1)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3x}{\left(2x+6\right)x}-\frac{x-6}{2x^2+6x}\\ =\frac{3x}{2x^2+6x}-\frac{x-6}{2x^2+6x}=\frac{3x-\left(x-6\right)}{2x^2+6x}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

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 \(x = \frac{{ - 9}}{{10}}\).

\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)=\frac{4}{3}.\frac{-1}{3}=\frac{-4}{9}\)

 k nha

\(\frac{-4}{9}\)k mk nha

\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{6}-\frac{3}{6}\right)\)

\(=\frac{4}{3}.\frac{-2}{6}\)

\(=\frac{4}{3}.\frac{-1}{3}\)

\(=\frac{-4}{9}\)

x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)

x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)

=>x.(1/2-2/3+3/4)=1/4

=>x.7/12=1/4

=>x=1/4:7/12

=>x=1/4.12/7

=>x=3/7

Đây là những bài cơ bản mà bạn!

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

\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)

\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)

\(< =>10x-4=15-9x\)

\(< =>10x+9x=15+4\)

\(< =>19x=19< =>x=1\)

\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)

\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)

\(< =>30x+9=36+24+32x\)

\(< =>32x-30x=9-36-24\)

\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

\(\begin{array}{l}a)\frac{{{x^2} + 4{\rm{x}} + 4}}{{{x^2} - 4}} + \frac{x}{{2 - x}} + \frac{{4 - x}}{{5{\rm{x}} - 10}}\\ = \frac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} - \frac{x}{{x - 2}} + \frac{{4 - x}}{{5\left( {x - 2} \right)}}\\ = \frac{{x + 2}}{{x - 2}} - \frac{x}{{x - 2}} + \frac{{4 - x}}{{5\left( {x - 2} \right)}}\\ = \frac{{5\left( {x + 2} \right) - 5x + 4 - x}}{{5\left( {x - 2} \right)}} = \frac{{ - x + 14}}{{5\left( {x - 2} \right)}}\end{array}\)

\(\begin{array}{l}b)\frac{x}{{{x^2} + 1}} - \left( {\frac{3}{{x + 6}} + \frac{{x - 2}}{{x + 4}}} \right) + \left[ {\frac{3}{{x + 6}} - \left( {\frac{1}{{{x^2} + 1}} - \frac{{x - 2}}{{x + 4}}} \right)} \right]\\ = \frac{x}{{{x^2} + 1}} - \frac{3}{{x + 6}} - \frac{{x - 2}}{{x + 4}} + \frac{3}{{x + 6}} - \frac{1}{{{x^2} + 1}} + \frac{{x - 2}}{{x + 4}}\\ = \frac{x}{{{x^2} + 1}} - \frac{1}{{{x^2} + 1}} = \frac{{x - 1}}{{{x^2} + 1}}\end{array}\)