giúp mình với hoàn thành trước 5h ngày hôm nay
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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
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)
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) \(9x^2-6x+1\)
\(=\left(3x\right)^2-2\cdot3x\cdot1+1\)
\(=\left(3x-1\right)^2\)
\(a,A=\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\)
\(=\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{9+\dfrac{9}{7}-\dfrac{9}{11}+\dfrac{9}{1001}-\dfrac{9}{13}}\)
\(=\dfrac{3\cdot\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\cdot\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\)
\(=\dfrac{3}{9}\)
\(=\dfrac{1}{3}\)
\(---\)
\(b,B=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{7\cdot2^{29}\cdot3^{18}-5\cdot2^{28}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{2^{28}\cdot3^{18}\cdot\left(7\cdot2-5\right)}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(14-5\right)}\)
\(=\dfrac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-1\right)}{2^{28}\cdot3^{18}\cdot9}\)
\(=\dfrac{2\cdot\left(10-1\right)}{9}\)
\(=\dfrac{2\cdot9}{9}\)
\(=2\)
\(---\)
\(c,C=\dfrac{5\cdot2^{30}\cdot3^{18}-4\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{2^{28}\cdot3^{18}\cdot\left(5\cdot3-7\cdot2\right)}\)
\(=\dfrac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\cdot\left(15-14\right)}\)
\(=\dfrac{2\cdot\left(10-9\right)}{1}\)
\(=2\)
\(---\)
\(d,D=\dfrac{15^{15}\cdot7^{16}}{6\cdot3^{14}\cdot35^{15}-15^8\cdot35^7\cdot7\cdot21^7}\)
\(=\dfrac{\left(3\cdot5\right)^{15}\cdot7^{16}}{2\cdot3\cdot3^{14}\cdot\left(5\cdot7\right)^{15}-\left(3\cdot5\right)^8\cdot\left(5\cdot7\right)^7\cdot7\cdot\left(3\cdot7\right)^7}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{2\cdot3^{15}\cdot5^{15}\cdot7^{15}-3^8\cdot5^8\cdot5^7\cdot7^7\cdot7\cdot3^7\cdot7^7}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{2\cdot3^{15}\cdot5^{15}\cdot7^{15}-3^{15}\cdot5^{15}\cdot7^{15}}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{3^{15}\cdot5^{15}\cdot7^{15}\cdot\left(2-1\right)}\)
\(=\dfrac{7}{1}\)
\(=7\)
#\(Toru\)
uk giải giúp mình vói áp dụng tính chát chia hết của một tổng