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11 tháng 8 2017
\(3^{x+2}+3^x=90\Leftrightarrow3^x.3^2+3^x=90\Leftrightarrow3^x\left(3^2+1\right)=90\Leftrightarrow3^x.10=90\Leftrightarrow3^x=9\Leftrightarrow3^x=3^2\Leftrightarrow x=2\)
Vậy ...
\(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{6}\)
Vậy ...
N
11 tháng 8 2017
1.
a) \(3^{x+2}+3^x=90\)
\(\Leftrightarrow3^x\left(3^2+1\right)=90\)
\(\Leftrightarrow3^x.10=90\)
\(\Leftrightarrow3^x=9=3^2\)
\(\Leftrightarrow x=2\)
vậy...
b) \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{2}{3}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{3}\\x+\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)
Vậy...
tik mik nha !!!
17 tháng 8 2017
a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)
\(\Leftrightarrow3\left(x-2\right)=5x\)
\(\Leftrightarrow3x-6=5x\)
\(\Leftrightarrow5x-3x=6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)
\(\Leftrightarrow4x+92=2x+80\)
\(\Leftrightarrow4x-2x=80-92\)
\(\Leftrightarrow2x=-12\)
\(\Leftrightarrow x=-6\)
c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)
d, \(B=1+2+2^2+........+2^{2017}\)
\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)
\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)
\(\Leftrightarrow B=2^{2018}-1\)
DH
17 tháng 8 2017
\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)
\(< =>3x-6=5x=>x=-3\)
\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)
\(4x+92=3x+120=>x=28\)
14 tháng 5 2022
a: =>5x=3x-6
=>2x=-6
hay x=-3
b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)
=>x-3=10 hoặc x-3=-10
=>x=13 hoặc x=-7
c: \(\left|x^3+1\right|+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-1
NM
1
17 tháng 8 2017
\(\dfrac{x-2}{5}=\dfrac{x}{3}\)
\(\Leftrightarrow\left(x-2\right)3=5x\)
\(\Leftrightarrow3x-6=5x\)
\(\Leftrightarrow5x-3x=-6\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
Vậy .....
b, \(B=1+2+2^2+..........+2^{2017}\)
\(\Leftrightarrow2B=2+2^2+.......+2^{2018}\)
\(\Leftrightarrow2B-B=\left(2+2^2+......+2^{2018}\right)-\left(1+2+......+2^{2017}\right)\)
\(\Leftrightarrow B=2^{2018}-1\)
c, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)
\(\Leftrightarrow4x+92=3x+120\)
\(\Leftrightarrow4x-3x=120-92\)
\(\Leftrightarrow x=28\)
28 tháng 3 2017
a) M =1+3+32+33+......+3118+3119
M = ( 1+3+32 ) +...+ ( 3117 + 3118+3119 )
M = 1. ( 1+3+32 ) + ... + 3117 . ( 3117 + 3118+3119 )
M = ( 1+3+32 ) .( 1 + ... + 3117 )
M = 13 . ( 1 + ... + 3117 ) \(⋮\) 13 (đpcm )
28 tháng 3 2017
b) Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2009^2}< \dfrac{1}{2008.2009}\)
\(\dfrac{1}{2010^2}< \dfrac{1}{2009.2010}\)
=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\) (1)
Biến đổi vế trái:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2010}\)
= \(1-\dfrac{1}{2010}\)
= \(\dfrac{2009}{2010}< 1\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < 1 hay:
N < 1
16 tháng 5 2018
3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
16 tháng 5 2018
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
31 tháng 7 2017
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
30 tháng 7 2017
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
DV
2
14 tháng 10 2017
\(\left(\dfrac{3}{5}\right)^5.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3\)
\(\Rightarrow x=\left(\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=\dfrac{27}{125}\)
LN
14 tháng 10 2017
Bạn sai rồi nhé như thế này mới đúng :
Lời giải :
\(\left(\dfrac{3}{5}\right)^5.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)
\(\left(\dfrac{3}{5}\right)^2.\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3.\left(\dfrac{3}{5}\right)^2\)
Vì : \(\left(\dfrac{3}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\)
Nên : \(\left(\dfrac{3}{5}\right)^3.x=\left(\dfrac{9}{25}\right)^3\)
\(\Rightarrow x=\left(\dfrac{9}{25}\right)^3:\left(\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=\left(\dfrac{9}{25}:\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=\left(\dfrac{3}{5}\right)^3\)
\(\Rightarrow x=\dfrac{3.3.3}{5.5.5}\)
\(\Rightarrow x=\dfrac{3^3}{5^3}\)
\(\Rightarrow x=\left(3:5\right)^3\)
\(\Rightarrow x=0,6^3\)
\(\Rightarrow x=0,216\)
Vậy \(x=0,216\)
25 tháng 4 2018
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
25 tháng 4 2018
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)
*Nếu \(x^2-4=0\)
⇒ x2 = 4
⇒ x ∈ {2 ; -2}
*Nếu \(6-2x=0\)
⇒2x = 6
⇒ x = 6 : 2 = 3
Vậy x ∈ { -2 ; 2 ; 3 }