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\(9^9*9^8*9^7*9^6*9^5*9^4*9^3*9^2*9^1*9^0=\)99+8+7+6+5+4+3+2+1+0 = 945
\(\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{2005}{9}\right)\)
\(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{9}{9}\right)...\left(1-\dfrac{2005}{9}\right)\)
\(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{2}{9}\right)\cdot...\cdot0\cdot...\cdot\left(1-\dfrac{2005}{9}\right)=0\)
\(\left(1-\dfrac{1}{9}\right).\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{2005}{9}\right)\)
\(=\left(1-\dfrac{1}{9}\right).\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{9}{9}\right)...\left(1-\dfrac{2005}{9}\right)\)
\(=\left(1-\dfrac{1}{9}\right).\left(1-\dfrac{2}{9}\right)...0...\left(1-\dfrac{2005}{9}\right)=0\)
\(I=\left(1-\frac{1}{9}\right)\left(1-\frac{2}{9}\right)\left(1-\frac{3}{9}\right)...\left(1-\frac{2005}{9}\right)\)
\(I=\left(1-\frac{1}{9}\right)\left(1-\frac{2}{9}\right)...\left(1-\frac{9}{9}\right)...\left(1-\frac{2005}{9}\right)\)
\(I=\left(1-\frac{1}{9}\right)\left(1-\frac{2}{9}\right)...\left(1-1\right)...\left(1-\frac{2005}{9}\right)\)
\(I=\left(1-\frac{1}{9}\right)\left(1-\frac{2}{9}\right)...0...\left(1-\frac{2005}{9}\right)\)
I = 0
=> I = 0
23.3\(x\) = 9
3\(^x\) = 9 : 23
3\(^x\) \(\approx\)0,39
Lớp 6 chưa học logarit nên em xem lại đề bài
Còn lên cấp ba thì giải tiếp như sau:
3\(^x\) \(\approx\) 0,39
\(x\) \(\approx\) \(log_30,39\) \(\approx\) -0,875
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