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20x +(1+6+11+.....+96)=990
20x + 970=990
20x=990-997
20x=20
x=1
(x+1)+(x+6)+...+(x+96)=990
\(\Leftrightarrow\)20x+(1+6+11+...+96) = 990
\(\Leftrightarrow\)20x + \(\frac{\left(1+96\right)\times\left[\left(96-1\right):5+1\right]}{2}\)= 990
\(\Leftrightarrow\)20x+970=990
\(\Leftrightarrow\)20x=990-970=20
\(\Leftrightarrow\)x=1
Vậy x= 1
( x + 1 ) + ( x + 6 ) + ( x + 11 ) + ... + ( x + 96 ) = 990
=> ( x + x + x + ... + x ) + ( 1 + 6 + 11 + ... + 96 ) = 990
20 số x
=> x . 20 + 970 = 990
=> x . 20 = 990 - 970
=> x . 20 = 20
=> x = 20 : 20
=> x = 1
Chúc bạn học tốt nha !!!
Dãy số 1; 6; 11; ...; 96 có: ( 96 -1 ) : 5 + 1 = 20 số
(x + 1 ) + (x + 6 ) +(x + 11) +...+ ( x +96) =1010
(x + x + x +...+ x ) + ( 1 + 6 + 11+...+ 96 ) = 1010
\(20\times x+\left(96+1\right).20:2=1010\)
\(20\times x+970=1010\)
\(20\times x=1010-970\)
\(20\times x=40\)
\(x=2\)
Ta có : (x+1)+(x+6)+(x+11)+...+(x+96) =1010
=>x+1+x+6+x+11+...+x+96=1010
=>(x+x+x+...+x)+(1+6+11+...+96)=1010
=>20×x+(1+96)×20÷5=1010
=>20×x+388=1010
=>20×x=622
=>x=622÷20
=>x=31,1
Câu hỏi của Nguễn Thị Thu Trang - Toán lớp 5 - Học toán với OnlineMath
Bài làm:
Ta có: \(\left(x+1\right)+\left(x+6\right)+\left(x+11\right)+...+\left(x+96\right)=990\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+6+11+...+96\right)=990\)
\(\Leftrightarrow20\times x+\frac{\left(96+1\right)\times20}{2}=990\)
\(\Leftrightarrow20\times x+970=990\)
\(\Leftrightarrow20\times x=20\)
\(\Leftrightarrow x=1\)
Vậy x = 1
Học tốt!!!!
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 : \(x=\frac{45}{44}\)
`[ ( 2 xx x - 11 ) : 3 + 1 ] xx 5 = 20`
`( 2 xx x - 11 ) : 3 + 1=20:5`
`( 2 xx x - 11 ) : 3 + 1=4`
`( 2 xx x - 11 ) : 3 =4-1`
`( 2 xx x - 11 ) : 3 =3`
`2xx x -11=3xx3`
`2xx x -11=9`
`2xx x =9+11`
`2 xx x=20`
`x=20:2`
`x=10`
Vậy `x=10`
`b, x - 96 = ( 443 - x ) - 15`
`x-96=443-x-15`
` x+x=443-15+96`
`2x=524`
`x=524:2`
`x= 262`
Vậy `x=262`
\(#Nqoc\)
`a)`
\([ ( 2 \times x - 11 ) \div 3 + 1 ] \times 5 = 20\)
`(2 \times x - 11) \div 3 + 1 = 20 \div 5`
`(2 \times x - 11) \div 3 + 1 = 4`
`(2 \times x - 11) \div 3 = 4 - 1`
`(2 \times x - 11) \div 3 = 3`
`2 \times x - 11 = 3 \times 3`
`2 \times x - 11 = 9`
`2 \times x = 9 + 11`
`2 \times x = 20`
`x = 20 \div 2`
`x = 10`
Vậy, `x = 10`
`b)`
\(x - 96 = ( 443 - x ) - 15\)
`x - 96 = 443 - x - 15`
`x - 96 = 428 - x`
`x = 428 - x + 96`
`x = 524 - x`
`x - 524 + x = 0`
`(x + x) - 524 = 0`
`2x - 524 = 0`
`2x = 524`
`x = 524 \div 2`
`x = 262`
Vậy, `x = 262.`
(x x 6 + x x 7) x 1 = x x (6 + 7) x 1 = x x 13 x 1 = x x 13
(x x 3 + x x 20) x 1 = x x (3 + 20) x 1 = x x 23 x 1 = x x 23
Vì 13 ≠ 23 nên x x 13 ≠ x x 23
Vậy 2 vế trên không thể bằng nhau được.
Góp ý: Bạn hãy sửa đề lại nhé, đề hơi sai 😅😅😅
Chúc bạn học tốt.
😁😁😁
Nhân thấy : từ 1---> 96 có : (96-1):5+1=20 ( số )
(x+1)+(x+6)+(x+11)+...+(x+96)=990
x+1+x+6+x+11+...+x+96 =990
20.x + ( 1+6+11+..+96) =990
20.x + 970 = 990
20.x = 990-970
20x =20
=> x=1
vậy x=1
x=1
tk nha!