\(a)\) \(2x-5=21\)

\(\Leftrightarrow\) \(2x=21+5\)

\(\Leftrightarrow\) \(2x=26\)

\(\Leftrightarrow\) \(x=26:2\)

\(\Leftrightarrow\) \(=13\)

\(b)\) \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{5}{6}-\frac{3}{4}\)

\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{1}{12}\)

\(\Leftrightarrow\) \(x=\frac{1}{3}\)

\(c)\) \(\frac{7}{8}-\frac{1}{2}x=\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{2}x=\frac{7}{8}-\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{2}x=\frac{1}{24}\)

\(\Leftrightarrow\) \(x=\frac{1}{12}\)

Bài 1:

Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)

\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)

\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)

\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)

\(A=\frac{2x19}{20}=\frac{19}{10}\)

Bài 2:

Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)

Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)

\(Bx100=\frac{9}{10}x100=90\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)

=>  \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)

bài 1 đáp án là:19/10

2:147/50

đáp án bài 1 là: 19/10

đáp án bài 2 là 147/50

\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{103}{50}\right)\right]\cdot2=89\)

\(\left(1-\frac{1}{10}\right)\cdot100-\frac{5}{2}:\left(x+\frac{103}{50}\right)\cdot2=89\)

\(\frac{9}{10}\cdot100-\frac{5}{2}\cdot2:\left(x+\frac{103}{50}\right)=89\)

\(90-5\cdot\left(x+\frac{103}{50}\right)=89\)

\(5\cdot\left(x+\frac{103}{50}\right)=1\)

\(x+\frac{103}{50}=\frac{1}{5}\)

\(x=-\frac{93}{50}\)

1/1*2+1/2*3+1/3*4+...+1/9*10

=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

=1-1/10

=9/10

nho k cho minh voi nhe

\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ......... + \(\frac{1}{7.8}\)+ \(\frac{1}{8.9}\)+ \(\frac{1}{9.10}\)

\(=\)\(1\)\(-\)\(\frac{1}{10}\)

\(=\)\(\frac{9}{10}\)

9/10 nha

Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)

đặt A=1/1×2+1/2×3+1/3×4+

bn đặt ra đi sau đó rồi tính vd 1/1x2=1/1-1/2

= 9/10

k nha