Tìm x biết :
a, \(\left(2.x-7\right)^5=243\)
Tìm số tự nhiên x , biết
\(2\cdot\left(x-1\right)^2=8\)
\(\left(2x+1\right)^3=125\)
\(\left(x-2\right)^5=243\)
\(5\left(x-4\right)^2-7=13\)
\(221-\left(3x+2\right)^3=96\)
Tìm x ∈ N biết :
a) \(8< 2^x\le2^9.2^{-5}\)
b)\(27< 81^3:3^x< 243\)
c)\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-3}{5}\right)^2\)
\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
Tìm \(x\) biết:
\(\left(\sqrt{3}\right)^x=243\)
\(0,1^x=1000\)
\(\left(\dfrac{1}{2}\right)^x=1024\)
\(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
\(5^{x-1}+5^{x+2}=3\)
a: \(\left(\sqrt{3}\right)^x=243\)
=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)
=>\(\dfrac{1}{2}\cdot x=5\)
=>x=10
b: \(0,1^x=1000\)
=>\(\left(\dfrac{1}{10}\right)^x=1000\)
=>\(10^{-x}=10^3\)
=>-x=3
=>x=-3
c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
=>\(\left(0,2\right)^{x+3}< 0,2\)
=>x+3>1
=>x>-2
d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)
=>2x+1<-2
=>2x<-3
=>\(x< -\dfrac{3}{2}\)
e: \(5^{x-1}+5^{x+2}=3\)
=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)
=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)
=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)
bài 1: Tìm x,y biết rằng:
\(x+(-\frac{31}{12})^2=\left(\frac{49}{12}\right)^2-x=y^2\)
bài 2: tìm x biết:
a.\(5^x.\left(5^3\right)^2=625\) b.\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\) c.\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
d.\(172x^2-7^9:98^3=2^{-3}\)
Bài 3: Tìm x \(\varepsilon\)N biết:
a.\(8< 2^x\le2^9\times2^{-5}\) b.\(27< 81^3:3^x< 243\) \(\left(\frac{2}{5}\right)^x>\left(\frac{5}{2}\right)^{-3}\times\left(-\frac{2}{5}\right)^2\)c.
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
\(\text{a) 3^{2x} * 9^{2x}=3^{12} b) 7^{3x}= 7^5}:7^3c25x^2=25d\left(x-\frac{\left(2\right)}{3}\right)^5=\frac{1}{243}\)
Tìm x \(\in\) N biết :
a)\(8< 2^x\le2^9.2^{-5}\)
b) \(27< 81^3:3^x< 243\)
c)\(\left(\frac{2}{5}\right)^x>\left(\frac{5}{2}\right)^{-3}.\left(\frac{-2}{5}\right)^2\)
a,\(8< 2^x\le2^9.2^{-5}\)
\(2^3< 2^x\le2^4\)
\(\Rightarrow x=4\)
b, \(27< 81^3.3^x< 243\)
\(3^3< 3^{12-x}< 3^5\)
\(\Rightarrow3< 12-x< 5\)
12-x=4
x=8
c,\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^3.\left(\frac{2}{5}\right)^2\)
\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^5\)
\(\Rightarrow x>5\)
x=6;7;8........
CÁC BÀI TẬP VỀ LŨY THỪA SỐ HỮU TỈ
Tìm số hữu tỉ x biết:
a, \(8< 2^x< \frac{2^9}{2^5}\)
b, \(27< 81^3:3^x< 243\)
c, \(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^3.\left(-\frac{2}{5}\right)^2\) CÁC BN NHỚ GIẢI THEO CÁCH CỦA LỚP 7 NHÉ!!! =^.^=
tìm x biết \(\left(x-1\right)^5=-243\)
(x-1)5= -243 (x-1)5=-35 => x-1=-3
x= -3+1
x=-2 k cho mik nha
mik co het suc viet do
Tìm x, biết :
a) \(\left(x-\frac{1}{5}\right)^5\)=\(\frac{1}{243}\)
b) | 2x - 1 | - x = 1
c)\(\left|\frac{3}{5}-\frac{1}{2}x\right|\)>\(\frac{2}{5}\)
a.
\(\left(x-\frac{1}{5}\right)^5=\frac{1}{243}\)
\(x-\frac{1}{5}=\sqrt[5]{\frac{1}{243}}\)
\(x-\frac{1}{5}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{5}\)
\(x=\frac{8}{15}\)
b.
|2x-1|-x=1
\(\Leftrightarrow\orbr{\begin{cases}2x-1-x=1\\-2x+1-x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
Vậy x= 0 hoặc x=2
c. \(\left|\frac{3}{5}-\frac{1}{2}x\right|>\frac{2}{5}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{5}-\frac{1}{2}x>\frac{2}{5}\\-\frac{3}{5}+\frac{1}{2}x>\frac{2}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x< \frac{1}{5}\\\frac{1}{2}x>\frac{-1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x< \frac{2}{5}\\x>\frac{-2}{5}\end{cases}}\)
Vậy....
Bài giải
a, \(\left(x-\frac{1}{5}\right)^5=\frac{1}{243}\)
\(\left(x-\frac{1}{5}\right)^5=\left(\frac{1}{2}\right)^5\)
\(x-\frac{1}{5}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{5}\)
\(x=\frac{7}{10}\)