1 . Tìm x :
a ) \(\frac{1}{2}x+\frac{9}{4}-\frac{8}{20}=\frac{77}{20}\)
b ) \(\frac{28}{100}x+\frac{4}{25}+\frac{10}{4}=\frac{203}{50}\)
c ) \(\frac{100}{50}=\frac{1}{2}x\)
Tìm x :
a ) \(\frac{1}{2}x+\frac{9}{4}-\frac{8}{20}=\frac{77}{20}\)
b ) \(\frac{28}{100}x+\frac{4}{25}+\frac{10}{4}=\frac{203}{50}\)
c ) \(\frac{100}{58}=\frac{1}{2}x\)
AI NHANH NHẤT + RÕ RÀNG NHẤT + ĐÚNG NHẤT = 2 LIKE
b)\(\frac{28}{100}x+\frac{4}{25}+\frac{10}{4}=\frac{203}{50}\)
<=>\(\frac{28}{100}x=\frac{203}{50}-\frac{10}{4}-\frac{4}{25}=\frac{406-250-16}{100}\)
<=>\(\frac{28}{100}x=\frac{140}{100}\)
<=>\(x=\frac{140}{100}:\frac{28}{100}=\frac{140}{100}.\frac{100}{28}\)
<=>x=5
a) \(\frac{1}{2}x+\frac{9}{4}-\frac{8}{20}=\frac{77}{20}\)
=> \(\frac{1}{2}x=\frac{77}{20}+\frac{8}{20}-\frac{9}{4}=2\)
=> \(\frac{x}{2}=2\Rightarrow x=2\cdot2=4\)
vậy x = 4
b) \(\frac{28}{100}x+\frac{4}{25}+\frac{10}{4}=\frac{203}{50}\)
=> \(\frac{28}{100}x=\frac{203}{50}-\frac{10}{4}-\frac{4}{25}=1\frac{2}{5}=\frac{7}{5}\)
=> \(\frac{28x}{100}=\frac{7}{5}\Rightarrow28x=\frac{100\cdot7}{5}=140\Rightarrow x=140:28=5\)
vậy x = 5
c) \(\frac{100}{58}=\frac{1}{2}x\)
=> \(\frac{100}{58}=\frac{x}{2}\Rightarrow x=\frac{100\cdot2}{58}=3\frac{13}{29}=\frac{100}{29}\)
vậy x = 100/29
Tìm X:
\(\frac{50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}}{\left(2x-1\right)}\)=11
Tìm x :
a ) ( 9 - 8x ) 2 = 100
b ) ( 20 - x ) 2 = 60
c ) \(\frac{1}{2}x+\frac{8}{4}=\frac{100}{25}\)
d ) ( 90 - x ) 10 = 1000
a, (9-8x)x2=100
9-8x =100:2
9-8x =50
8x =9-50
8x =-41
x =-41:8
x = \(\frac{-41}{8}\)
b, (20-x)x2=60
20-x =60:2
20-x =30
x =20-30
x =-10
c, \(\frac{1}{2}x+\frac{8}{4}=\frac{100}{25}\)
\(\frac{1}{2}x+2=4\)
\(\frac{1}{2}x=4-2\)
\(\frac{1}{2}x=2\)
x =2:\(\frac{1}{2}\)
x =4
d, (90-x)10=1000
90-x =1000:10
90-x =100
x =90-100
x=-10
****
d)(90-x)10=1000
<=>90-x=1000:10
<=>90-x=100
<=>x=90-100
<=>x=-10
Bài 1:Tìm x,biết
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
b)\(x+\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}=\frac{2}{3}\)
c)\(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}=\frac{5}{24}\)
d)\(8-\frac{8-\frac{8}{5}+\frac{8}{25}-\frac{8}{125}}{9-\frac{9}{5}+\frac{9}{25}-\frac{9}{125}}:\frac{161616}{151515}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{1}{23}}\)
a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)
Tính nhanh :
a) 17 x 8 + 51 x 4
b) 2 x 2 x 3 x 5 x 19
c) 54 x 275 + 825 x 15 + 275
d) 100 - 99 + 98 - 97 + 96 - 95 + 94 - 93 + ... + 4 - 3 + 2
e) \(\frac{167x198+98}{198x168-100}\)
g) \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2019x2020}\)
h)\(\frac{4}{2x4}+\frac{4}{4x6}+\frac{4}{6x8}+...+\frac{4}{16x18}\)
k) 1,5 + 2,5 + 3,5 + 4,5 + 5,5 + 6,5 + 7,5 + 8,5
m) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
n) \(\frac{13}{50}+9\%+\frac{14}{100}+24\%\)
\(17.8+51.4=34.4+51.4=4\left(51+34\right)=4.84=336\) \(2.2.3.5.19=\left(2.5\right).\left(3.19\right).2=10.2.57=570.2=1140\) \(54.275+825.15+275=54.275+45.275+275=275\left(54+45+1\right)=100.275=27500\) \(\frac{167.198+98}{198.168-100}=\frac{167.198+98}{198.167+198-100}=\frac{167.198+98}{167.198+98}=1\)
\(\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)
a) 17 x 8 + 51 x 4
= 17 x 4 x 2 + 17 x 3 x 4
= 17 x 4 x ( 2 + 3 )
= 14 x 4 x 5
= 14 x 20
= 280
b) 2 x 2 x 3 x 5 x 19
= ( 2 x 5 ) x ( 3 x 19 ) x 2
= 10 x 57 x 2
= 570 x 2
= 1140
c) 54 x 275 + 825 x 15 + 275
= 54 x 275 + 275 x 3 x 15 + 275 x 1
= 54 x 275 + 275 x 45 + 275 x 1
= 275 x ( 54 + 45 + 1 )
= 275 x 100
= 27500
d) 100 - 99 + 98 - 97 + 96 - 95 + 94 - 93 + ... + 4 - 3 + 2
= (100 - 99) + (98 - 97) + (96 - 95) + (94 - 93) + ... + (4 - 3) + 2
= (1 + 1 + ... + 1) + 2
( 49 số 1 )
= 49 + 2
= 51
k) 1,5 + 2,5 + 3,5 + 4,5 + 5,5 + 6,5 + 7,5 + 8,5
= ( 1,5 + 8,5 ) + ( 2,5 + 7,5 ) + ( 3,5 + 6,5 ) + ( 4,5 + 5,5 )
= 10 + 10 + 10 + 10
= 40
m) Đề bn xem ở trên
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
Tìm số tự nhiên x, biết:
a, \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-....-\frac{20}{53-55}=\frac{3}{11}\)
b, \(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)
c,\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
Bài 1 :Tìm a,b biết :
\(a+b=3.\left(a-b\right)=\)\(2\frac{a}{b}\)
Bài 2 :
\(A=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{5}+\frac{100}{67}+...+\frac{100}{98.99}+\frac{1}{99}\)
Bài 1 :
\(a+b=3.\left(a-b\right)=\)\(2\frac{a}{b}\)
\(\Rightarrow a+b=3.\left(a-b\right)\)
\(\Rightarrow a+b=3a-3b\)
\(\Rightarrow3a-3b-a-b=0\)
\(\Rightarrow2a-4b=0\)
\(\Rightarrow2.\left(a-2b\right)=0\)
\(\Rightarrow\hept{\begin{cases}a-2b=0\\a=2b\end{cases}}\)
Ta có : \(a+b=\frac{2a}{b}\)
Thay \(a=2b\) vào
\(\Rightarrow2b+b=\frac{2.23}{b}\)
\(\Rightarrow3b=\frac{4b}{b}\Rightarrow3b=4\)
\(\Rightarrow b=\frac{4}{3}\Rightarrow a=2.\frac{4}{3}=\frac{8}{3}\)
Vậy \(a=\frac{8}{3}\) và \(b=\frac{4}{3}\)
Chúc bạn học tốt ( -_- )
Bài 2 :
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{5}+\frac{100}{6.7}+...+\)\(\frac{100}{98.99}+\frac{1}{99}\)
\(B=\frac{100}{2}+\frac{100}{6}+\frac{100}{12}+\frac{100}{20}+\frac{100}{30}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{9900}\)
\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)
\(B=100.\frac{100}{2}+\frac{100}{2}-\frac{1}{3}+\frac{100}{3}-\frac{100}{4}+\frac{100}{4}-\frac{100}{5}+\frac{100}{5}-\frac{100}{6}+\frac{100}{6}\)\(-\frac{100}{7}+...+\frac{100}{98}+\frac{100}{99}+\frac{100}{99}-1\)
\(B=100-1\)
\(B=99\)
Chúc bạn học tốt ( -_- )
Tính
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
Tính tổng :
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)
\(B=100\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(B=100\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(B=100\left(1-\frac{1}{100}\right)\)
\(B=100.\frac{99}{100}=99\)