1.Tìm x
a.0,5 x - 2/3 x=7/12
b.x÷4 1/3=-2,5
c.( 3x/7+1) ÷ (-4)=-1/28
d.x + 30% x= -1,3
2.Tính bằng cách hợp lý
a.3/5.7 + 3/7.9 + ............+ 3/2019.2021
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Khách
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15 tháng 7 2021
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
15 tháng 7 2021
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
7 tháng 7 2017
G=\(\frac{3}{2.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{2015.2017}\)
G=\(3.\left(\frac{1}{2.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}\right)\)
G=\(3.\left(\frac{1}{2}.\frac{1}{5}+\frac{1}{5}.\frac{1}{7}+\frac{1}{7}.\frac{1}{9}+...+\frac{1}{2013}.\frac{1}{2015}+\frac{1}{2015}.\frac{1}{2017}\right)\)
G=\(3.\left(\frac{1}{2}+\frac{1}{2017}\right)\)
G=1.5
Anh ko bik có đúng ko nữa lâu quá rồi. Em thông cảm nhé
LH
1
11 tháng 7 2018
Ta tách ra làm 2 ý nhé:
* \(1,11+0,19-1,32-\left(\frac{1}{2}+\frac{1}{3}\right):2\)
\(=1,3-1,32-\frac{5}{6}=-0,02-\frac{5}{6}=-\frac{1}{50}-\frac{5}{6}=-\frac{64}{75}\)
* \(\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}\)
\(\left(3\frac{5}{8}-\frac{1}{2}\right):2\frac{23}{26}=3\frac{1}{8}:2\frac{23}{26}=1\frac{1}{12}=\frac{13}{12}\)
VẬY TA CÓ: \(\frac{-64}{75}< x< \frac{13}{12}\)mà \(\left(x\inℕ\right)\)
\(\Rightarrow\)ta có x là số nguyên nằm giữa -0.8533... và 1,0833...
vậy ta có x là các số nguyên 0 và 1
MK KO CHẮC CHO LẮM NÊN BN CÓ THỂ THAM KHẢO Ý KIẾN MẤY BN KHÁC NHÉ.
K MK NHA. CHÚC BN HỌC TỐT. ^_^
XO
4 tháng 8 2020
Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
=> \(\frac{7}{x}=\frac{7}{15}\)
=> x = 15 (tm)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)
=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)
=> \(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
=> 2x = 90
=> x = 45
31 tháng 3 2021
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
1)a, \(0,5x-\frac{2}{3}x=\frac{7}{12}\) b) \(x:4\frac{1}{3}=-\frac{2}{5}\)
\(x\left(0,5-\frac{2}{3}\right)=\frac{7}{12}\) \(x=-\frac{2}{5}.4\frac{1}{3}\)
\(-\frac{1}{6}x=\frac{7}{12}\) \(x=-\frac{26}{15}\)
\(x=\frac{-7}{2}\)
c) \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\) d) \(x+30\%x=-1,3\)
\(\frac{3x}{7}+1=-\frac{1}{28}.\left(-4\right)\) \(x\left(1+30\%\right)=-1,3\)
\(\frac{3x}{7}+1=\frac{1}{7}\) \(1,3x=-1,3\)
\(\frac{3x}{7}=-\frac{6}{7}\) \(x=-1,3:1,3\)
\(x=-6.7:7:3\) \(x=-1\)
\(x=-2\)
Bài 2:
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{2019.2021}=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2019.2021}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2019}-\frac{1}{2021}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{2021}\right)\)
\(=\frac{3}{2}.\frac{2016}{10105}\)
\(=\frac{3024}{10105}\)