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a) \(3,4+x\cdot3=11,53+0,8\)
\(3,4+x\cdot3=12,33\)
\(x\cdot3=12,33-3,4\)
\(x\cdot3=8,93\)
\(x=8,93:3=\frac{893}{300}\)
b) \(x:2,1-3,8=3,2-1,8\)
\(x:2,1-3,8=1,4\)
\(x:2,1=1,4+3,8\)
\(x:2,1=5,2\)
\(x=5,2\cdot2,1=10,93\)
a) \(3,4+3x=11,53+0,8\)
\(\Leftrightarrow3x=11,53+0,8-3,4=8,93\)
\(\Leftrightarrow x=\frac{8,93}{3}=2,97\left(6\right)\)
Vậy \(x=2,97\left(6\right)\)
b) \(\frac{x}{2,1}-3,8=3,2-1,8\)
\(\Leftrightarrow\frac{x}{2,1}=3,2-1,8+3,8=5.2\)
\(\Leftrightarrow x=5,2\cdot2,1=10.92\)
Vậy \(x=10,92\)
Học tố nhé ^3^
Bài 1 : Tính :
a)\(\frac{6}{5}\)x 4,7 - 3,2
= 5,64 - 3,2
= 2,44
b) 17,6 x 6,7 + 3,2
= 117,92 + 3,2
= 121,12
Bài 2 :
a) 0,8 x 2,5 x 1,25
= ( 0,8 x 1,25 ) x 2,5
= 1 x 2,5
= 2,5
b) 4 x 5,6 x 2,5
= ( 4 x 2,5 ) x 5,6
= 10 x 5,6
= 56
c) 4,65 x 5,6 + 5,6 + 5,6 x 4,35
= 4,65 x 5,6 + 5,6 x 1 + 5,6 x 4,35
= 5,6 x ( 4,65 + 1 + 4,35 )
= 5,6 x 10
= 56
1.
\(\frac{6}{5}\)x \(4,7-3,2\)\(=5,64-3,2=2,44\)
\(17,6\)x \(6,7+3,2\)\(=117,92+3,2=121,12\)
2.
\(0,8\)x \(2,5\)x \(1,25\)\(=\)\(1,25\)x \(0,8\)x \(2,5\)\(=\)\(1\)x \(2,5\)\(=2,5\)
\(4\)x \(5,6\)x \(2,5\)\(=\)\(4\)x \(2,5\)x \(5,6\)\(=\)\(10\)x \(5,6\)\(=\)\(56\)
\(4,65\)x \(5,6\)+ \(5,6\)+ \(5,6\)x \(4,35\)= \(5,6\)x \(\left(4,65+1+4,35\right)\)= \(5,6\)x \(10\)= \(56\)
Nếu đúng thì cho mình 1 k mình nhé :)
Chúc bạn học tốt :)
Trả lời
\(0,8\cdot70+3,2\cdot6+1,6\cdot3\)
\(=56+19,2+4,8\)
\(=56+\left(19,2+4,8\right)\)
\(=56+24\)
\(=80\)
0,8 x 70 + 3,2 x 6 + 1,6 x 3
= 0,8 x 70 + 0,8 x 4 x 6 + 0,8 x 2 x 3
= 0,8 x (70 + 4 x 6 + 2 x 3)
= 0,8 x (70 + 24 + 6)
= 0,8 x (94 + 6)
= 0,8 x 100
= 80
9,45 - 2,37 + 1,37 = 8,45
2,3 : 3,2 x 0,8 = 0,575
Nếu đúng thì tk nha
a 9.45-2.37-1.37=9.45-1=8.45
b 2.3*0.8chia3.2*0.8=0.575
dung thi cho k
Gợi ý :
a) tìm khoảng cách -> số số hạng -> tổng
b) và c) : đặt nhân tử chung rồi tính nốt phần còn lại
a) x x 0,8 = 1,2 x 4,5
x x 0,8 = 5,4
x = 5,4 : 0,8
x = 6,75
Vậy x = 6,75.
b) 45,54 : x = 18 : 5
45,54 : x = 3,6
x = 45,54 : 3,6
x = 12,65
Vậy x = 12,65.
\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
b) \(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{10050}\)
Bài 1:
a)\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot42,9-11,7\cdot0,25+0,25\cdot0,8\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot\left(42,9-11,7+0,8\right)\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot32\)
\(\frac{2\cdot x-4,36}{0,125}=8\)
\(2\cdot x-4,36=8\cdot0,125\)
\(2\cdot x-4,36=1\)
\(2\cdot x=1+4,36\)
\(2\cdot x=5,36\)
\(x=\frac{5,36}{2}=2,68\)
b) \(N=\frac{1}{1\cdot5}+\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+\frac{1}{15\cdot20}+...+\frac{1}{2005\cdot2010}\)
\(4N=\frac{4}{1\cdot5}+\frac{4}{5\cdot10}+\frac{4}{10\cdot15}+\frac{4}{15\cdot20}+...+\frac{4}{2005\cdot2010}\)
\(4N=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(4N=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(N=\frac{2009}{2010}\div4=\frac{2009}{8040}\)
Bài 2:
a) ( x + 5,2 ) : 3,2 = 4,7 ( dư 0,5 )
\(x+5,2=4,7\cdot3,2+0,5\)
\(x+5,2=15,54\)
\(x=15,54-5,2=10,34\)
b)\(A=\frac{4047991-2010\cdot2009}{4050000-2011\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4050000-2009-2010\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4047991-2010\cdot2009}=1\)
Bài 3:
a) \(104,5\cdot x-14,1\cdot x+9,6\cdot x=25\)
\(x\cdot\left(104,5-14,1+9,6\right)=25\)
\(x\cdot100=25\)
\(x=\frac{25}{100}=\frac{1}{4}=0,25\)
b) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
`a, 3,2 \div x=0,8`
`x=3,2 \div 0,8`
`x=4`
`b, 5 \times x = 37,5`
`x=37,5 \div 5`
`x=7,5`
`@` `\text {_vdn_}`