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1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4
a) 2^x.2^4=128
=>2^x.2^2=2^7
=>2^x=2^7:2^2
=>2^x=2^5
=>x=5
b)x^15=x
=>x^15-x=0
=>x(x^16-x)=0
=>2 trượng hợp:x=0 và x^16-1=0(x^16-1=0 cx 2 th nha)
b),d),e) như nhau nha!
c) dễ rồi
\(a)2^x\cdot4=128\)
\(\Rightarrow2^x=\frac{128}{4}\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(b)x^{15}=x\)
\(\Rightarrow x^{15}-x=0\)
\(\Rightarrow x(x^{14}-1)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x^{14}=1\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=1\end{cases}}\)
\(c)(2x+1)^3=125\)
\(\Rightarrow(2x+1)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2=2\)
\(d)(x-5)^4=(x-5)^6\)
\(\Rightarrow(x-5)^6-(x-5)^4=0\)
\(\Rightarrow(x-5)^4\cdot\left[(x-5)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}(x-5)^4=0\\(x-5)^2-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=6\end{cases}}\)
\(e)(2x-15)^5=(2x-15)^3\)
\(\Rightarrow(2x-15)^5-(2x-15)^3=0\)
\(\Rightarrow(2x-15)^3-\left[(2x-15)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}(2x-15)^3=0\\(2x-15)^2-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\varnothing\\x=8\end{cases}}\)
Chúc bạn hoc tốt :>
\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)
<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)
<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x>\frac{1}{3}\\x>\frac{-2}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}x< \frac{1}{3}\\x< \frac{-2}{5}\end{cases}}\)
<=>\(x>\frac{1}{3}\)hoặc \(x< \frac{-2}{5}\)
câu c tương tự nha
học tốt
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
`a)(x+5)^3=-64`
`(x+5)^3=(-4)^3`
`x+5=-4`
`x=-4-5=-9`
Vậy `x=-9`
`2)(2x-3)^3=8`(9 không được)
`(2x-3)^3=2^3`
`2x-3=2`
`2x=5`
`x=5/2`
Vậy `x=5/2`
a)\(\left(x+5\right)^3=64\)
\(\left(x+5\right)^3=4^3\)
\(x+5=4\)
\(x=4-5\)
\(x=-1\)
b) \(\left(2x-3\right)^3=9\)
\(\left(2x-3\right)^3=3^3\)
\(2x-3=3\)
\(2x=3+3\)
\(2x=6\)
\(x=\dfrac{6}{2}\)
\(x=3\)