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Số số hạng là: (2x+1-1):2+1=2x:2+1=x+1(số hạng)
Ta có: 1+3+5+…+(2x+1)=400
=> (1+2x+1).(x+1):2=400
=> (2+2x).(x+1):2=400
=> 2.(x+1)(x+1):2=400
=> (x+1)2=202
=> x+1=20
=> x=19
Vậy x=19
a) \(215+x=400\)
\(\Rightarrow x=400-215\)
\(\Rightarrow x=185\)
b) \(12,5-2x=\dfrac{1}{5}\)
\(\Rightarrow2x=12,5-\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{123}{10}\)
\(\Rightarrow x=\dfrac{123}{10}:2\)
\(\Rightarrow x=\dfrac{123}{20}\)
Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)
=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)
=> x = 9
Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)
=> \(\frac{15}{16}:x=\frac{11}{12}\)
=> \(x=\frac{45}{44}\)
Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)
=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
=> \(\frac{1}{x+1}=\frac{1}{800}\)
=> x = 799
Bài 2 :
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)
Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)
\(=1-\frac{1}{12}=\frac{11}{12}\) (2)
Thay (1) và (2) vào biểu thức (*) ta được :
\(\frac{15}{16}:x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)
\(\Leftrightarrow x=\frac{45}{44}\)
Vậy : \(x=\frac{45}{44}\)
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\left(\text{Quy đồng nhé !}\right)\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{399}{800}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{800}\)
=> x + 1 = 800
<=> x = 799
1/3+1/6+1/10+...+1/x(x+1):2=399/400
2.[1/3.2+1/6.2+1/10.2+...+1/x(x+1)]=399/400
2.[1/6+1/12+1/20+...+1/x(x+1)]=399/400
2.[1/2.3+1/3.4+1/4.5+...+1/x(x+1)]=399/400
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=399/800
1/2-1/x+1=399/800
1/x+1=1/800
=> x+1=800
=> x=799
(\(x+2\)) + ( 2\(\times\) \(x\) + 4) + ( 3 \(\times\) \(x\) + 6) = 400 \(\times\)0,12
\(x\) + 2 + 2 \(\times\) \(x\) + 4 + 3 \(\times\) \(x\) + 6 = 48
\(x\) \(\times\) ( 1 + 2 + 3) + ( 2 + 4 + 6) = 48
\(x\) \(\times\) 6 + 12 = 48
\(x\) \(\times\) 6 = 48 - 12
\(x\) \(\times\) 6 = 36
\(x\) = 36 : 6
\(x\) = 6
1+3+5+...+(x-2)+x=400
=>\(\dfrac{\left(x-1\right):2+1\cdot\left(1+x\right)}{2}\)=400
=>(\(\dfrac{x}{2}\)-\(\dfrac{1}{2}\)+1)\(\cdot\)(1+x)=800
=>(\(\dfrac{x}{2}\)+\(\dfrac{1}{2}\))\(\cdot\)(1+x)=800
=>(x+1):2\(\cdot\)(1+x)=800
=>(x+1)2=800x2=1600=402
=>x=40-1=39
Vậy x=39.