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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
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)
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
Bài 2:
a) \(\frac{4}{9}+x=\frac{-5}{3}\)
\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)
\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)
Vậy: \(x=\frac{-19}{9}\)
b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)
\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)
\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)
c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)
\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-5\right\}\)
\(17.8+51.4=34.4+51.4=4\left(51+34\right)=4.84=336\) \(2.2.3.5.19=\left(2.5\right).\left(3.19\right).2=10.2.57=570.2=1140\) \(54.275+825.15+275=54.275+45.275+275=275\left(54+45+1\right)=100.275=27500\) \(\frac{167.198+98}{198.168-100}=\frac{167.198+98}{198.167+198-100}=\frac{167.198+98}{167.198+98}=1\)
\(\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)
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
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