Tính.(giải chi tiết giúp mình nhé)
K
Khách
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NT
3
26 tháng 2 2017
Đặt A = 1.3.5 + 3.5.7 + 5.7.9 + ...... + 47.49.51
8A = 1.3.5.8 + 3.5.7.8 + ...... + 47.49.51.8
= 1.3.5(7 + 1) + 3.5.7.(9 - 1) + ...... + 47.49.51(53 - 45)
= 1.3.5.7 + 1.3.5 + 3.5.7.9 - 1.3.5.7 + ......... + 47.49.51.53 - 47.47.49.51
= 1.3.5 + 47.49.51.53
=> A = \(\frac{1.3.5+47.49.51.53}{8}=778128\)
SH
3
A
23 tháng 10 2016
\(0,0\left(3\right)+0,0\left(16\right)\)\(=\frac{1}{30}+\frac{16}{990}\)\(=\)\(\frac{33}{990}+\frac{16}{990}=\frac{49}{990}\)
VN
0
4 tháng 1 2017
Lấy (32,43 + 85,94)x 0,01-0,0136
118,37 x0.01 - 0,0136
1.1837 - 0,0136
1,701
LS
3
30 tháng 8 2015
(-4,9)+5,5+4,9+(-5,5)
=[(-4,9)+4,9]+[5,5+(-5,5)]
=(4,9-4,9)+(5,5-5,5)
=0+0
=0
25 tháng 7 2021
Bài 18
a, Với \(a>0;a\ne1;4\)
\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b, Thay a = 9 => căn a = 3
\(A=\dfrac{3-2}{3.3}=\dfrac{1}{9}\)
c, Ta có : \(A.B=\dfrac{\sqrt{a}-2}{3\sqrt{a}}.\dfrac{3\sqrt{a}}{\sqrt{a}+1}=\dfrac{\sqrt{a}-2}{\sqrt{a}+1}< 0\)
Vì \(\sqrt{a}+1>\sqrt{a}-2\)
\(\left\{{}\begin{matrix}\sqrt{a}+1>0\\\sqrt{a}-2< 0\end{matrix}\right.\Leftrightarrow a< 4\)
Kết hợp với đk vậy \(0< a< 4;a\ne1\)
25 tháng 7 2021
Bài 18:
1) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
2) Thay a=9 vào B, ta được:
\(B=\dfrac{3\cdot3}{3+1}=\dfrac{9}{4}\)
25 tháng 7 2021
a, \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)ĐK : \(x>0;x\ne1\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b, \(A=\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\Rightarrow3\sqrt{x}-3=\sqrt{x}\Leftrightarrow2\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\)
c, \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=\dfrac{\sqrt{x}-1-9x}{\sqrt{x}}\)
\(=1-\dfrac{1}{\sqrt{x}}-9\sqrt{x}\)Đặt \(\sqrt{x}=t^2\left(t>0\right)\)
\(1-t-9t^2=-\left(9t^2-t-1\right)=-\left(9t^2-2.3.\dfrac{1}{6}.t+\dfrac{1}{36}-\dfrac{37}{36}\right)\)
\(=-\left(3t-\dfrac{1}{6}\right)+\dfrac{37}{36}\le\dfrac{37}{36}\)
Dấu ''='' xảy ra khi t = 1/18 => t^2 = 1/324 => \(\sqrt{x}=\dfrac{1}{324}\Rightarrow x=\dfrac{1}{104876}\)
Vậy GTLN P là 37/36 khi x = 1/104876
N
Giải chi tiết giúp mình nhé
3
NT
27 tháng 8 2021
d. \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Rightarrow x^2+5x-14=x^2+3x-4\)
\(\Rightarrow x^2+5x-x^2-3x=-4+14\)
\(\Rightarrow2x=10\) \(\Rightarrow x=\dfrac{10}{3}\) \(\Rightarrow x=5\)
KK
27 tháng 8 2021
\(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
⇔ \(\dfrac{\left(x-2\right)\left(x+7\right)}{\left(x-1\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x-1\right)}{\left(x+7\right)\left(x-1\right)}\)
⇔ (x - 2)(x + 7) = (x + 4)(x - 1)
⇔ x2 + 7x - 2x - 14 = x2 - x + 4x - 4
⇔ x2 - x2 + 7x - 2x + x - 4x = 14 - 4
⇔ 2x = 10
⇔ x = 10/2 = 5
13 tháng 7 2021
\(\dfrac{\sqrt{3}-3}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}-3\right)\left(\sqrt{3}-1\right)}{2}=\dfrac{3-\sqrt{3}-3\sqrt{3}+3}{2}=\dfrac{6-4\sqrt{3}}{2}=3-2\sqrt{3}\)
Ta có:\(P=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)...\left(1-\dfrac{1}{1+2+...+2014}\right)\)
\(P=\dfrac{2}{1+2}\cdot\dfrac{2+3}{1+2+3}\cdot...\cdot\dfrac{2+3+...+2014}{1+2+3+...+2014}\)
\(P=\dfrac{\dfrac{1\cdot4}{2}}{\dfrac{2\left(2+1\right)}{2}}\cdot\dfrac{\dfrac{2\left(3+2\right)}{2}}{\dfrac{3\left(3+1\right)}{2}}\cdot...\cdot\dfrac{\dfrac{2013\left(2014+2\right)}{2}}{\dfrac{2014\left(2014+1\right)}{2}}\)
\(P=\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot...\cdot\dfrac{2013\cdot2016}{2014\cdot2015}\)
\(P=\dfrac{1\cdot4\cdot2\cdot5\cdot...\cdot2013\cdot2016}{2\cdot3\cdot3\cdot4\cdot...\cdot2014\cdot2015}\)
\(P=\dfrac{\left(1\cdot2\cdot...\cdot2013\right)\left(4\cdot5\cdot...\cdot2016\right)}{\left(2\cdot3\cdot\cdot...\cdot2014\right)\left(3\cdot4\cdot...\cdot2015\right)}\)
\(P=\dfrac{2016}{2014\cdot3}\)
\(P=\dfrac{336}{1007}\)
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