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KT
14 tháng 8 2018
\(\frac{437^2-363^2}{537^2-463^2}\)
\(=\frac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)
\(=\frac{74.800}{74.1000}\)
\(=\frac{80}{1000}=\frac{2}{25}\)
14 tháng 8 2018
\(\frac{437^2-363^2}{537^2-463^2}\)
\(=\frac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)( Áp dụng hằng đẳng thức \(A^2-B^2=\left(A-B\right)\left(A+B\right)\))
\(=\frac{74\cdot800}{74\cdot1000}\)
\(=\frac{4}{5}\)
8 tháng 10 2017
\(a.\)
\(A=5\dfrac{4}{23}.27\dfrac{3}{47}+5\dfrac{4}{23}.\left(-4\dfrac{3}{47}\right)\)
\(A=5\dfrac{4}{23}\left(27\dfrac{3}{47}-4\dfrac{3}{47}\right)\)
\(A=5\dfrac{4}{23}\left(27-4\right)\)
\(A=5\dfrac{4}{23}.23\)
\(A=119\)
\(b.\)
\(B=2^3+3.1-2^{-2}.4+\left(-2^2:\dfrac{1}{2}\right).8\)
\(B=2^3+3-\dfrac{1}{4}.4+\left(-8\right).8\)
\(B=2^3+3-1-64\)
\(B=-54\)
23 tháng 10 2017
Ta có: \(A=\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
\(A>\dfrac{1}{100.101}+\dfrac{1}{101.102}+\dfrac{1}{102.103}+\dfrac{1}{103.104}+\dfrac{1}{104.105}\)\(A>\dfrac{1}{100}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{102}+\dfrac{1}{102}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{104}+\dfrac{1}{104}-\dfrac{1}{105}\)\(A>\dfrac{1}{100}-\dfrac{1}{105}\)
\(A>\dfrac{1}{2100}\)
Mà \(B=\dfrac{1}{2^2.3.5^2.7}\)=\(\dfrac{1}{2100}\)
=> \(A>B\)
Vậy \(A>B\)
28 tháng 10 2017
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=t\) \(\Rightarrow a=bt\);\(c=dt\)
rồi bạn thế vào điều phải chứng minh là ra
8 tháng 7 2017
Thực hiện phép tính
a) \(1-\dfrac{1}{2}:\dfrac{-3}{2}+\dfrac{2}{3}=1-\dfrac{1}{2}.\dfrac{2}{-3}+\dfrac{2}{3}=\)
\(=1+\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3+1+2}{3}=\dfrac{6}{3}=2\)
b) \(\dfrac{-3}{4}-\dfrac{-7}{2}+\dfrac{2}{-3}=\dfrac{-3}{4}+\dfrac{7}{2}+\dfrac{-2}{3}=\)
\(=\dfrac{-9+42-8}{12}=\dfrac{25}{12}\)
c) \(0,25-3\dfrac{1}{2}:\dfrac{1}{2}+\dfrac{-3}{4}.\dfrac{2}{-3}=0,25-\dfrac{7}{2}.2+\dfrac{1}{2}=\)
\(=0,25-7+0,5=-\dfrac{25}{4}\)
22 tháng 7 2017
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
20 tháng 1 2018
a,\(\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{31}{35}\)
b,\(\dfrac{-4}{3}:\dfrac{2}{15}=\dfrac{-60}{6}=-10\)
c,\(\dfrac{3}{7}.\dfrac{2}{9}+\dfrac{7}{9}.\dfrac{3}{7}=\dfrac{3}{7}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{3}{7}\)
d,\(\left(\dfrac{-2}{3}\right)^3.9^2+\left(\dfrac{-3}{4}\right)^2.32\)
\(=\dfrac{\left(-2\right)^3}{3^3}.3^4+\dfrac{\left(-3\right)^2}{4^2}.2^5\)
\(=\left(-8\right).3+\dfrac{3^2}{4^2}.2^5\)
\(=\left(-24\right)+2.9\)
\(=\left(-24\right)+18\)
\(=-6\)
HN
10 tháng 7 2017
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
10 tháng 7 2017
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
12 tháng 7 2022
a: \(\Leftrightarrow\left(3x-2\right):\dfrac{7}{5}=\dfrac{17}{7}:\dfrac{13}{5}=\dfrac{85}{91}\)
\(\Leftrightarrow3x-2=\dfrac{85}{91}\cdot\dfrac{7}{5}=\dfrac{17}{13}\)
=>3x=43/13
hay x=43/39
b: \(\Leftrightarrow9x+207=121-8x\)
=>19x=-86
hay x=-86/19
c: \(\Leftrightarrow x^2-9=16\)
=>x2=25
=>x=5 hoặc x=-5
d: \(\Leftrightarrow\left|x\right|=\dfrac{1.64\cdot3.11}{8.51}\simeq0,6\)
=>x=0,6 hoặc x=-0,6
a: \(\dfrac{63^2-47^2}{215^2-105^2}=\dfrac{\left(63-47\right)\cdot\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)
\(=\dfrac{16\cdot110}{110\cdot320}=\dfrac{16}{320}=\dfrac{1}{20}\)
b: \(\dfrac{437^2-363^2}{537^2-463^2}=\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)
\(=\dfrac{74\cdot800}{74\cdot1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)