Câu 37: trang 22- sgk toán 7 tập 1
Tìm giá trị của biểu thức sau :
a. $\frac{4^{2}.4^{3}}{2^{10}}$
b. $\frac{(0,6^{5})}{(0,2)^{6}}$
c. $\frac{2^{7}.9^{3}}{6^{5}.8^{2}}$
d. $\frac{6^{3}+3.6^{2}+3^{3}}{-13}$
Bài Làm:
a. $\frac{4^{2}.4^{3}}{2^{10}}$
= $\frac{4^{2+3}}{2^{2.5}}$
= $\frac{4^{5}}{4^{5}}=1$
Vậy $\frac{4^{2}.4^{3}}{2^{10}}=1$
b. $\frac{(0,6^{5})}{(0,2)^{6}}$
= $\frac{(0,2.3)^{5}}{(0,2)^{6}}$
= $\frac{0,2^{5}.3^{5}}{(0,2)^{5}.0,2}$
= $\frac{3^{5}}{0,2}$
= $\frac{243}{0,2}=1215$
Vậy $\frac{(0,6^{5})}{(0,2)^{6}}=1215$
c. $\frac{2^{7}.9^{3}}{6^{5}.8^{2}}$
= $\frac{2^{2}.2^{5}.(3^{2})^{3}}{(2.3)^{5}.(2^{3})^{2}}$
= $\frac{2^{2}.2^{5}.3^{6}}{2^{5}.3^{5}.2^{6}}$
= $\frac{3}{2^{4}}=\frac{3}{16}$
Vậy $\frac{2^{7}.9^{3}}{6^{5}.8^{2}}=\frac{3}{16}$
d. $\frac{6^{3}+3.6^{2}+3^{3}}{-13}$
= $\frac{(2.3)^{3}+3.(2.3)^{2}+3^{3}}{-13}$
= $\frac{2^{3}.3^{3}+3.2^{2}.3^{2}+3^{3}}{-13}$
= $\frac{3^{3}(2^{3}+2^{2}+1)}{-13}$
= $\frac{3^{3}.13}{-13}$
= $-3^{3}=-27$
Vậy $\frac{6^{3}+3.6^{2}+3^{3}}{-13=-27$
