7.4 mũ x-1 + 4 mũ x-1 = 23
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\(7\cdot4^x=112\)
\(\Rightarrow4^x=\dfrac{112}{7}\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
_____
\(2\cdot5^{x-3}=250\)
\(\Rightarrow5^{x-3}=\dfrac{250}{2}\)
\(\Rightarrow5^{x-3}=125\)
\(\Rightarrow5^{x-3}=5^3\)
\(\Rightarrow x-3=3\)
\(\Rightarrow x=3+3\)
\(\Rightarrow x=6\)
____
\(12:\left\{400:\left[500-\left(5^3+5^2\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+25\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+175\right)\right]\right\}+10\)
\(=12:\left[400:\left(500-300\right)\right]+10\)
\(=12:\left(400:200\right)+10\)
\(=12:2+10\)
\(=6+10\)
\(=16\)
\(7\cdot4^x=112\)
\(\Rightarrow4^x=112:7\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
\(---\)
\(2\cdot5^{x-3}=250\)
\(\Rightarrow5^{x-3}=250:2\)
\(\Rightarrow5^{x-3}=125\)
\(\Rightarrow5^{x-3}=5^3\)
\(\Rightarrow x-3=3\)
\(\Rightarrow x=3+3\)
\(\Rightarrow x=6\)
Vậy \(x=6.\)
\(---\)
\(12:\left\{400:\left[500-\left(5^3+5^2\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+175\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-300\right]\right\}+10\)
\(=12:\left\{400:200\right\}+10\)
\(=12:2+10\)
\(=6+10\)
\(=16\)
#\(Toru\)
Vẽ đồ thị hàm số:
1, y = 1/4x mũ 2
2, y = -1/4 x mũ 2
3, y = -2 x mũ 2
4, y = -1/2 x mũ 2
5, y = 3 x mũ 2


a: \(2^x=2^3\)
nên x=3
c: \(11^x=1331\)
nên x=3
d: \(2^x+4=12\)
nên \(2^x=8\)
hay x=3

\(23+3x=5^6:5^3\) \(9^{x-1}=9\) \(x^4=16\) \(2^x:2^5=1\)
\(23+3x=5^2\) => x-1= 0 \(x^4=2^4\) \(2^x=1.2^5\)
\(23+3x=25\) x= 0+1 => x= 2 \(2^x=2^5\)
3x= 25-23 x= 1 => x= 5
3x= 2
x= \(\frac{2}{3}\)

\(-22x^3-\left(-21x^3+19x^2+23^0\right)-\left(-x^3-18x^2\right)+\left(x^2-23^1\right)\)
\(=-22x^3+21x^3-19x^2-1+x^3+18x^2+x^2-23\)
\(=\left(-22x^3+21x^3+x^3\right)+\left(-19x^2+18x^2+x^2\right)+\left(-1-23\right)\)
\(=0x^3+0x^2-24\)
\(=-24\)
Vậy biểu thức trên có giá trị không phụ thuộc vào biến.
\(7\cdot4^{x-1}+4^{x-1}=23\)
=>\(8\cdot4^{x-1}=23\)
=>\(2^3\cdot2^{2x-2}=23\)
=>\(2^{2x-2+3}=23\)
=>\(2^{2x+1}=23\)
=>\(2x+1=\log_223\)
=>\(2x=\log_223-1\)
=>\(x=\frac12\left(\log_223-1\right)\)