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a, \(\left(3x^2-5\right)+3^4+6^0=5^3\)
\(\left(3x^2-5\right)+81+1=125\)
\(3x^2-5=43\)
\(3x^2=48\)
\(x^2=16\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=4\end{cases}}\)
Vậy ...
b,\(3x+2x\left(2^3.5-3^2.4\right)+5^2=4^4\)
\(3x+2x\left(8.5-9.4\right)+25=256\)
\(3x+2x.9=231\)
\(21x=231\)
x=11
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!
\(16^x< 128^4\)
=> \(\left[2^4\right]^x< \left[2^7\right]^4\)
=> \(2^{4x}< 2^{28}\)
=> 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
\(1,2x+3x-4x=\left(-2\right)^3\)
<=>\(x=-8\)
\(2,x-2x=4^2+4^0\)
<=>\(-x=16+1\)
<=>\(-x=17\)
<=>\(x=-17\)
\(3,2^3x-3^2x=|12-21|\)
<=>\(-x=9\)
<=>\(x=-9\)
\(4,x-45=2x+54\)
<=>\(x-2x=54+45\)
<=>\(-x=99\)
<=>\(x=-99\)
\(5,5x-12+23=6^7:6^5\)
<=>\(5x+11=6^2\)
<=>\(5x+11=36\)
<=>\(5x=25\)
<=>\(x=5\)
a,71.2-6.(2x+5)=10^5:10^3
142-6.(2x+5)=10^2
142-6.(2x+5)=100
6.(2x+5)=142-100
6.(2x+5)=42
2x+5=42:6
2x+5=7
2x=7-5
2x=2
x=1
Vậy x=1
a)3x-1=100+10 b)5^(2x-3)-50=75 c)(3x-16).7^5=2.7^6 d)(x-1)^5+3^2.5=72
3x=111 5^2x-3=75+50 3x-16=7^6:7^5.2 (x-1)^2=72-9.5
x=111:3 5^2x-3=125=5^3 3x-16=14 (x-1)^2=27
x=37 =>2x-3=3 3x=16+14 làm thành 2 trường hợp
2x=6 3x=30 .. .....
x=3 x=10