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\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)
\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)
\(x\) + 1 = 16
\(x\) = 16 - 1
\(x\) = 15
a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)
\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)
b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x-1}{3}=\frac{y-1}{4}=\frac{z+2}{5}=\frac{z-1+y-1+z+2}{3+4+5}=\frac{-36}{12}=-3\)
=> \(\hept{\begin{cases}\frac{x-1}{3}=-3\\\frac{y-1}{4}=-3\\\frac{z+2}{5}=-3\end{cases}}\) => \(\hept{\begin{cases}x-1=-9\\y-1=-12\\z+2=-15\end{cases}}\) => \(\hept{\begin{cases}x=-8\\x=-11\\x=-13\end{cases}}\)
Vậy ...
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 × [1/6×7 + 1/7×8 + 1/8×9 + ... + 1/x(x + 1)] = 2/9
1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/x + 1 = 2/9 : 2
1/6 - 1/x + 1 = 2/9 × 1/2 = 1/9
1/x + 1 = 1/6 - 1/9
1/x + 1 = 1/18
=> x + 1 = 18
=> x = 18 - 1 = 17
\(\Rightarrow\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{x.\left(x+1\right)}=2.\left(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2.\left(\frac{1}{5}-\frac{1}{x+1}\right)=\frac{2}{5}-\frac{2}{x+1}=\frac{3}{10}\)
=> \(\frac{2}{x+1}\)= \(\frac{1}{10}=\frac{2}{20}\)
=> x +1 = 20 => x = 19
bạn trên sai rồi, nếu đã nhân đôi lên tất cả thì cx phải nhân luôn con cuối chứ
Ta có:
\(A=\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow A=\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow A=\dfrac{1.2}{2.3.7}+\dfrac{1.2}{2.4.7}+\dfrac{1.2}{2.4.9}+...+\dfrac{1.2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow A=2\left(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow A=2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow A=2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=17\)
Vậy \(x=17\)
Ta có các TH:
+/ x-1\(\ge\)0 => x\(\ge\)1=> Ix-1I=x-1 và I1-xI=x-1
Phương trình tương đương: 2016(x-1)+(x-1)2=2015(x-1)
<=> (x-1)+(x-1)2=0 <=> (x-1)(1+x-1)=0
<=> x(x-1)=0 => x=0 (Loại) và x=1 (Chọn)
+/ x-1< 0 => x<1=> Ix-1I=1-x và I1-xI=1-x
Phương trình tương đương: 2016(1-x)+(x-1)2=2015(1-x)
<=> (1-x)+(x-1)2=0 <=> (x-1)(-1+x-1)=0
<=> (x-1)(x-2)=0 => x=1 (Loại) và x=2 (Loại) vì x<1
ĐS: x=1
Suy ra 2016 . |x-1| - 2015. |1-x| + ( x-1 )^2 =0 ( chuyển vế)
suy ra |x-1| (2016-2015) + (x-1)^2 =0 ( đổi |1-x| thành |x-1| rồi phân phối)
suy ra |x-1| . 1 + (x-1)^2 =0
Suy ra |x-1| + (x-1)^2 =0
Vì | x-1| >=0, mọi x
(x-1)^2 >=0, mọi x
suy ra |x-1| + (x-1)^2 >= 0, mọi x
dấu ' = ' xảy ra <=> (x-1) =0 hoặc (x-1)^2 =0
Tính ra thì cả 2 kết quả đều ra x=1
vậy x=1
Ko tránh khỏi thiếu sót, nếu sai ai đo sửa lại nhé. thắc mắc gì thì cứ hỏi
_Hết_
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{2}{9}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
\(\frac{1}{x+1}=\frac{1}{18}\Rightarrow18.1=1\left(x+1\right)\)
\(\Rightarrow18=x+1\Rightarrow x=18-1=17\)
Cảm ơn bạn nha!