\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
Những câu hỏi liên quan
Tìm x thuộc Z biết \(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
Tìm số nguyên x biết
\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
\(\Rightarrow\frac{7}{6}< |x-\frac{2}{3}|< \frac{26}{9}\)
\(\Rightarrow\frac{21}{18}< |x-\frac{2}{3}|< \frac{52}{18}\)
Rùi tự thay vào
Đúng 0
Bình luận (0)
\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
\(\Leftrightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)
\(\Leftrightarrow\frac{7}{6}< 2\le\left|x-\frac{2}{3}\right|\le2< \frac{26}{9}\)
\(\Leftrightarrow\left|x-\frac{2}{3}\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=2\\x-\frac{2}{3}=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=--\frac{4}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{8}{3};-\frac{4}{3}\right\}\)
1.Tìm x, biết:\(\left|\frac{5}{3}-x\right|\)- \(\left|\frac{-5}{6}\right|\)= \(\frac{-5}{9}\)
2.Tìm số nguyên x, biết: \(\frac{\sqrt{49}}{6}\) < \(\left|x-\frac{2}{3}\right|\)< \(\frac{26}{\sqrt{81}}\)
Tìm số nguyên x biết: \(\frac{\sqrt{49}}{6}< |x-\frac{2}{3}|< \frac{26}{\sqrt{81}}\)
tìm số nguyên x biết
\(\frac{\sqrt{49}}{6}< |x-\frac{2}{3}|< \frac{26}{\sqrt{81}}\)
Tính
a) \(2\sqrt{\frac{25}{16}}-3\sqrt{\frac{49}{36}}+4\sqrt{\frac{81}{64}}\)
b) \(\left(3\sqrt{2}\right)^2-\left(4\sqrt{\frac{1}{2}}\right)^2+\frac{1}{16}.\left(\sqrt{\frac{3}{4}}\right)^2\)
c) \(\frac{2}{3}\sqrt{\frac{81}{16}}-\frac{3}{4}\sqrt{\frac{64}{9}}+\frac{7}{5}.\sqrt{\frac{25}{196}}\)
a) = \(\frac{7}{2}\)
b) = \(\frac{643}{64}\)
c) = 0
Đúng 0
Bình luận (0)
Tìm số nguyên x biết: \(\frac{\sqrt{49}}{6}< Ix-\frac{2}{3}I< \frac{26}{\sqrt{81}}\)
- Ta có : \(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
=> \(\left\{{}\begin{matrix}\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|\\\left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{7}{6}< \left|x-\frac{2}{3}\right|\\\left|x-\frac{2}{3}\right|< \frac{26}{9}\end{matrix}\right.\)
- TH1 : \(x-\frac{2}{3}\ge0\left(x\ge\frac{2}{3}\right)\)
=> \(\left\{{}\begin{matrix}\frac{7}{6}< x-\frac{2}{3}\\x-\frac{2}{3}< \frac{26}{9}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{11}{6}< x\\x< \frac{32}{9}\end{matrix}\right.\)
=> \(\frac{11}{6}< x< \frac{32}{9}\)
Mà x là số nguyên .
=> \(x\in\left\{2,3\right\}\)
- TH2 : \(x-\frac{2}{3}< 0\left(x< \frac{2}{3}\right)\)
=> \(\left\{{}\begin{matrix}\frac{7}{6}< \frac{2}{3}-x\\\frac{2}{3}-x< \frac{26}{9}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{2}< -x\\-x< \frac{20}{9}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-\frac{1}{2}>x\\x>-\frac{20}{9}\end{matrix}\right.\)
=> \(-\frac{1}{2}>x>-\frac{20}{9}\)
Mà x là số nguyên .
=> \(x\in\left\{-1,-2\right\}\)
I là tham số à bạn
\(\frac{\sqrt{49}}{6}< |x-\frac{2}{3}|< \frac{26}{\sqrt{81}}\)
Làm giúp mik vs ai nhanh mik tick cho nha!
\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
\(\Rightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)
\(\Rightarrow\frac{21}{18}< \left|x-\frac{12}{18}\right|< \frac{52}{18}\)
còn lại cậu tự tính nha
\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)
\(\frac{7}{6}< x-\frac{2}{3}< \frac{26}{9}\)
\(\frac{11}{6}< x< \frac{32}{9}\)
Tính
\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)