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VN
Võ Ngọc Phương
VIP
1 tháng 10 2023
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
1 tháng 10 2023
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
AD
2
AD
22 tháng 7 2023
\(1,\left[\left(-\dfrac{2}{5}\right)+\dfrac{1}{3}\right]-\left(\dfrac{3}{5}-\dfrac{1}{3}\right)=-\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{1}{3}\\ =\left(-\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{1}{3}+\dfrac{1}{3}\right)\\ =\dfrac{-2-3}{5}+\dfrac{1+1}{3}\\ =-\dfrac{5}{5}+\dfrac{2}{3}\\ =-1+\dfrac{2}{3}\\ =\dfrac{-3+2}{3}=-\dfrac{1}{3}\\ b,\left(\dfrac{3}{2}-\dfrac{3}{4}\right)-\left(0,25+\dfrac{1}{2}\right)\\ =\left(\dfrac{3}{2}-\dfrac{3}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{2}\right)\\ =\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}-\dfrac{1}{2}\\ =\left(\dfrac{3}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{3}{4}-\dfrac{1}{4}\right)\\ =\dfrac{3-1}{2}+\dfrac{-3-1}{4}\\ =\dfrac{2}{2}-\dfrac{4}{4}=1-1=0\)
22 tháng 7 2023
1: =-2/5+1/3-3/5+1/3
=-1+2/3=-1/3
2: =3/2-3/4-1/4-1/2
=1-1=0
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
CC
1
5 tháng 7 2023
1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
H
5 tháng 11 2023
\(\sqrt{0,36}-\left|-0,75\right|:\left(-1\dfrac{3}{4}\right)\\ =0,6-0,75:\dfrac{-7}{4}\\ =\dfrac{3}{5}-\dfrac{3}{4}\cdot\dfrac{-4}{7}\\ =\dfrac{3}{5}-\dfrac{-12}{28}\\ =\dfrac{3}{5}+\dfrac{3}{7}\\ =\dfrac{36}{35}\)
\(1-\left(\dfrac{5}{9}-\dfrac{2}{3}\right)^2:\dfrac{4}{27}\\ =1-\left(\dfrac{5}{9}-\dfrac{6}{9}\right)^2\cdot\dfrac{27}{4}\\ =1-\left(-\dfrac{1}{9}\right)^2\cdot\dfrac{27}{4}\\ =1-\dfrac{1}{81}\cdot\dfrac{27}{4}\\ =1-\dfrac{1}{12}\\ =\dfrac{11}{12}\)
18 tháng 12 2021
Câu 1:
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)
18 tháng 12 2021
\(1,\Leftrightarrow\left(x+\dfrac{1}{3}\right)^2=\dfrac{4}{9}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\\ 2,=15:\left(\dfrac{2}{3}\right)^4\cdot\left(\dfrac{2}{3}\right)^6:\left(\dfrac{2}{3}\right)^9=15\cdot\left(\dfrac{2}{3}\right)^{-7}=15\cdot\dfrac{3^7}{2^7}=15\cdot\dfrac{2187}{128}=\dfrac{32805}{128}\)
= \(\dfrac{19}{27}\)