K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
LH
2
H
14 tháng 10 2023
\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\\ \Rightarrow\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{2}{15}=\dfrac{2}{5}\\ \Rightarrow x=\dfrac{2}{5}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{6}{15}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{8}{15}\\ \left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\\ \Rightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\\ \Rightarrow2x+5=4\\ \Rightarrow2x=4-5\\ \Rightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
ND
Nguyễn Đăng Nhân
CTVHS
14 tháng 10 2023
\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)
\(\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(x-\dfrac{2}{15}=\dfrac{2}{5}\)
\(x=\dfrac{2}{5}+\dfrac{2}{15}\)
\(x=\dfrac{8}{15}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)
\(2x+5=4\)
\(2x=-1\)
\(x=-0,5\)
M
10 tháng 10 2018
a, \(125^3:5^7=\left(5^3\right)^3:5^7=5^9:5^7=5^2\)
b, \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2^2}{7^2}\right)^5:\left(\dfrac{2^3}{7^3}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left[\left(\dfrac{2}{7}\right)^2\right]^5:\left[\left(\dfrac{2}{7}\right)^3\right]^2\)
=\(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2}{7}\right)^{10}:\left(\dfrac{2}{7}\right)^6\)
= \(\left(\dfrac{2}{7}\right)^{18-10-6}=\left(\dfrac{2}{7}\right)^2\)
c, \(3-\left(\dfrac{-7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)
= 3 - 1 +\(\left[\left(\dfrac{1}{3}\right)^5.3^5\right]\)
= 2 + 1=3
d, \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(9.5\right)^{10}.5^{20}}{\left(25.3\right)^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\)
= \(\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)
30 tháng 11 2017
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....0......\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=0\)
15 tháng 2 2021
a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)
\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)
\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)
\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)
\(=-\dfrac{891}{100}\)
b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)
\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)
\(=\dfrac{58}{8}+\dfrac{100}{8}\)
\(=\dfrac{158}{8}=\dfrac{79}{4}\)
c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)
\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)
\(=4-1-\dfrac{2}{5}\)
\(=3-\dfrac{2}{5}\)
\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)
e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)
\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}+\dfrac{28}{15}\)
\(=\dfrac{-25}{60}+\dfrac{112}{60}\)
\(=\dfrac{87}{60}=\dfrac{29}{20}\)
f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)
\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{8}\)
\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)
g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)
\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)
\(=\left(\dfrac{1}{2}\right)^{55}\)
\(=\dfrac{1}{2^{55}}\)
h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)
\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)
\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)
\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)
\(=\dfrac{1}{800000}\)
AH
Akai Haruma
Giáo viên
20 tháng 10 2023
Lời giải:
Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$
$\Rightarrow x=2018a; y=2019a; z=2020a$
$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$
Mặt khác:
$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$
Từ $(1); (2)$ ta có đpcm.
TT
10 tháng 9 2017
\(A=\left(-\dfrac{5}{7}+\dfrac{8}{5}\right):\dfrac{91}{8}+\left(-\dfrac{2}{7}-\dfrac{3}{5}\right):\dfrac{91}{8}\)
\(=\dfrac{31}{35}:\dfrac{91}{8}+\dfrac{-31}{35}:\dfrac{91}{8}\)
\(=\dfrac{248}{3185}+\dfrac{-248}{3185}\)
= 0
\(B=\dfrac{13}{15}:\left(\dfrac{4}{5}-\dfrac{3}{7}\right)+\dfrac{13}{15}:\left(\dfrac{2}{5}-\dfrac{1}{9}\right)\)
\(=\dfrac{13}{15}:\dfrac{13}{35}+\dfrac{13}{15}:\dfrac{13}{45}\)
\(=\dfrac{7}{3}+3\)
\(=\dfrac{16}{3}\)
3 tháng 12 2021
\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)
\(\left(z-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)
\(\left(z-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(\Rightarrow z-\dfrac{2}{15}=\dfrac{2}{5}\)
\(z=\dfrac{2}{5}+\dfrac{2}{15}\)
\(z=\dfrac{8}{15}\)
Ta có: \(\left(z-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)
\(\Leftrightarrow\left(z-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(\Leftrightarrow z-\dfrac{2}{15}=\dfrac{2}{5}\)
\(\Leftrightarrow z=\dfrac{2}{5}+\dfrac{2}{15}=\dfrac{6}{15}+\dfrac{2}{15}\)
hay \(z=\dfrac{8}{15}\)
Vậy: \(z=\dfrac{8}{15}\)