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CTVHS
1 tháng 2 2023
Đặt : \(\dfrac{x}{5}=\dfrac{y}{3}=k\)
`=>x=5k,y=3k`
Ta có : \(x^2-y^2=4=>\left(5k\right)^2-\left(3k\right)^2=4\\ =>25k^2-9k^2=4\\ =>16k^2=4\\ =>k^2=\dfrac{1}{4}\\ =>k=\pm\dfrac{1}{2}\)
\(=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
16 tháng 3 2020
\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
\(\Rightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)
\(\Rightarrow\frac{21}{18}< \left|x-\frac{12}{18}\right|< \frac{52}{18}\)
còn lại cậu tự tính nha
16 tháng 3 2020
\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
\(\frac{7}{6}< x-\frac{2}{3}< \frac{26}{9}\)
\(\frac{11}{6}< x< \frac{32}{9}\)
8 tháng 9 2021
\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)
\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)
\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)
\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))
8 tháng 9 2021
\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)
23 tháng 10 2018
a)Với x>=0
\(\frac{5}{11}\sqrt{x}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
\(\sqrt{x}=\frac{1}{2}:\frac{5}{11}=\frac{11}{10}\)
\(x=\frac{11^2}{10^2}=\frac{121}{100}\)(thỏa mãn)
b) x=0
c) \(x=\pm\sqrt{3}\)vì x<0 => \(x=-\sqrt{3}\)
d) x=1 hoặc -1
e) \(x=\pm\sqrt{2}\)
T
23 tháng 10 2018
\(a,\frac{5}{11}\sqrt{x}-\frac{1}{3}=\frac{1}{6}.\)
\(\frac{5}{11}\sqrt{x}=\frac{1}{6}+\frac{1}{3}\)
\(\frac{5}{11}\sqrt{x}=\frac{1}{2}\)
\(\sqrt{x}=\frac{1}{2}:\frac{5}{11}\)
\(\sqrt{x}=\frac{11}{10}\)
\(\Rightarrow x=\frac{121}{100}\)
\(b.x^2=0\)
\(\Leftrightarrow x=0\)
\(c.x^2=3\left(x< 0\right)\)
\(\Leftrightarrow x=-\sqrt{3}\)
\(d.x^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
\(e.x^2=2\)
\(\Leftrightarrow x=\sqrt{2}\)
24 tháng 4 2018
Xét \(\frac{1}{\sqrt{13}}>\frac{1}{\sqrt{14}}\Rightarrow\frac{1}{\sqrt{13}}-1< \frac{1}{\sqrt{14}}+1\)
Mà \(\sqrt{225}< \sqrt{289}\)
\(\Rightarrow\sqrt{225}-\left(\frac{1}{\sqrt{13}}-1\right)< \sqrt{289}-\left(\frac{1}{\sqrt{14}}+1\right)\)
Vậy....................
9 tháng 10 2019
\(a,\frac{x+8}{3}+\frac{x+7}{2}=-\frac{x}{5}\)
\(\Leftrightarrow\frac{10\cdot\left(x+8\right)}{30}+\frac{15\left(x+7\right)}{30}=\frac{-6x}{30}\)
\(\rightarrow10x+80+15x+105=-6x\)
\(\Leftrightarrow31x+185=0\)
\(\Leftrightarrow x=-\frac{185}{31}\)
b,\(b,\frac{x-8}{3}+\frac{x-7}{4}=4+\frac{1-x}{5}\)
\(\Leftrightarrow\frac{20\left(x-8\right)}{60}+\frac{15\left(x-7\right)}{60}=\frac{240}{60}+\frac{12\left(1-x\right)}{60}\)
\(\rightarrow20x-160+15x-105=240+12-12x\)
\(\Leftrightarrow47x-517=0\)\(\Leftrightarrow x=11\)
bình phương lên
x=1/9
x= (1/3)^2
x=1/9