Gải phương trình\(\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}=\frac{2}{1+\sqrt{x}}\)
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 2 2020
ĐKXĐ: \(x\ge0\)
Ta có:
\(VT=\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}\le\sqrt{\frac{2}{x+3}+\frac{2}{3x+1}}=\sqrt{\frac{8\left(x+1\right)}{\left(x+3\right)\left(3x+1\right)}}=\frac{2}{\left(1+\sqrt{x}\right)}\sqrt{\frac{2\left(1+\sqrt{x}\right)^2\left(x+1\right)}{\left(x+3\right)\left(3x+1\right)}}\)
Mà \(\left(x+3\right)\left(3x+1\right)-2\left(1+\sqrt{x}\right)^2\left(x+1\right)=x^2-4x\sqrt{x}+6x-4\sqrt{x}+1=\left(\sqrt{x}-1\right)^4\ge0\)
\(\Rightarrow\left(x+3\right)\left(3x+1\right)\ge2\left(1+\sqrt{x}\right)^2\left(x+1\right)>0\)
\(\Rightarrow\frac{2\left(1+\sqrt{x}\right)^2\left(x+1\right)}{\left(x+3\right)\left(3x+1\right)}\le1\)
\(\Rightarrow VT\le\frac{2}{1+\sqrt{x}}\)
Dấu "=" xảy ra khi và chỉ khi \(x=1\)
5 tháng 12 2016
Dk: x\(\ge0\)
lien hop
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=2\Rightarrow x=1\)
28 tháng 8 2017
\(\left(\sqrt{x^2+16}-5\right)\)\(-3\left(x-3\right)-\left(\sqrt{x^2+7}-4\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x^2+16}-5\right)\left(\sqrt{x^2+16}+5\right)}{\sqrt{x^2+16}+5}\)\(-3\left(x-3\right)-\frac{\left(\sqrt{x^2+7}-4\right)\left(\sqrt{x^2+7}+4\right)}{\sqrt{x^2+7}+4}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x^2+16}+5}-3-\frac{1}{\sqrt{x^2+7}+4}\right)=0\)
ben trong ngoac bn tu xu li nhe
\(\Rightarrow x=3\)
MC
18 tháng 10 2018
a, \(\sqrt{8}+\sqrt{18}-\sqrt{\frac{1}{2}}=2\sqrt{2}+3\sqrt{2}-\frac{1}{2}\sqrt{2}\)
\(=\frac{9}{2}\sqrt{2}\)
b, \(\frac{3-\sqrt{3}}{\sqrt{3}}+\frac{2\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}+\frac{2\sqrt{2}}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)
\(=\sqrt{3}-1+\frac{2\sqrt{2}}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)
\(=\frac{2\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+1\right)\) \(=\frac{2\sqrt{2}-\left(\sqrt{2}+1\right)^2}{\sqrt{2}+1}\)
\(=\frac{2\sqrt{2}-2-2\sqrt{2}-1}{\sqrt{2}+1}=-\frac{2+1}{\sqrt{2}+1}\)
c, PT xác định với mọi x nha!
\(\sqrt{x^2-2x+1}=3\) \(\Rightarrow x^2-2x+1=9\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(2x-8\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy...
bạn tự kl
28 tháng 3 2020
a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)
hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)
\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)
\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)
\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)
b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)
Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)
Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được
\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)
(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)
\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)
\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)
(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)
15 tháng 8 2017
\(\sqrt{x^2-6x+9}\) \(-\frac{\sqrt{3}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=0\)
\(\Leftrightarrow\left|x-3\right|-\sqrt{3}=0\)
\(\Leftrightarrow\left|x-3\right|=\sqrt{3}\)
th1 \(x\ge3\Rightarrow x-3=\sqrt{3}\Rightarrow x=3+\sqrt{3}\)
th2 \(x< 3\Rightarrow3-x=\sqrt{3}\Rightarrow x=3-\sqrt{3}\)
WR
23 tháng 6 2019
Đkxđ: \(\hept{\begin{cases}x\ge-\frac{1}{4}\\y\ge2\end{cases}}\)
\(\Leftrightarrow2+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=y\Leftrightarrow2+\frac{1}{2}+\sqrt{x+\frac{1}{2}}=y\Leftrightarrow\sqrt{x+\frac{1}{2}}+\frac{5}{2}=y\)
do x,y nguyên dương nên \(\sqrt{x+\frac{1}{2}}+\frac{5}{2}\)nguyên dương\(\Leftrightarrow\sqrt{x+\frac{1}{2}}=\frac{k}{2}\)(K là số nguyên lẻ, \(k>1\))
\(\Rightarrow x=\frac{k^2-2}{4}\)
do \(k^2\)là số chính phương chia 4 dư 0,1 \(\Rightarrow x=\frac{k^2-2}{4}\notin Z\)
=> ko tồn tại cặp số nguyên dương x,y tmđkđb
\(Đkxđ:x\ge0\)
Ta có: Bất phương trình tương đương với:
\(\left(1+\sqrt{x}\right)\left(\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}\right)=2\)
Áp dụng BĐT Cô - si ta có:
\(\frac{1}{\sqrt{3x+1}}=\sqrt{\frac{1}{x+1}.\frac{x+1}{3x+1}}\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{x+1}{3x+1}\right)\)
\(\sqrt{\frac{x}{3x+1}}=\sqrt{\frac{1}{2}.\frac{2x}{3x+1}}\le\frac{1}{2}\left(\frac{1}{2}+\frac{2x}{3x+1}\right)\)
\(\Rightarrow\frac{1+\sqrt{x}}{\sqrt{3x+1}}\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{2}+1\right)=\frac{1}{2}\left(\frac{1}{x+1}+\frac{3}{2}\right)\left(1\right)\)
\(\frac{1}{\sqrt{x+3}}=\sqrt{\frac{1}{2}.\frac{2}{x+3}}\le\frac{1}{2}\left(\frac{1}{2}+\frac{2}{x+3}\right)\)
\(\frac{\sqrt{x}}{\sqrt{x+3}}=\sqrt{\frac{x}{x+1}.\frac{x+1}{x+3}}\le\frac{1}{2}\left(\frac{x}{x+1}+\frac{x+1}{x+3}\right)\)
\(\Rightarrow\frac{1+\sqrt{x}}{\sqrt{x+3}}\le\frac{1}{2}\left(\frac{x}{x+1}+\frac{3}{2}\right)\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow\left(1+\sqrt{x}\right)\left(\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}\right)\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{x}{x+1}+3\right)=2\)
Đẳng thức xảy ra \(\Leftrightarrow x=1\)
Vậy nghiệm của pt là \(x=1\)