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1/21 = 2/42 = 1/7-1/8
1/28 = 2/56 = 1/8-1/9
.....
1/(x*(x+1)) = 1/x -1/(x+1)
cộng lại với nhau ta sẽ được 1/7 - 1/(x+1) = 2/9
suy ra 1/(x+1) = -5/62 :D
Đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\)
=> \(A=\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x\left(x+1\right)}\)
\(\frac{A}{2}=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\)
=> \(\frac{A}{2}=\frac{1}{6}-\frac{1}{x+1}=\frac{x+1-6}{6\left(x+1\right)}=\frac{x-5}{6\left(x+1\right)}\) => \(A=\frac{x-5}{3\left(x+1\right)}=\frac{2}{9}\)
<=> 3(x-5)=2(x+1) <=> 3x-15=2x+2 <=> x=17
Đáp số: x=17
* ĐK: \(x\ne0\)
Đề ra ...<=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{1}{9}\)
<=> \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{6}-\frac{1}{x+1}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{x+1}\left(1-\frac{1}{x}\right)=\frac{1}{6}-\frac{1}{9}\)
<=> \(\frac{x-1}{x\left(x+1\right)}=\frac{1}{36}\)
<=> \(\frac{x-1}{x\left(x-1\right)}=\frac{x-1}{36.\left(x-1\right)}\)
=> x(x-1) = 36. (x-1) => x =36
\(\frac{2}{2}.\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x+\left(x+1\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)
\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
x+1=18
x=18-1
x=17
a)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+2\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+2}=\frac{2}{9}:2\)
\(\frac{1}{x+2}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+2}=\frac{1}{18}\)
=>x+2=18
=>x=16
b tương tự nhân nó với 1/2
nhân 2 cả tử và mẫu lên=> ta có 2/6.7 + 2/7.8........
tự làm nốt nhé ;-)
1/21+1/28+1/36+...+2/[x.(x+1)]=2/9
=>2/42+2/56+2/72+...+2/[x.(x+1)]=2/9
=>2.(1/42+1/56+1/72+...+1/[x.(x+1)])=2/9
=>2.(1/6-1/7+1/7-1/8+1/8-1/9+...+1/x-1/(x+1))=2/9
=>1/6-1/(x+1)=1/9
=>1/(x+1)=1/18
=>x+1=18
=>x=17
Đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{21.2}+\frac{1}{28.2}+\frac{1}{36.2}...+\frac{2}{x.\left(x+1\right).2}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6.4}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}-\frac{1}{x+1}\)
\(\Rightarrow A=\left(\frac{1}{6}-\frac{1}{x+1}\right):\frac{1}{2}\)
Theo bài ra ta có :
\(\left(\frac{1}{6}-\frac{1}{x+1}\right):\frac{1}{2}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{3}{18}-\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
Vậy x = 17
Ta có :
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\) ( cái đề hình như có 1 phân số \(\frac{2}{9}\) đúng không bạn )
\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=1:\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=18\)
\(\Leftrightarrow\)\(x=18-1\)
\(\Leftrightarrow\)\(x=17\)
Vậy \(x=17\)
Chúc bạn học tốt ~
fuck you