(x-3)^4=16/81
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\(9^7.81:9^5=9^7.9^2:9^5=9^{7+2-5}=9^4\\ x^{12}:x.x^8=x^{12-1+8}=x^{19}\\ 16.2^4:8=2^4.2^4:2^3=2^{4+4-3}=2^5\\ 64.4^5:16=4^3.4^5:4^2=4^{3+5-2}=4^6\\ 3^{12}.3:3^8=3^{12+1-8}=3^5\\ 7^9.7^{12}:2015^0=7^{9+12}:1=7^{19}\)
\(2^x\cdot2=128\)
\(\Rightarrow2^x=128:2\)
\(\Rightarrow2^x=64\)
\(\Rightarrow2^x=2^6\)
\(\Rightarrow x=6\)
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\(4^x:4^3=16\)
\(\Rightarrow4^{x-3}=16\)
\(\Rightarrow4^{x-3}=4^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=3+2=5\)
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\(2^x:3=48\)
\(\Rightarrow2^x=48:3\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
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\(3^{x-5}=81\)
\(\Rightarrow3^{x-5}=3^4\)
\(\Rightarrow x-5=4\)
\(\Rightarrow x=4+5\)
\(\Rightarrow x=9\)
2ˣ⁻² = 128
2ˣ⁻² = 2⁷
x - 2 = 7
x = 7 + 2
x = 9
Các câu còn lại xem bài bạn Phong
`@` `\text {Ans}`
`\downarorw`
`2^x \div 2 = 128?`
`=> 2^x = 128 * 2`
`=> 2^x = 256`
`=> 2^x = 2^8`
`=> x = 8`
Vậy, `x = 8`
______
`4^x \div 4^3 = 16`
`=> 4^x = 16 * 4^3`
`=> 4^x = 4^2 * 4^3`
`=> 4^x = 4^5`
`=> x = 5`
Vậy, ` x = 5`
________
`12^x \div 3 = 48 ?`
`=> 12^x = 48 * 3`
`=> 12^x = 144`
`=> 12^x = 12^2`
`=> x = 2`
Vậy, `x = 2`
________
`3^x - 5 = 81`
`3^x = 81 + 5`
`3^x = 86`
Bạn xem lại đề
`@` `\text {Kaizuu lv uuu}`
Ai biết trả lời nhanh giúp mình nha mình đang cần luôn rất rất rất rất gấp.
Ta có:
(x - 3)4 = \(\frac{16}{81}\)
=> (x - 3)4 = \(\left(\frac{2}{3}\right)^4\)
=> \(\left[\begin{matrix}x-3=\frac{2}{3}\\x-3=\frac{-2}{3}\end{matrix}\right.\)
=> \(\left[\begin{matrix}x=\frac{2}{3}+3\\x=\frac{-2}{3}+3\end{matrix}\right.\)
=> \(\left[\begin{matrix}x=\frac{11}{3}\\x=\frac{7}{3}\end{matrix}\right.\)
Vậy \(\left[\begin{matrix}x=\frac{11}{3}\\x=\frac{7}{3}\end{matrix}\right.\).
a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)
\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)
\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)
\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)
b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)
\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)
\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)
\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)
\(\Leftrightarrow x=\frac{1}{-3}\)
c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)