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\(\left(\frac{1}{2}\right)^x=\frac{1}{64}\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^6\)
=> x=6
\(\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\dfrac{36}{49}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\left(\dfrac{6}{7}\right)^2\\ \Rightarrow\dfrac{1}{2}-\dfrac{x}{3}=\pm\dfrac{6}{7}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-\dfrac{x}{3}=\dfrac{6}{7}\\\dfrac{1}{2}-\dfrac{x}{3}=-\dfrac{6}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{5}{14}\\\dfrac{x}{3}=\dfrac{19}{14}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{14}\times3\\x=\dfrac{19}{14}\times3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{14}\\x=\dfrac{57}{14}\end{matrix}\right.\)
\(\left(3-\dfrac{2}{3}x\right)^3=-\dfrac{1}{64}\\ \Rightarrow\left(3-\dfrac{2}{3}x\right)^3=\left(-\dfrac{1}{4}\right)^3\\ \Rightarrow3-\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=3-\left(-\dfrac{1}{4}\right)\\ \Rightarrow\dfrac{2}{3}x=\dfrac{13}{4}\\ \Rightarrow x=\dfrac{13}{4}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{13}{4}\times\dfrac{3}{2}\\ \Rightarrow x=\dfrac{39}{8}\)
Hic 2 câu em làm dr xong tự nhiên thử lung tung rồi lại xóa bài ;-;
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
( 2x - 5 )^5 - 64 = 960
( 2x - 5 )^5 = 1024
( 2x - 5 )^5 = 4^5 = (-4)^5
(+) Th1: 2x - 5 = 4 (+) Th2: 2x - 5 = -4
2x = 9 2x = 1
x = 4,5 x = 0,5
Vậy x E { 4,5 ; 0,5 }
\(\left(2x-5\right)^5-64=960\)
\(\Rightarrow\left(2x-5\right)^5=960+64\)
\(\Rightarrow\left(2x-5\right)^5=1024\)
\(\Rightarrow\left(2x-5\right)^5=4^5\)
\(\Rightarrow2x-5=4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(\left(x-7\right)^2=64\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^2=8^2\\\left(x-7\right)^2=\left(-8\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=8\\x-7=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=15\\x=-1\end{cases}}\)
Vậy ...
câu 1; \(-154+\left(x-9-18\right)=40\)
\(\Leftrightarrow-154+x-9-18=40\)
\(\Leftrightarrow x=40+154+9+18\)
\(\Leftrightarrow x=221\)
Câu 2: \(\left|9-x\right|=64+\left(-7\right)\)
\(\Leftrightarrow\left|9-x\right|=57\)
\(\Leftrightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=9-57\\x=-57-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=-48\\x=-66\end{cases}}}\)
Vậy...
hok tốt!!
a: \(99_{10}=1100011_2\)
b: \(64_{10}=1000000_2\)
c: \(218_{10}=11011010_2\)
d: \(255_{10}=11111111_2\)
218 - 2| x + 1 | = 64
=> 2| x + 1 | = 154
=> | x + 1 | = 77
=> x + 1 = 77 hoặc x + 1 = -77
=> x = 76 hoặc x = -78