(\Rightarrow M\ge\frac{a^2+c^2-b^2}{\sqrt{2}b}+\frac{a^2+b^2-c^2}{\sqrt{2}c}+\frac{c^2+b^2-a^2}{\sqrt{2}a}=\frac{1}{\sqrt{2}}\left(\frac{a^2}{b}+\frac{c^2}{b}-b+\frac{a^2}{c}+\frac{b^2}{c}-c+\frac{c^2}{a}+\frac{b^2}{a}-a\right)\)\(\ge\frac{1}{\sqrt{2}}\left(\frac{4\left(a+b+c\right)^2}{2\left(a+b+c\right)}-a-b-c\right)=\frac{1}{\sqrt{2}}\left(a+b+c\right)=\frac{6}{\sqrt{2}}\)
S
13 tháng 7 2019
ĐKXĐ: \(x\ge1;y\ge25\)
\(D=\frac{1}{x}\sqrt{\frac{x-1}{\left(x-2\right)^2+25}}+\frac{1}{y}\sqrt{\frac{y-25}{\left(y-50\right)^2+1}}\)
Vì x>=1,y>=25 => x-1>=0,y-25>=0
=> D >= 0
Dấu "=" xảy ra <=> x=1,y=25
Vậy MinD=0 khi x=1,y=25
Ta có: \(\left(x-2\right)^2+25\ge25;\left(y-50\right)^2+1\ge1\)
=>\(\frac{1}{x}\sqrt{\frac{x-1}{\left(x-2\right)^2+25}}\le\frac{1}{x}\sqrt{\frac{x-1}{25}};\frac{1}{y}\sqrt{\frac{y-25}{\left(y-50\right)^2+1}}\le\frac{1}{y}\sqrt{y-25}\)
=>\(D\le\frac{1}{x}\sqrt{\frac{x-1}{25}}+\frac{1}{y}\sqrt{y-25}\)
Vì x>=1 => x-1>=0. Áp dụng bđt cosi với 2 số dương x-1 và 1 ta có:
\(\sqrt{x-1}=\sqrt{\left(x-1\right).1}\le\frac{x-1+1}{2}=\frac{x}{2}\)
=>\(\frac{1}{x}\sqrt{\frac{x-1}{25}}\le\frac{1}{x}\cdot\frac{x}{2}\cdot\frac{1}{\sqrt{25}}=\frac{1}{10}\)
Vì y>=25 => y-25>=0. ÁP dụng bđt cô si cho 2 số dương 25 và y-25 ta có:
\(\sqrt{y-25}=\frac{\sqrt{25\left(y-25\right)}}{5}\le\frac{25+y-25}{2.5}=\frac{y}{10}\)
=>\(\frac{1}{y}\sqrt{y-25}=\frac{1}{y}\cdot\frac{y}{10}=\frac{1}{10}\)
Suy ra \(D\le\frac{1}{10}+\frac{1}{10}=\frac{1}{5}\)
Dấu "=" xảy ra <=> x=2,y=50
Vậy MaxD = 1/5 khi x=2,y=50
27 tháng 7 2018
Ukm
It's very hard
l can't do it
Sorry!
11 tháng 7 2019
\(đkxđ\Leftrightarrow x\ge4\)
\(P=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}}\)
\(=\frac{\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}}{\sqrt{\frac{4^2}{x^2}-2.\frac{4}{x}+1}}\)
\(=\frac{\sqrt{\left(x-4+2\right)^2}+\sqrt{\left(x-4-2\right)^2}}{\sqrt{\left(\frac{4}{x}-1\right)^2}}\)
\(=\frac{|x-2|+|x-6|}{|\frac{4}{x}-1|}=\frac{x-2+|x-6|}{|\frac{4}{x}-1|}\)
Dùng bảng xét dấu nha
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2020
\(a=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{9+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{9+\sqrt{x+7}}}{x-1}=\frac{29}{36}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(2-\frac{3}{x}\right)^2.x^3\left(4+\frac{7}{x}\right)^3}{x^3\left(3+\frac{1}{x^3}\right).x^2\left(10+\frac{9}{x^2}\right)}=\frac{2.4}{3.10}=\frac{4}{15}\)
\(c=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\left(2x+1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\frac{\frac{-4x^2}{\sqrt{1+4x}+2x+1}+\frac{8x^3+12x^2}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\left(\frac{-4}{\sqrt{1+4x}+2x+1}+\frac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=\frac{-4}{1+1}+\frac{12}{1+1+1}=2\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2020
\(d=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}\left(\sqrt{1+4x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\frac{4x\sqrt{1+6x}}{x\left(\sqrt{1+4x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{6x}{x\left(\sqrt{1+6x}+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\frac{4\sqrt{1+6x}}{\sqrt{1+4x}+1}+\lim\limits_{x\rightarrow0}\frac{6}{\sqrt{1+6x}+1}=\frac{4}{1+1}+\frac{6}{1+1}=5\)
\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}\left(\sqrt{1+2x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\frac{2x\sqrt[3]{1+4x}}{x\left(\sqrt{1+2x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{4x}{x\left(\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\frac{2\sqrt[3]{1+4x}}{\sqrt{1+2x}+1}+\lim\limits_{x\rightarrow0}\frac{4}{\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1}=\frac{2}{1+1}+\frac{4}{1+1+1}=\frac{7}{3}\)