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LG
26 tháng 11 2017
a) \(\sqrt{x+1}=7\Rightarrow x+1=49\Rightarrow x=48\)
b) \(\left(x-2\right).\left(x+\dfrac{2}{3}\right)>0\)
\(\Rightarrow\left(x-2\right).\left(x+\dfrac{2}{3}\right)\) cùng dấu
\(\Rightarrow\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>2\\x>-\dfrac{2}{3}\end{matrix}\right.\Rightarrow x>2\)
Với \(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 2\\x< -\dfrac{2}{3}\end{matrix}\right.\Rightarrow x< -\dfrac{2}{3}\)
Vậy \(\left[{}\begin{matrix}x>2\\x< -\dfrac{2}{3}\end{matrix}\right.\)
c) \(\left(\dfrac{2}{3}x-1\right).\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{3}{4}x+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Chúc bạn học tốt!!!!
26 tháng 11 2017
a, \(\sqrt{x+1}=7\\ \Rightarrow x+1=49\\ \Rightarrow x=48\)
b,TH1:
\(\left\{{}\begin{matrix}x-2>0\\x +\dfrac{2}{3}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x>\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow x>2\)
TH2:
\(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x< \dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow x< \dfrac{-2}{3}\)
=> Vậy 2<x< \(\dfrac{-2}{3}\)
c, TH1:
\(\dfrac{2}{3}x-1=0\\ \Rightarrow\dfrac{2}{3}x=1\\ \Rightarrow x=\dfrac{3}{2}\)
TH2:
\(\dfrac{3}{4}x+\dfrac{1}{2}=0\\ \Rightarrow\dfrac{3}{4}x=\dfrac{-1}{2}\\ \Rightarrow x=\dfrac{-2}{3}\)
Vậy x = \(\dfrac{3}{2};\dfrac{-2}{3}\)
28 tháng 1 2019
a) ⇒ \(\dfrac{5}{3}x\) \(=\) \(\dfrac{5}{6}+\dfrac{1}{4}\)
⇒ \(\dfrac{5}{3}x=\dfrac{13}{12}\)
⇒ \(x=\dfrac{13}{12}:\dfrac{5}{3}\)
⇒\(x=\dfrac{13}{20}\)
3 tháng 10 2017
\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}\)
\(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\\\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\\.............\\\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\end{matrix}\right.\)
Suy ra:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{99}}>\dfrac{99}{\sqrt{100}}\)
\(linh=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.....+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{99}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}\)
\(\)\(linh>10\left(đpcm\right)\)
Bài này ko phải 100 nhé
14 tháng 4 2018
19) \(\sqrt{19-x}=19\)
\(\Rightarrow\sqrt{19-x}=\sqrt{19^2}\)
\(\Rightarrow19-x=19^2\)
\(\Rightarrow19-19^2=x\)
\(\Rightarrow x=19\left(1-19\right)=-19.18=-342\)
21) \(\sqrt{x-1}=\dfrac{1}{3}\)
\(\Rightarrow\sqrt{x-1}=\sqrt{\left(\dfrac{1}{3}\right)^2}\)
\(\Rightarrow x-1=\dfrac{1}{3^2}\)
\(x=\dfrac{1+9}{9}=\dfrac{10}{9}\)
24)\(\sqrt{2x+\dfrac{5}{4}}=\dfrac{3}{2}\)
\(\Rightarrow\sqrt{2x+\dfrac{5}{4}}=\sqrt{\left(\dfrac{3}{2}\right)^2}\)
\(\Rightarrow2x+\dfrac{5}{4}=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow2x=\dfrac{9-5}{4}=1\)
\(\Rightarrow x=0,5\)
25) \(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\)
\(\Rightarrow\sqrt{\dfrac{2x-7}{6}}=\sqrt{\left(\dfrac{1}{6}\right)^2}\)
\(\Rightarrow\dfrac{2x-7}{6}=\left(\dfrac{1}{6}\right)^2=\dfrac{1}{36}\)
\(\Rightarrow\dfrac{12x-42}{36}=\dfrac{1}{36}\)
\(\Rightarrow12x-42=1\)
\(\Rightarrow12x=43\)
\(\Rightarrow x=\dfrac{43}{12}\)
29 tháng 3 2018
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
.............................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.........+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+.....+\dfrac{1}{10}=\dfrac{100}{10}=10\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+......+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right)\)
2 tháng 9 2022
Bài 1:
a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)
=>2 căn x=6
=>căn x=3
=>x=9
b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)
=>x=1
24 tháng 9 2017
a) \(7-\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}=7\)
\(\Rightarrow x=\left(\sqrt{7}\right)^2\)
b) \(5\sqrt{x}+1=40\)
\(\Rightarrow5\sqrt{x}=39\)
\(\Rightarrow\sqrt{x}=7,8\)
\(\Rightarrow x=\left(\sqrt{7,8}\right)^2\)
c) \(\dfrac{5}{12}\sqrt{x}-\dfrac{1}{6}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{12}\sqrt{x}=\dfrac{1}{2}\)
\(\Rightarrow\sqrt{x}=1,2\)
\(\Rightarrow x=\left(\sqrt{1,2}\right)^2\)
d) \(4x^2-1=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\Rightarrow x=0,5\\2x+1=0\Rightarrow x=-0,5\end{matrix}\right.\)
e) \(\sqrt{x+1}-2=0\)
\(\Rightarrow\sqrt{x+1}=2\)
\(\Rightarrow x+1=1,414\)
\(\Rightarrow x=0,414\)
f) \(2x^2+0,82=1\)
\(\Rightarrow2x^2=0,18\)
\(\Rightarrow x^2=0,09\)
\(\Rightarrow x=\pm0,3\)
g) Không có kết quả
19 tháng 11 2022
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
22 tháng 7 2017
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
27 tháng 5 2022
a: \(\left(2x+3\right)\left(3x-5\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5\ge0\\2x+3\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>=\dfrac{5}{3}\\x< =-\dfrac{3}{2}\end{matrix}\right.\)
b: \(\dfrac{x}{3-x}>-1\)
\(\Leftrightarrow\dfrac{x}{3-x}+1>0\)
\(\Leftrightarrow\dfrac{x+3-x}{3-x}>0\)
=>3-x>0
hay x<3
c: \(\dfrac{x-1}{x+5}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x-1}{x+5}-\dfrac{3}{2}\ge0\)
\(\Leftrightarrow\dfrac{2x-2-3x-15}{2\left(x+5\right)}>=0\)
\(\Leftrightarrow\dfrac{x+17}{2\left(x+5\right)}< =0\)
=>-17<=x<-5
d: \(\dfrac{7}{4x^2-1}\ge0\)
=>4x2-1>0
=>(2x-1)(2x+1)>0
=>x>1/2 hoặc x<-1/2
\(\dfrac{x}{\sqrt{x}-1}>1\) (ĐK: \(x\ge0,x\ne1\))
\(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\sqrt{x}-1>0\) (vì \(x-\sqrt{x}+1=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\))
\(\Leftrightarrow\sqrt{x}>1\)
\(\Leftrightarrow x>1\) (thỏa mãn)
Vậy \(x>1\).