Question: d) 6^15\cdot 9^{10} \over 3^{34}\cdot 2^{13}
e
e) 9^2\cdot 2^{11} \over 16^2\cdot 6^3
g) 27^4\cdot 4^3 \over 9^5\cdot 8^2
h) 3^{29}\cdot 4^{16} \over 27^9\cdot 8^{11}
f) 4^5\cdot 9^4 \over 8^3\cdot 27^3
i) 4^{20}\cdot 3^{35} \over 2^{37}\cdot 27^{12}
Bài 2.2 Thực hiện phép tính
a) \left(-\frac{1}{7}\right)^0 - 2\ \frac{4}{9} + \left(-\frac{2}{3}\right)^2
b) \frac{1}{2} - \left(\frac{2}{3}\right)^9 : \left(\frac{2}{3}\right)^7 + \frac{5}{6}
c) \left(\frac{3}{7}\right)^3 \cdot \left(\frac{7}{6}\right)^3 + \frac{2}{3} : \left(\frac{4}{3}\right)^2
d) 25^{10}\cdot \left(\frac{1}{5}\right)^{20} + \left(-\frac{3}{4}\right)^8 \cdot \left(-\frac{4}{3}\right)^8 - 2011^0
e) 4\cdot \left(\frac{1}{2}\right)^3 + 3\cdot \left(\frac{1}{2}\right)^2 - 2\cdot \left(\frac{1}{2}\right)^0
f) 2^3 - 3\cdot \left(\frac{1}{2}\right)^0 - \frac{1}{2^2} + \left[\left(-2\right)^2 : \frac{1}{2}\right] : 8
Dạng 2. Tìm giá trị chưa biết (Tìm x)
Bài 3. Tìm x, biết:
a) x : \left(-\frac{1}{2}\right)^3 = -\frac{1}{2}
b) (x-1)^3 = \frac{1}{8}
c) (x+1)^2 = \frac{4}{49}
d) \left(\frac{4}{7} - \frac{1}{2}x\right)^2 = \frac{4}{9}
e) \left(2x-\frac{1}{5}\right)^2 + \frac{16}{25} = 1
f) \left(-\frac{1}{3}\right)^{3x+1} = \frac{1}{81}
g) \left(-\frac{6}{7}\right)^{3.x+4} = \frac{36}{49}
h) \left(x-\frac{1}{3}\right)^3 = \left(\frac{1}{5}\right)^6
i) \left(\frac{1}{3}\right)^{x+3} + \left(\frac{1}{3}\right)^{x+2} = \frac{4}{27}
Dạng 3. So sánh hai số hữu tỉ
Bài 4. So sánh
a) \frac{5}{2^3} và \frac{5}{3^2}
b) \frac{-7}{2^5} và \frac{-7}{5^2}
c) \left(\frac{-3}{2^{300}}\right)^0 và \left(\frac{0}{300}\right)^1
Bài 5. So sánh
a) (1)^{24} và (1)^{36}
b) (1)^{50} và (1)^{75}
c) \left(-1\right)^{300} và \left(-\frac{1}{2}\right)^{200}
1. **Problem:** Simplify the expression $$\frac{6^{15}\cdot 9^{10}}{3^{34}\cdot 2^{13}}$$.
2. **Rewrite bases as powers of primes:**
- $6 = 2 \cdot 3$
- $9 = 3^2$
So,
$$6^{15} = (2 \cdot 3)^{15} = 2^{15} \cdot 3^{15}$$
$$9^{10} = (3^2)^{10} = 3^{20}$$
3. Substitute back:
$$\frac{2^{15} \cdot 3^{15} \cdot 3^{20}}{3^{34} \cdot 2^{13}} = \frac{2^{15} \cdot 3^{35}}{3^{34} \cdot 2^{13}}$$
4. Simplify powers by subtracting exponents of like bases:
$$= 2^{15-13} \cdot 3^{35-34} = 2^2 \cdot 3^1 = 4 \cdot 3 = 12$$
**Final answer:** $12$.
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