\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)

Bài b tương tự

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b

\(\frac{-2}{9}\le\frac{x}{12}\le\frac{-1}{18}\)

\(\Leftrightarrow\frac{-8}{36}\le\frac{3x}{36}\le\frac{-2}{36}\)

\(\Leftrightarrow-8\le3x\le-2\)

\(\Leftrightarrow-2\le x\le-1\)

Vậy \(x\in\left\{-1;-2\right\}\)

\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)

\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)

2 câu sau tự làm nha

\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)

\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)

=> \(x\in\left\{-2;-1;0;1;2\right\}\)

C =22−3 x 4− x =12−3 x +104− x =3(4− x )4− x +104− x =3+104− x ... +199.100 ). x =201251 +201252 +. .... =12√3 (4+2√3√3+1+4−2√3√3−1 )=12√3 .4√3−4+6−2√3+4√3+4−6−2√3(√3−1)(√3+1).

\(\frac{1}{4}+\frac{8}{9}< \frac{x}{36}< 1-\frac{3}{8}+\frac{5}{6}\)

\(=\frac{41}{36}< \frac{x}{36}< \frac{35}{24}\)

=>\(x=35< x< 41\)

\(e,\frac{22}{15}-x=-\frac{8}{27}\)

=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)

=> \(x=\frac{22}{15}+\frac{8}{27}\)

=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)

\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)

=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)

=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)

=> \(\frac{2x-5}{5}=-\frac{5}{4}\)

=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)

=> \(2x=-\frac{5}{4}\)

=> \(x=-\frac{5}{8}\)

\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)

=> \(-\frac{9}{4}x+\frac{37}{4}=20\)

=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)

=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)

\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)

=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)

=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)

=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)

=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)

=> \(-10\le x\le-\frac{13}{7}\)

Đến đây tìm x

\(VT=\sum\frac{x}{x\left(x+y+z\right)+yz}=\sum\frac{x}{\left(x+y\right)\left(x+z\right)}=\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(VT=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(VT=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y+z\right)\left(xy+yz+zx\right)-xyz}=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+\frac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)-xyz}\)

\(VT\le\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+\frac{1}{9}3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}-xyz}\)

\(VT\le\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+xyz-xyz}=\frac{9}{4}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

giúp mình với , mình cần gấp , ai nhanh mình k nha

Câu a là thế này nha  \(\frac{-36}{9}\) \(\le\) x     \(\le\) \(\frac{-15}{5}\)