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1.a) \(\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5=\frac{3^{15}}{5^{15}}.\frac{5^{10}}{3^{10}}=\frac{3^5}{5^5}=\left(\frac{3}{5}\right)^5\)
b)\(\left(\frac{2}{3}\right)^{10}:\left(\frac{4}{9}\right)^4=\frac{2^{10}}{3^{10}}.\frac{3^8}{2^8}=\frac{2^2}{3^2}=\left(\frac{2}{3}\right)^2\)
2.
a)\(2^x=4\Rightarrow2^x=2^2\Rightarrow x=2\)
b)\(x^3=-27\Rightarrow x^3=-3^3\Rightarrow x=-3\)
c)\(x^2=16\Rightarrow x=\pm4\)
d)\(\left(x+1\right)^2=9\Rightarrow\hept{\begin{cases}x+1=3\Rightarrow x=2\\x+1=-3\Rightarrow x=-4\end{cases}}\)
c) \(5x-7=3x+9\)
d) \(5x-\left|9-7x\right|=3\)
e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
h) \(5^{-1}.25^x=125\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
g) \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow6=1x\)
\(\Rightarrow x=6\left(TM\right).\)
Vậy \(x=6.\)
Chúc bạn học tốt!
a)
\(\left(\frac{x-1}{4}\right)^2=\frac{4}{9}\)
⇒ \(\frac{x-1}{4}=\frac{2}{3}\)
\(\Rightarrow\frac{\left(x-1\right).3}{12}=\frac{8}{12}\)
=> (x - 1) = 8/3
=> x = 8/3 - 1
=> x = 5/3
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
a. 6,5 -9/4:/x+1/3\=/-2\
6,5-9/4:/x+1/3\=2
9/4:/x+1/3\=6,5-2
9/4:/x+1/3\=4,5
/x+1/3\=9/4:4,5
/x+1/3\=1/2
x+1/3=1/2 hoặc x+1/3= -1/2
x= 1/2-1/3 x= -1/2-1/3
x= 1/6 x= -5/6
Vậy x=1/6 hoặcx= -5/6
b. 2-/3/2x-1/4\ = /-5/4\
2-/3/2x-1/4\=5/4
/3/2x-1/4\=2-5/4
/3/2x-1/4\=3/4
3/2x-1/4=3/4 hoặc 3/2x-1/4= -3/4
3/2x=3/4+1/4 3/2x= -3/4+1/4
3/2x=1 3/2x= -1/2
x=1:3/2 x= -1/2:3/2
x=2/3 x= -1/3
Vậy x=2/3 hoặc x= -1/3
a: =>2x+5=4
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left(3x-4\right)^2\cdot\left[\left(3x-4\right)^2-1\right]=0\)
=>(3x-4)(3x-5)(3x-3)=0
hay \(x\in\left\{1;\dfrac{4}{3};\dfrac{5}{3}\right\}\)
c: \(\Leftrightarrow3^{x+1}=3^{2x}\)
=>2x=x+1
=>x=1
d: \(\Leftrightarrow2^{2x+3}=2^{2x-10}\)
=>2x+3=2x-10
=>0x=-13(vô lý)
\(\left(\frac{1}{4}\right)^3\cdot4^3=\left(\frac{1}{4}\cdot4\right)^3=1^3=1\)
\(\frac{1000^4}{250^4}=4^4=256\)
\(2^2\cdot9\cdot\frac{1}{54}\cdot\left(\frac{4}{9}\right)^2=2^2\cdot3^2\cdot2\cdot3^3\cdot\left(\frac{4}{9}\right)^2=\left[\left(2\cdot3\cdot\frac{4}{9}\right)^2\right]\cdot2\cdot3^3=\frac{64}{9}\cdot2\cdot27=384\)
2. a) 2x = 9 => x không thỏa mãn
b) x2 = 9 => x = \(\pm\)3
c) (x + 1)2 = 4 => (x + 1)2 = \(\pm\)22
=> \(\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
Bài 1 :
\(a,\left(\frac{1}{4}\right)^3.4^3\)
\(=\frac{1}{4^3}.4^3\)
\(=1\)
\(b,\frac{1000^4}{250^4}=\frac{\left(250.4\right)^4}{250^4}=\frac{250^4.4^4}{250^4}=4^4=256\)
\(d,2^2.9.\frac{1}{54}.\left(\frac{4}{9}\right)^2\)
\(=36.\frac{1}{54}.\frac{4^2}{9^2}\)
\(=\frac{18.2.16}{18.3.81}\)
\(=\frac{32}{243}\)
Bài 2 :
\(a,2^x=9\)
\(\Rightarrow\)x không thỏa mãn
\(b,x^2=9\)
\(\Rightarrow x^2=3^2\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(c,\left(x+1\right)^2=4\)
\(\Rightarrow\left(x+1\right)^2=2^2\)
\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
Học tốt