\(\frac{1}{2}:2x=-\frac{1}{3}\)

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

\(x=-\frac{3}{4}\)

\(\frac{1}{2}:2x=\frac{-1}{3}\)

       \(2x=\frac{1}{2}:\frac{-1}{3}\)

       \(2x=-\frac{3}{2}\)

        \(x=-\frac{3}{4}\)

Vậy x = -3/4

\(\frac{1}{2}:2x=-\frac{1}{3}\)

 \(\Leftrightarrow2x=\frac{1}{2}:-\frac{1}{3}\)

\(\Leftrightarrow x=-\frac{3}{2}:2\)

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

Vậy \(x=-\frac{3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x+1}{2}=\frac{y+3}{4}\)\(=\frac{z+5}{6}\)\(=\frac{2.\left(x+1\right)+3.\left(y+3\right)+4.\left(z+5\right)}{2.2+3.4+4.6}\)

\(=\frac{2x+2+3y+9+4z+20}{4+12+24}\)\(=\frac{\left(2x+3y+4z\right)+\left(2+9+20\right)}{40}\)

\(=\frac{9+31}{40}=\frac{40}{40}=1\)

Cứ thế là tìm x+1 rồi tìm x

                    y+3           y

                    x+5           z

Từ gt,suy ra : (x - 2)(2 - x) = -16.4

-(x - 2)² = -64

(x - 2)² = 64

\(\Rightarrow\orbr{\begin{cases}x-2=-8\\x-2=8\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=10\end{cases}}}\)

\(x=-16.4=-64\)

\(x^2=-8^2\)

Vay: x=-8

Ma  theo de bai x-2

Nen ta lay x+2

x+2=-8+2=-6

=>\(x=-6\)

x-2/4=-16/2-x

=>x-2/4=-16/-(x-2)=16/x-2

=> (x-2)^2=16*4=64

=>x-2=8                  hoặc                               x-2=-8

=>x=10                   hoặc                           =>x=-6

KL

(1/2-3/4) x - 7/3 = -5/9

-1/4 x  =16/9

x= -64/9

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

\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)

\(\left(2x-1\right)=-\frac{4}{63}\)

2x= -4/63 + 1

2x = 59/63

x = 59/63 : 2

x = 59/126

1/3:(2.x-1)=-5-1/4

1/3:(2.x-1)=-21/4

2.x-1=1/3:-21/4

2.x-1=-4/63

2.x=-4/63+1

2.x=\(3\frac{59}{63}\)

x=\(3\frac{59}{63}\):2

x=\(1\frac{61}{63}\)

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

\(\frac{1}{3}\div\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-20}{4}+\frac{-1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-21}{4}\)

\(2x-1=\frac{1}{3}\div\frac{-21}{4}\)

\(2x-1=\frac{1}{3}.\frac{-4}{21}\)

\(2x-1=\frac{-4}{63}\)

\(2x=\frac{-4}{63}+1\)

\(2x=\frac{-4}{63}+\frac{63}{63}\)

\(2x=\frac{59}{63}\)

\(x=\frac{59}{63}\div2\)

\(x=\frac{59}{63}.\frac{1}{2}\)

\(x=\frac{59}{126}\)

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

\(\Leftrightarrow\frac{3\left(x-2\right)-2\left(1+x\right)}{6}=\frac{4-3x-4}{4}\)

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

\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow4x-32=-18x\)

\(\Rightarrow x=\frac{16}{11}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)

\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(x=2002\)

Vậy x = 2002

Bài này lớp 6 thật à bạn. 

Mik mới lớp 5 thui à.  Sorry nha. 

a) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Leftrightarrow2x=18\)

\(\Leftrightarrow x=9\)

b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)

\(\Leftrightarrow\left(\frac{1}{9}\right)^x=\left(\frac{1}{3}\right)^{66}\)

\(\Leftrightarrow x=66\)

b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)

\(\left(\frac{1}{3}\right)^{2.x}=\left(\frac{1}{3}\right)^{3.22}\)

\(2x=3.22\)

\(x=33\)