129 -5[29-5]2 , yêu cầu là tính
tìm x
4(x-3)=72-110
ai nhanh thì mink tick
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a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
5/6 + 11/12 + 19/20 + ... + 109/110
= (1 - 1/6) + (1 - 1/12) + (1 - 1/20) + ... + (1 - 1/110)
= (1 - 1/2×3) + (1 - 1/3×4) + (1 - 1/4×5) + ... + (1 - 1/10×11)
= (1 + 1 + 1 + ... + 1) - (1/2×3 + 1/3×4 + 1/4×5 + ... + 1/10×11)
( có 9 số 1)
= 1 x 9 - (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/10 - 1/11)
= 9 - (1/2 - 1/11)
= 9 - (11/22 - 2/22)
= 9 - 9/22
= 198/22 - 9/22
= 189/22
☆☆☆☆☆
1a) \(\frac{12}{17}+\frac{19}{17}+\frac{8}{17}\)
\(=\left(\frac{12}{17}+\frac{8}{17}\right)+\frac{19}{17}\)
\(=\frac{20}{17}+\frac{19}{17}\)
\(=\frac{39}{17}\)
b) \(\frac{2}{5}+\frac{7}{12}+\frac{13}{12}\)
\(=\frac{2}{5}+\frac{5}{3}\)
\(=\frac{31}{15}\)
2) \(\frac{25}{3}-x=\frac{5}{6}\)
\(x=\frac{25}{3}-\frac{5}{6}\)
\(x=\frac{45}{6}\)
ủng hộ tớ nha
a, Tính :
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}+\frac{109}{110}\)
\(A=\frac{1}{2}+\frac{4}{6}+\frac{1}{6}+\frac{10}{12}+\frac{1}{12}+\frac{18}{20}+\frac{1}{20}+\frac{28}{30}+\frac{1}{30}+\frac{40}{42}+\frac{1}{42}+\frac{54}{56}+\frac{1}{56}\)
\(+\frac{70}{72}+\frac{1}{72}+\frac{88}{90}+\frac{1}{90}+\frac{108}{110}+\frac{1}{110}\)
24 + 3 x (5 - x) = 27
27 x (5 - x) = 27
5 - x = 27 : 27
5 - x = 1
x = 5 - 1
x = 4
24 + 3. ( 5 - x ) = 27
3 . ( 5 - x ) = 27 - 24
3 . ( 5 - x ) = 3
5 - x = 3 : 3
5 - x = 1
x = 5 - 1
x = 4
1,a) 695- [200+ (11- 12)]
= 695- [200+ (11- 1)]
= 695- [200+ 10]
= 695- 210
= 485
b) (519: 517+ 3): 7
= (52+ 3): 7
= (25+ 3): 7
= 28: 7
= 4
c) 129- 5[29- (6- 12)]
= 129- 5[29- (6- 1)]
= 129- 5[29- 5]
= 129- 5. 24
= 129- 120
= 9
3,a) 2x- 49= 5. 32
2x- 49= 5. 9
2x- 49= 45
2x = 45+ 49
2x = 94
x = 94: 2
x = 47
c) 2x- 15= 17
2x = 17+ 15
2x = 32
2x = 25
=> x = 5
Câu 3b bạn tự làm nhé, xin loiosxn vì không giúp được cả bài.
CHÚC BẠN HỌC GIỎI !!!
MÌNH TÌM RA CÁCH LÀM CÂU 3b RỒI !!!
5x+ 2x= 45+ 20: 15
5x+ 2x= 45+ \(\frac{4}{3}\)
5x+ 2x= \(\frac{139}{3}\)
(5+ 2)x=\(\frac{139}{3}\)
7x =\(\frac{139}{3}\)
x =\(\frac{139}{3}\): 7
x =\(\frac{139}{21}\)
CHÚC BẠN HỌC GIỎI !!!
\(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}\)
\(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{4}-\frac{1}{12}=\frac{3-1}{12}=\frac{2}{12}=\frac{1}{6}\)
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{110}+\frac{1}{132}\)
\(\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+\frac{1}{8\times9}+...+\frac{1}{11\times12}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{4}-\frac{1}{12}=\frac{3-1}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)
129 - 5[29-5]2 = 4
x = -6,5
129-5(29-5)^2= 129 - 5*(24)^2= 129-2880=2751
4(x -3)= 72 -110
4(x-3)=-38
x-3=-38/4
x-3=-9.5
x= -9.5+3= -6.5