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CN
4
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 11 2023
1, (0,25 - \(x\)) : - \(\dfrac{3}{5}\) = - \(\dfrac{3}{4}\)
0,25 - \(x\) = - \(\dfrac{3}{4}\) x (- \(\dfrac{3}{5}\))
0,25 - \(x\) = \(\dfrac{9}{20}\)
\(x\) = 0,25 - 0,45
\(x\) = - 0,2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 11 2023
2, - \(\dfrac{3}{8}\)\(x\) - 0,75 = - 1\(\dfrac{1}{2}\)
- \(\dfrac{3}{8}\)\(x\) - 0,75 = -1,5
\(\dfrac{3}{8}\)\(x\) = - 0,75 + 1,5
\(\dfrac{3}{8}\)\(x\) = 0,75
\(x\) = 0,75 : \(\dfrac{3}{8}\)
\(x\) = 2
AD
3
H9
HT.Phong (9A5)
CTVHS
23 tháng 7 2023
\(\left(-0,75\right)-\left(-1+\dfrac{2}{3}\right):0,5+\left(-\dfrac{1}{4}\right)\)
\(=\left(-0,75\right)-\left(-1-\dfrac{2}{3}\right)\cdot\dfrac{1}{2}-0,25\)
\(=\left(-0,75-0,25\right)+\dfrac{5}{6}\)
\(=-1+\dfrac{5}{6}\)
\(=-\dfrac{11}{6}\)
_________________
\(\left[\left(-\dfrac{3}{2}\right)+\dfrac{2}{3}\right]^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)
\(=\left(-\dfrac{9}{6}+\dfrac{4}{6}\right)^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)
\(=\left(\dfrac{-5}{6}\right)^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)
\(=\dfrac{25}{36}\cdot\dfrac{24}{25}-\dfrac{1}{5}\)
\(=\dfrac{2}{3}-\dfrac{1}{5}\)
\(=\dfrac{7}{15}\)
AD
23 tháng 7 2023
\(a,\left(-0,75\right)-\left(-1+\dfrac{2}{3}\right):0,5-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(\dfrac{-3+2}{3}\right):\dfrac{1}{2}-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(-\dfrac{1}{3}\right):\dfrac{1}{2}-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(-\dfrac{1}{3}\right)\times2-\dfrac{1}{4}\\ =-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{4}\\ =\left(-\dfrac{3}{4}-\dfrac{1}{4}\right)+\dfrac{2}{3}\\ =-\dfrac{4}{4}+\dfrac{2}{3}\\ =-1+\dfrac{2}{3}\\ =\dfrac{-3+2}{3}=-\dfrac{1}{3}\)
\(b,\left[\left(-\dfrac{3}{2}\right)+\dfrac{2}{3}\right]^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\left(\dfrac{-3\times3+2\times2}{6}\right)^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\left(-\dfrac{5}{6}\right)^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\dfrac{25}{36}\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\dfrac{2}{3}-\dfrac{1}{5}\\ =\dfrac{2\times5-3}{15}=\dfrac{7}{15}\)
NT
2
3 tháng 4 2020
a) Ta có: \(x-\frac{2}{3}=0,75\)
\(\Leftrightarrow x-\frac{2}{3}=\frac{3}{4}\)
hay \(x=\frac{3}{4}+\frac{2}{3}=\frac{17}{12}\)
Vậy: \(x=\frac{17}{12}\)
b) Ta có: \(\frac{1}{3}+\frac{2}{3}x=-1\)
\(\Leftrightarrow\frac{2}{3}x=-1-\frac{1}{3}=-\frac{4}{3}\)
hay \(x=\frac{-4}{3}:\frac{2}{3}=\frac{-4}{3}\cdot\frac{3}{2}=-2\)
Vậy: x=-2
c) Ta có: \(\left|2x-3\right|-\frac{3}{4}=4,25\)
\(\Leftrightarrow\left|2x-3\right|=\frac{17}{4}+\frac{3}{4}=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy: x∈{-1;4}
3 tháng 4 2020
a, \(x-\frac{2}{3}=0.75\)
\(x=0.75-\frac{2}{3}\)
\(x=\frac{1}{12}\)
Vậy...
b, \(\frac{1}{3}+\frac{2}{3}\cdot x=-1\)
\(\frac{2}{3}\cdot x=-1-\frac{1}{3}=-\frac{4}{3}\)
\(x=-\frac{4}{3}:\frac{2}{3}=-2\)
Vậy x = -2
c, \(|2x-3|-\frac{3}{4}=4.25\)
\(|2x-3|=4.25-\frac{3}{4}=\frac{7}{2}\)
=> \(2x-3=\frac{7}{2}hay2x-3=-\frac{7}{2}\)
2x = \(\frac{7}{2}+3\) 2x = \(-\frac{7}{2}+3\)
2x = \(\frac{13}{2}\) 2x = \(-\frac{1}{2}\)
x = \(\frac{12}{2}:2=\frac{13}{4}\) x = \(-\frac{1}{2}:2\) = \(-\frac{1}{4}\)
Vậy...
AH
Akai Haruma
Giáo viên
27 tháng 11 2023
Lời giải:
$\frac{1}{4}-3x+\frac{3}{2}=-0,75$
$3x=\frac{1}{4}+\frac{3}{2}-(-0,75)=2,5$
$\Rightarrow x=2,5:3=\frac{5}{6}$
15 tháng 4 2019
Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{2019}{3^{2019}}\)
=>\(3A=1+\frac{2}{3}+...+\frac{2019}{3^{2018}}\)
=>\(2A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{2018}}\)
=>\(2B=3-\frac{1}{3^{2018}}\)=>\(B=\frac{3-\frac{1}{3^{2018}}}{2}\)
=>\(2A=\frac{3-\frac{1}{3^{2018}}}{2}-\frac{2019}{3^{2019}}=\frac{\frac{3^{2019}-1}{3^{2018}}}{2}-\frac{2019}{3^{2019}}\)
\(=\frac{3^{2019}-1}{3^{2018}.2}-\frac{2019}{3^{2019}}=\frac{3\left(3^{2019}-1\right)-2019.2}{3^{2019}.2}\)
15 tháng 4 2019
Nhầm tí
dòng thứ 2 từ dưới lên cm bé hơn 0,75 luôn nhá
LG
1
DH
18 tháng 2 2020
Đặt: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2019}{3^{2019}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2019}{3^{2018}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2018}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2018}}}{2}\)
Thay vào \(2A\Rightarrow2A=1+\frac{\left(1-\frac{1}{3^{2018}}\right)}{2}-\frac{2019}{3^{2019}}\)
\(=1+\frac{1}{2}-\frac{1}{2.3^{2018}}-\frac{2019}{3^{2019}}< 1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow A< 0,75\left(đpcm\right)\)
H
5 tháng 11 2023
\(\sqrt{0,36}-\left|-0,75\right|:\left(-1\dfrac{3}{4}\right)\\ =0,6-0,75:\dfrac{-7}{4}\\ =\dfrac{3}{5}-\dfrac{3}{4}\cdot\dfrac{-4}{7}\\ =\dfrac{3}{5}-\dfrac{-12}{28}\\ =\dfrac{3}{5}+\dfrac{3}{7}\\ =\dfrac{36}{35}\)
\(1-\left(\dfrac{5}{9}-\dfrac{2}{3}\right)^2:\dfrac{4}{27}\\ =1-\left(\dfrac{5}{9}-\dfrac{6}{9}\right)^2\cdot\dfrac{27}{4}\\ =1-\left(-\dfrac{1}{9}\right)^2\cdot\dfrac{27}{4}\\ =1-\dfrac{1}{81}\cdot\dfrac{27}{4}\\ =1-\dfrac{1}{12}\\ =\dfrac{11}{12}\)
KV
5 tháng 11 2023
1/9 - 0,3 . 5/9 + 1/3 + 5
= 1/9 - 3/10 . 5/9 + 1/3 + 5
= 1/9 - 1/6 + 16/3
= -1/18 + 16/3
= 95/18
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