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a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2
=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)
=> \(-2x=\frac{-85}{4}\)
=> \(x=\frac{-85}{4}:\left(-2\right)\)
=> \(x=\frac{85}{8}\)
b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)
=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)
=> \(\frac{-2}{3}x=\frac{15}{29}\)
=> x = \(\frac{15}{29}:\frac{-2}{3}\)
=> x = \(\frac{-45}{58}\)
a) (1,5 . 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
=> (2,85 - x - 0,5) : 0,25 = 60
=> (2,85 - 0,5) - x = 60 . 0,25
=> 2,35 - x = 15
=> x = 2,35 - 15
=> x = -12,65
Vậy x = -12,65
b) \(1-\left(5\frac{2}{9}+x-7\frac{7}{18}\right)\div2\frac{1}{6}=0\)
\(\Rightarrow\left(5\frac{2}{9}-7\frac{7}{18}+x\right)\div2\frac{1}{6}=1-0\)
\(\Rightarrow\left(\frac{47}{9}-\frac{133}{18}+x\right)\div2\frac{1}{6}=1\)
\(\Rightarrow\frac{-13}{6}+x=2\frac{1}{6}\)
\(\Rightarrow x=2\frac{1}{6}-\frac{-13}{6}\)
\(\Rightarrow x=\frac{13}{6}+\frac{13}{6}\)
\(\Rightarrow x=\frac{26}{6}\)
\(\Rightarrow x=\frac{13}{3}\)
Vậy \(x=\frac{13}{3}\)
c) \(35\left(2\frac{1}{5}-x\right)=32\)
\(\Rightarrow2\frac{1}{5}-x=32\div35\)
\(\Rightarrow\frac{11}{5}-x=\frac{32}{35}\)
\(\Rightarrow x=\frac{11}{5}-\frac{32}{35}\)
\(\Rightarrow x=\frac{9}{7}\)
Vậy \(x=\frac{9}{7}\)
d) \(\frac{4}{3}+\left(x\div2\frac{2}{3}-0,5\right).1\frac{35}{55}=0,6\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{3}{5}-\frac{4}{3}\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{-11}{15}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-11}{15}\div\frac{18}{11}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-121}{270}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{-121}{270}+\frac{1}{2}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{7}{135}\)
\(\Rightarrow x=\frac{7}{135}.\frac{8}{3}\)
\(\Rightarrow x=\frac{56}{405}\)
Vậy \(x=\frac{56}{405}\)
e) \(1\frac{1}{3}.2\frac{2}{4}\div\frac{5}{6}.1\frac{1}{11}=11-5\div x\)
\(\Rightarrow\frac{4}{3}.\frac{5}{2}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow\frac{10}{3}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow4.\frac{12}{11}=11-5\div x\)
\(\Rightarrow11-5\div x=\frac{48}{11}\)
\(\Rightarrow5\div x=11-\frac{48}{11}\)
\(\Rightarrow5\div x=\frac{73}{11}\)
\(\Rightarrow x=5\div\frac{73}{11}\)
\(\Rightarrow x=\frac{55}{73}\)
Vậy \(x=\frac{55}{73}\)
a) (1,5 * 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
(2,85 - x - 0,5) : 0,25 = 60
(2,85 - x - 0,5) = 60 x 0,25
(2,85 - x - 0,5) = 15
2,35 - x = 15
x = 2,35 - 15
x = -12,65
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
a)
\(x.\frac{7}{9}=\frac{2}{3}+2\frac{1}{2}\)
\(x.\frac{7}{9}=\frac{19}{6}\)
\(x=\frac{19}{6}:\frac{7}{9}\)
\(x=\frac{57}{14}\)
b) \(\frac{5}{7}+x:\frac{9}{4}=\frac{4}{3}\)
\(x:\frac{9}{4}=\frac{4}{3}-\frac{5}{7}\)
\(x:\frac{9}{4}=\frac{13}{21}\)
\(x=\frac{13}{21}.\frac{9}{4}\)
\(x=\frac{39}{28}\)
1)
a) \(x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=5\)
\(x+\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}=5\)
\(x+\frac{127}{128}=5\)
\(x=5-\frac{127}{128}=\frac{513}{128}\)
b) \(x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}=3\)
\(x+\frac{729}{2187}+\frac{243}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}=3\)
\(x+\frac{2186}{2187}=3\)
\(x=3-\frac{2186}{2187}=\frac{4375}{2187}\)
2)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
b) \(5\frac{1}{2}+3\frac{5}{6}+\frac{2}{3}\)
\(=\left(5+3\right)+\left(\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\right)\)
\(=8+\left(\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\right)\)
\(=8+2=10\)
c) \(7\frac{7}{8}+1\frac{4}{6}+3\frac{3}{5}\)
\(=\left(7+1+3\right)+\left(\frac{7}{8}+\frac{2}{3}+\frac{3}{5}\right)\)
\(=11+\left(\frac{105}{120}+\frac{80}{120}+\frac{72}{120}\right)\)
\(=11+\frac{257}{120}=\frac{1577}{120}\)
3) Gọi số đó là x. Theo đề ta có :
\(\frac{16-x}{21+x}=\frac{5}{7}\)
\(7\left(16-x\right)=5\left(21+x\right)\)
\(112-7x=105+5x\)
\(112-105=7x-5x\)
\(7=2x\)
\(x=\frac{7}{2}=3,5\) ( vô lí )
Vậy không có số tự nhiên để thõa mãn điều kiện trên.
1.
a) 5/8 x 4/10 + 2/3 =
= 1/4+ 2/3 = 11/12
b)5/12 x 4/7+5/12 x3/7
=5/12 x (4/7 +3/7)
=5/12 x1 = 5/12
c)(4/5 + 3/10 - 1/5 ) x 6 : 4/7
= ( 8/10 + 3/10 + 2/10) x 6 x 7/4
=13/10 x 21/2
=273/20
2.
5/8 và 3/2
ta có 5/8 =10/16 ; 3/2 =24 /16
vì 24 /16 >10 /16 nên 3/2 > 5/8
b. tương tự như câu a nha
c 418/417 và 925 /926
418/417 > 1 ; 925 /926 < 1
vì 418 /417 >1 mà 925/926 < 1 nên 418 / 417 > 925 /926
chúc bạn học tốt nha !
tập trung sẽ bt làm