9:
a: x*(-4/7)^2=(-4/7)^5
=>x=(-4/7)^5:(-4/7)^2=(-4/7)^3
b: x:(-3/4)=(-3/4)^2
=>x=(-3/4)^2*(-3/4)=(-3/4)^3
c:x^4=256
=>x^4=4^4
=>x=4
d: =>x^5=(-2/3)^5
=>x=-2/3
e; =>(-2/3)^x=(-2/3)^5
=>x=5
g: =>x+1=3
=>x=2
9,
a) \(x\cdot\left(-\dfrac{4}{7}\right)^2=\left(-\dfrac{4}{7}\right)^5\)
\(\Rightarrow x=\left(-\dfrac{4}{7}\right)^5:\left(-\dfrac{4}{7}\right)^2\)
\(\Rightarrow x=\left(-\dfrac{4}{7}\right)^3\)
b) \(x:-\dfrac{3}{4}=\left(-\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)
\(x=\left(-\dfrac{3}{4}\right)^3\)
c) \(x^4=256\)
\(\Rightarrow x^4=4^4\)
\(\Rightarrow x=4\)
d) \(x^5=\dfrac{-32}{243}\)
\(\Rightarrow x^5=\left(-\dfrac{2}{3}\right)^5\)
\(\Rightarrow x=-\dfrac{2}{3}\)
e) \(\left(-\dfrac{2}{3}\right)^x=\dfrac{-32}{243}\)
\(\Rightarrow\left(-\dfrac{2}{3}\right)^x=\left(-\dfrac{2}{3}\right)^5\)
\(\Rightarrow x=5\)
g) \(\left(\dfrac{5}{7}\right)^{x+1}=\dfrac{125}{343}\)
\(\Rightarrow\left(\dfrac{5}{7}\right)^{x+1}=\left(\dfrac{5}{7}\right)^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
10,
a) \(\left[\left(-\dfrac{3}{5}\right)^3\right]^2=\left(-\dfrac{3}{5}\right)^6=\left(\dfrac{3}{2}\right)^6=\dfrac{729}{64}\)
\(\left[\left(-\dfrac{2}{5}\right)^3\right]^2=\left(-\dfrac{2}{5}\right)^6=\left(\dfrac{2}{5}\right)^6=\dfrac{64}{15625}\)
Bài 10:
\(a,\left[\left(-\dfrac{3}{5}\right)^3\right]^2=\left(-\dfrac{3}{5}\right)^{3.2}=\left(-\dfrac{3}{5}\right)^6=\dfrac{\left(-3\right)^6}{5^6}=\dfrac{729}{15625}\\ \left[\left(-\dfrac{2}{5}\right)^3\right]^2=\left(-\dfrac{2}{5}\right)^{3.2}=\left(-\dfrac{2}{5}\right)^6=\dfrac{\left(-2\right)^6}{5^6}=\dfrac{64}{15625}\)
\(b,\left(0,16\right)^2=\left[\left(0,4\right)^2\right]^2=\left(0,4\right)^4\\ \left(0,064\right)^3=\left[\left(0,4\right)^3\right]^3=\left(0,4\right)^9\)
`@` `\text {Ans}`
`\downarrow`
`9,`
`a)`
\(x\cdot\left(-\dfrac{4}{7}\right)^2=\left(-\dfrac{4}{7}\right)^5\)
`=> x = (-4/7)^5 \div (-4/7)^2`
`=> x = (-4/7)^3`
Vậy, `x= (-4/7)^3`
`b)`
\(x\div\dfrac{-3}{4}=\left(-\dfrac{3}{4}\right)^2\)
`=> x = (-3/4)^2*(-3/4)`
`=> x = (-3/4)^3`
`c)`
\(x^4=256\)
`=> x^4 = (+-4)^4`
`=> x = +-4`
Vậy, `x \in {4; -4}`
`d)`
\(x^5=\dfrac{-32}{243}\)
`=> x^5 = (-2/3)^5`
`=> x = -2/3`
Vậy, `x = -2/3`
`e)`
\(\left(\dfrac{-2}{3}\right)^x=\dfrac{-32}{243}\)
`=> (-2/3)^x = (-2/3)^5`
`=> x = 5`
Vậy, `x = 5`
`g)`
\(\left(\dfrac{5}{7}\right)^{x+1}=\dfrac{125}{343}\)
`=> (5/7)^(x+1) = (5/7)^3`
`=> x+1=3`
`=> x = 3 - 1`
`=> x=2`
Vậy, `x=2`
`10,`
`a)`
\(\left[\left(-\dfrac{3}{5}\right)^3\right]^2=\left(-\dfrac{3}{5}\right)^{3\cdot2}=\left(-\dfrac{3}{5}\right)^6\)
\(\left[\left(-\dfrac{2}{5}\right)^3\right]^2=\left(-\dfrac{2}{5}\right)^{3\cdot2}=\left(-\dfrac{2}{5}\right)^6\)
`@` `\text {Kaizuu lv uuu}`
\(9,\\ a,x\cdot\left(-\dfrac{4}{7}\right)^2=\left(-\dfrac{4}{7}\right)^5\\ =>x=\left(-\dfrac{4}{7}\right)^5:\left(-\dfrac{4}{7}\right)^2\\ =>x=\left(-\dfrac{4}{7}\right)^{5-2}\\ =>x=\left(-\dfrac{4}{7}\right)^3\)
\(b,x:\dfrac{-3}{4}=\left(-\dfrac{3}{4}\right)^2\\ =>x=\left(-\dfrac{3}{4}\right)^2\cdot\dfrac{-3}{4}\\ =>x=\left(-\dfrac{3}{4}\right)^{2+1}\\ =>x=\left(-\dfrac{3}{4}\right)^3\)
\(c,x^4=256\\ =>x^4=4^4\\ =>x=4\)
\(d,x^5=-\dfrac{32}{243}\\ =>x^5=\left(-\dfrac{2}{3}\right)^5\\ =>x=-\dfrac{2}{3}\)
\(e.\left(-\dfrac{2}{3}\right)^x=-\dfrac{32}{243}\\ =>\left(-\dfrac{2}{3}\right)^x=\left(-\dfrac{2}{3}\right)^5\\ =>x=5\)
\(g,\left(\dfrac{5}{7}\right)^{x+1}=\dfrac{125}{343}\\ =>\left(\dfrac{5}{7}\right)^x.\left(\dfrac{5}{7}\right)^1=\left(\dfrac{5}{7}\right)^3\\ =>\left(\dfrac{5}{7}\right)^x=\left(\dfrac{5}{7}\right)^3:\dfrac{5}{7}\\ =>\left(\dfrac{5}{7}\right)^x=\left(\dfrac{5}{7}\right)^2\\ =>x=2\)
Bài 9:
\(a,x.\left(-\dfrac{4}{7}\right)^2=\left(-\dfrac{4}{7}\right)^3\\ \Leftrightarrow x=\left(-\dfrac{4}{7}\right)^3:\left(-\dfrac{4}{7}\right)^2=\left(-\dfrac{4}{7}\right)^{3-2}\\ \Leftrightarrow x=-\dfrac{4}{7}\\ b,x:\dfrac{-3}{4}=\left(-\dfrac{3}{4}\right)^2\\ \Leftrightarrow x=\left(-\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^{2-1}\\ \Leftrightarrow x=\dfrac{-3}{4}\\ c,x^4=256\\ \Leftrightarrow\left(x^2\right)^2=256\\\sqrt{x^4}=x^2\sqrt{256}=16\\ \Leftrightarrow\left|x\right|=\sqrt{x^2}=\sqrt{16}=4\\ Vậy:x=4.hoặc.x=-4\)
\(Bài.9:\\ d,x^5=\dfrac{-32}{243}\\ \Leftrightarrow x=\sqrt[5]{-\dfrac{32}{243}}\\ \Leftrightarrow x=\sqrt[5]{\left(-\dfrac{2}{3}\right)^5}\\ \Leftrightarrow x=-\dfrac{2}{3}\\ e,\left(-\dfrac{2}{3}\right)^x=\dfrac{-32}{243}=\left(-\dfrac{2}{3}\right)^5\\ \Leftrightarrow x=5\\ g,\left(\dfrac{5}{7}\right)^{x+1}=\dfrac{125}{343}=\left(\dfrac{5}{7}\right)^3\\ \Rightarrow x+1=3\\ \Leftrightarrow x=2\)
