\(\left(4\frac{46}{65}+x\right).1\frac{1}{12}=5,75\)

\(\Rightarrow\frac{306}{65}+x.\frac{13}{12}=\frac{23}{4}\)

\(\Rightarrow\frac{51}{10}+\frac{13}{12}x=\frac{23}{4}\)

\(\Rightarrow306x=65x=345\)

\(\Rightarrow65x=39\)

\(\Rightarrow x=\frac{3}{5}\)

b, \(\frac{5}{4}-\left(\frac{3}{2}x+0,5\right)=1\frac{1}{4}\)

\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-0,5=\frac{5}{4}\)

\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-\frac{1}{2}=\frac{5}{4}\)

\(\Rightarrow\frac{3}{4}-\frac{3}{2}x=\frac{5}{4}\)

\(\Rightarrow3-6x=5\)

\(\Rightarrow-6x=2\)

\(\Rightarrow x=-\frac{1}{3}\)

Phần b) chị sai nhé ! Dấu [ ] là phần nguyên nâng cao của lớp 6 nhé.

\(a,\left(4\frac{46}{65}+x\right)\cdot1\frac{1}{12}=5,75\)

\( < =>\left(4\frac{46}{65}+x\right)=5,75:1\frac{1}{12}\)

\(< =>\left(4\frac{46}{65}+x\right)=\frac{69}{13}\)

\(< =>x=\frac{69}{13}-4\frac{46}{65}\)

\(< =>x=\frac{3}{5}\)

\(b,\frac{5}{4}-I\frac{3}{2}\cdot x+0,5I=1\frac{1}{4}\)

\(< = >I\frac{3}{2}\cdot x+0,5I=1\frac{1}{4}+\frac{5}{4}\)

\(< =>I\frac{3}{2}\cdot x+0,5I=\frac{5}{2}\)

\(< =>\left[\frac{3}{2}\cdot x+0,5\right]=\frac{5}{2}hoac\frac{-5}{2}\)

\(< =>\left[\frac{3}{2}\cdot x\right]=\frac{5}{2}-0,5hoac\frac{-5}{2}-0,5\)

\(< =>\left[\frac{3}{2}\cdot x\right]=2hoac-3\)

\(< =>\left[x\right]=2:\frac{3}{2}hoac-3:\frac{3}{2}\)

\(< =>\left[x\right]=\frac{4}{3}hoac-2\)

chuc ban hoc tot nhe :))

Dùng máy tính

Nếu ko có máy tính thì sao?

a)  \(=\frac{1}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}.\frac{6.6}{5.7}=\frac{6}{2.7}=\frac{3}{7}\)

B) \(=\frac{70}{11}+\frac{1}{9}-\frac{37}{11}-\frac{1}{9}=\left(\frac{70}{11}-\frac{37}{11}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)=\frac{33}{11}+0=3\)

BÀI 2:

A) \(\Leftrightarrow\frac{7}{2}x-\frac{x}{2}+\frac{2x}{2}=\frac{7}{2}.\frac{5}{6}\)

\(\Leftrightarrow\frac{7x-x+2x}{2}=\frac{35}{12}\)

\(\Leftrightarrow\frac{8x}{2}=\frac{35}{12}\)

\(\Leftrightarrow8x.12=35.2\Leftrightarrow96x=70\Leftrightarrow x=\frac{70}{96}=\frac{35}{48}\)

b) \(\left(x-\frac{3}{1.2}\right)+\left(x-\frac{3}{2.3}\right)+...+\left(x-\frac{3}{99.100}\right)=1\)

\(x-\frac{3}{1.2}+x-\frac{3}{2.3}+....x+\frac{3}{99.100}=1\)

\(\Leftrightarrow\left(x+x+x+...+x\right)-3\left(\frac{1}{1.2}+\frac{1}{1.3}+....+\frac{1}{99.100}\right)=1\)

ngoặc 1 có 99 số hạng x

\(\Leftrightarrow99x-3\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=1\)

\(\Leftrightarrow99x-3\left(1-\frac{1}{100}\right)=1\)

\(\Leftrightarrow99x-3.\frac{99}{100}=1\)

\(\Leftrightarrow99x=1+\frac{3.99}{100}\)

\(\Leftrightarrow99x=\frac{397}{100}\)

\(\Leftrightarrow x=\frac{397}{100.99}=\frac{397}{9900}\)

i don't know i mới học lớp 5

bn eie mik lớp 6 nha bn

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).