phân tích ra: 1313^2006=13^2006 x 101^2006

mà 13^4012=13^2006+2006 = 13^2006 x 13^2006

vì 101^2006>13^2006 nên 1313^2006 > 101^2006

OK XOG ROÀI ĐÓ BẠN

dư 0,7505 cô mình dạy nếu chia mãi không hết thì lấy đến 4 chữ số ở phần thập phân

hahaha tự đi mà làm có làm thì mới có ăn

\(=\text{0.75056179775}\)

Số hơi to

Còn thằng Việt kia ko làm thì im mồm

Âm 40 độ

Ta có: 0 độ C = 32 độ F

Ngày mai lạnh gấp đôi hôm nay thì ngày mai lạnh: 32 : 2 = 16 (độ F)

Vậy ngày mai lạnh 16 độ F(= -8,(8) độ C)

D

Chọn D

D.73cm

B2: a) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{1}{2}-x\right)\)

\(=-\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}\right)\)

\(=-x^2+\dfrac{1}{4}\)

b) \(\left(3x-2y\right)\left(3x+2y\right)\)

\(=\left(3x\right)^2-\left(2y\right)^2\)

\(=9x^2-4y^2\)

c) \(\left(x-3\right)\left(3+x\right)\)

\(=x^2-3^2\)

\(=x^2-9\)

d) \(x^2+6x+9\)

\(=x^2+2\cdot3\cdot x+3^2\)

\(=\left(x+3\right)^2\)

e) \(9x^2-6x+1\)

\(=\left(3x\right)^2-2\cdot3x\cdot1+1^2\)

\(=\left(3x-1\right)^2\)

f) \(x^2y^2+xy+\dfrac{1}{4}\)

\(=\left(xy\right)^2+2\cdot\dfrac{1}{2}\cdot xy+\left(\dfrac{1}{2}\right)^2\)

\(=\left(xy+\dfrac{1}{2}\right)^2\)

g) \(\left(x-y\right)^2+6\left(x-y\right)+9\)

\(=\left(x-y\right)^2+2\cdot3\cdot\left(x-y\right)+3^2\)

\(=\left(x-y+3\right)^2\)

h) \(x^2+8x+16\)

\(=x^2+2\cdot4\cdot x+4^2\)

\(=\left(x+4\right)^2\)

i) \(9x^2-24x+16\)

\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)

\(=\left(3x-4\right)^2\)

k) \(x^2-3x+\dfrac{9}{4}\)

\(=x^2-2\cdot\dfrac{3}{2}\cdot x+\left(\dfrac{3}{2}\right)^2\)

\(=\left(x-\dfrac{3}{2}\right)^2\)

l) \(4x^2y^4-4xy^3+y^2\)

\(=\left(2xy^2\right)^2-2\cdot2xy^2\cdot y+y^2\)

\(=\left(2xy^2-y\right)^2\)

m) \(9x^2-6x+1\)

\(=\left(3x\right)^2-2\cdot3x\cdot1+1\)

\(=\left(3x-1\right)^2\)

\(a,A=\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\)

\(=\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{9+\dfrac{9}{7}-\dfrac{9}{11}+\dfrac{9}{1001}-\dfrac{9}{13}}\)

\(=\dfrac{3\cdot\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\cdot\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\)

\(=\dfrac{3}{9}\)

\(=\dfrac{1}{3}\)

\(---\)

\(b,B=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{7\cdot2^{29}\cdot3^{18}-5\cdot2^{28}\cdot3^{18}}\)

\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{2^{28}\cdot3^{18}\cdot\left(7\cdot2-5\right)}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(14-5\right)}\)

\(=\dfrac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-1\right)}{2^{28}\cdot3^{18}\cdot9}\)

\(=\dfrac{2\cdot\left(10-1\right)}{9}\)

\(=\dfrac{2\cdot9}{9}\)

\(=2\)

\(---\)

\(c,C=\dfrac{5\cdot2^{30}\cdot3^{18}-4\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{2^{28}\cdot3^{18}\cdot\left(5\cdot3-7\cdot2\right)}\)

\(=\dfrac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\cdot\left(15-14\right)}\)

\(=\dfrac{2\cdot\left(10-9\right)}{1}\)

\(=2\)

\(---\)

\(d,D=\dfrac{15^{15}\cdot7^{16}}{6\cdot3^{14}\cdot35^{15}-15^8\cdot35^7\cdot7\cdot21^7}\)

\(=\dfrac{\left(3\cdot5\right)^{15}\cdot7^{16}}{2\cdot3\cdot3^{14}\cdot\left(5\cdot7\right)^{15}-\left(3\cdot5\right)^8\cdot\left(5\cdot7\right)^7\cdot7\cdot\left(3\cdot7\right)^7}\)

\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{2\cdot3^{15}\cdot5^{15}\cdot7^{15}-3^8\cdot5^8\cdot5^7\cdot7^7\cdot7\cdot3^7\cdot7^7}\)

\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{2\cdot3^{15}\cdot5^{15}\cdot7^{15}-3^{15}\cdot5^{15}\cdot7^{15}}\)

\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{3^{15}\cdot5^{15}\cdot7^{15}\cdot\left(2-1\right)}\)

\(=\dfrac{7}{1}\)

\(=7\)

#\(Toru\)