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Giải:
a) \(\left(3x^2-5\right)+3^4+6^0=5^3\)
\(\Leftrightarrow3x^2-5+3^4+6^0=5^3\)
\(\Leftrightarrow3x^2-5+81+1=125\)
\(\Leftrightarrow3x^2=125+5-81-1\)
\(\Leftrightarrow3x^2=48\)
\(\Leftrightarrow x^2=\dfrac{48}{3}=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy ...
b) \(3x+2x\left(2^3.5-3^2.4\right)+5^2=4^4\)
\(\Leftrightarrow3x+2x\left(8.5-9.4\right)+25=256\)
\(\Leftrightarrow3x+2x\left(40-36\right)+25=256\)
\(\Leftrightarrow3x+2x.4+25=256\)
\(\Leftrightarrow3x+8x+25=256\)
\(\Leftrightarrow11x+25=256\)
\(\Leftrightarrow11x=256-25=231\)
\(\Leftrightarrow x=\dfrac{231}{11}\)
\(\Leftrightarrow x=21\)
Vậy ...
Chúc bạn học tốt!
Bài làm :
\(1\text{)}...\Leftrightarrow\orbr{\begin{cases}\left(1+3x\right)^4=4^4\Leftrightarrow1+3x=4\Leftrightarrow3x=3\Leftrightarrow x=1\\\left(1+3x\right)^4=\left(-4\right)^4\Leftrightarrow1+3x=-4\Leftrightarrow3x=-5\Leftrightarrow x=-\frac{5}{3}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-\frac{5}{3}\end{cases}}\)
\(2\text{)}...\Leftrightarrow720\div\left(x-140\right)=-5\Leftrightarrow x-140=-144\Leftrightarrow x=-4\)
\(3\text{)}...\Leftrightarrow\left(2^4\right)^x.\left(2^2\right)^3=2^9\Leftrightarrow2^{4x}.2^6=2^9\Leftrightarrow2^{4x}=2^3\Leftrightarrow4x=3\Leftrightarrow x=\frac{3}{4}\)
\(4\text{)}...\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^4=3^4\Leftrightarrow2x-1=3\Leftrightarrow2x=4\Leftrightarrow x=2\\\left(2x-1\right)^4=\left(-3\right)^4\Leftrightarrow2x-1=-3\Leftrightarrow2x=-2\Leftrightarrow x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
a) \(3^{2x-1}=243\)
\(\Leftrightarrow3^{2x-1}=3^5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=5+1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
b) \(\left(3^x\right)^2:3^3=\dfrac{1}{243}\)
\(\Leftrightarrow3^{2x}:3^3=\dfrac{1}{3^5}\)
\(\Leftrightarrow3^{2x}:3^3=3^{-5}\)
\(\Leftrightarrow3^{2x-3}=3^{-5}\)
\(\Leftrightarrow2x-3=-5\)
\(\Leftrightarrow2x=-5+3\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{2}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
c) \(2^{3x+2}=4^{x+5}\)
\(\Leftrightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\)
\(\Leftrightarrow3x+2=2\left(x+5\right)\)
\(\Leftrightarrow3x+2=2x+10\)
\(\Leftrightarrow3x-2x=10-2\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
d) \(3^{x+1}=9^x\)
\(\Leftrightarrow3^{x+1}=\left(3^2\right)^x\)
\(\Leftrightarrow3^{x+1}=3^{2x}\)
\(\Leftrightarrow x+1=2x\)
\(\Leftrightarrow2x-x=1\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
\(16^x< 128^4\)
=> \(\left[2^4\right]^x< \left[2^7\right]^4\)
=> \(2^{4x}< 2^{28}\)
=> 4x < 28
=> x < 7
Đến đây tìm x được rồi
\(\left[3x^2-5\right]+3^4+6^0=5^3\)
=> \(\left[3x^2-5\right]=5^3-6^0-3^4=43\)
=> \(3x^2-5=43\)
=> \(3x^2=48\)
=> \(x^2=16\)
=> \(x=\pm4\)
\(3x+2x\left[2^3\cdot5-3^2\cdot4\right]+5^2=4^4\)
=> \(3x+2x\left[8\cdot5-9\cdot4\right]+25=256\)
=> \(3x+2x\cdot4+25=256\)
=> \(3x+2x\cdot4=231\)
Đến đây tìm x