1 , 1 , 2 , 3 , 4 , 9 , 8 , 27 , ... , ...
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\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) 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
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
a: =2/3+2/7=14/21+6/21=20/21
b: =17/18-33/18=-16/18=-8/9
c: =7/12-3/14=49/84-18/84=31/84
d: =-1/4-1/8=-3/8
a) 1/2 + 3/4 - (3/4 - 4 - 5)
= 1/2 + 3/4 - 3/4 + 4 + 5
= (3/4 - 3/4) + (4 + 5) + 1/2
= 0 + 9 + 1/2
= 19/2
b) [9/16 + 8/(-27)] - (19/27- 7/16 - 2)
= 9/16 - 8/27 - 19/27 + 7/16 + 2
= (9/16 + 7/16) + (-8/27 - 19/27) + 2
= 1 - 1 + 2
= 2
c) -5/8 . [4/9 + 7/(-12)]
= -5/8 . (-5/36)
= 25/288
d) 7/10 . (-3/5) + 7/10 . (-2/5) - (-3/10)
= 7/10 . (-3/5 - 2/5) + 3/10
= 7/10 . (-1) + 3/10
= -2/5
e) -3/7 . 5/9 + 4/9 . (-3/7) + 2 3/7
= -3/7 . (5/9 + 4/9) + 17/7
= -3/7 . 1 + 17/7
= 2
f) 8 2/7 - (3 4/9 + 4 2/7)
= 8 + 2/7 - 3 - 4/9 - 4 - 2/7
= (8 - 3 - 4) + (2/7 - 2/7) - 4/9
= 1 - 4/9
= 5/9
h) 3.(-1/2)² - (4/5 + 8/15) : 5/6
= 3.1/4 - 4/3 : 5/6
= 3/4 - 8/5
= -17/20
\(a,\left(-\dfrac{4}{9}\right)\times\dfrac{3}{8}+\dfrac{1}{18}=-\dfrac{1}{6}+\dfrac{1}{18}=\dfrac{-1.3}{6.3}+\dfrac{1}{18}=\dfrac{-3+1}{18}=-\dfrac{2}{18}=-\dfrac{1}{9}\)
xem lại đề câu b
\(c,\left(\dfrac{1}{2}-\dfrac{2}{3}\right)\times\dfrac{4}{4}=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3-4}{6}=-\dfrac{1}{6}\\ d,\dfrac{9}{7}\times\left(\dfrac{3}{7}-\dfrac{1}{2}\right)=\dfrac{9}{7}\times\left(\dfrac{3.2-7}{14}\right)=\dfrac{9}{7}\times\dfrac{-1}{14}=-\dfrac{9}{98}\)
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
Câu 1:
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)
\(1,\Leftrightarrow\left(x+\dfrac{1}{3}\right)^2=\dfrac{4}{9}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\\ 2,=15:\left(\dfrac{2}{3}\right)^4\cdot\left(\dfrac{2}{3}\right)^6:\left(\dfrac{2}{3}\right)^9=15\cdot\left(\dfrac{2}{3}\right)^{-7}=15\cdot\dfrac{3^7}{2^7}=15\cdot\dfrac{2187}{128}=\dfrac{32805}{128}\)
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