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28 tháng 1 2016
dễ
ai đi qua tick cho mình nha
ai tick thì may mắn trọn đời
, bạn nào làm đc mik tick cho
18 tháng 7 2017
Câu 1 :
\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)
\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)
\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)
18 tháng 7 2017
Câu 2 :
\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)
Sorry . Mình chỉ biết đến đây thôi
26 tháng 5 2022
b: \(\dfrac{2x+3}{3-x}\le0\)
\(\Leftrightarrow\dfrac{2x+3}{x-3}\ge0\)
=>x>3 hoặc x<=-3/2
c: \(\dfrac{x+5}{x+3}>1\)
\(\Leftrightarrow\dfrac{x+5-x-3}{x+3}>0\)
=>2/(x+3)>0
=>x+3>0
hay x>-3
A
3 tháng 8 2017
a) \(\left(x-2\right)^3=-27\)
\(\Rightarrow\left(x-2\right)^3=\left(-3\right)^3\)
\(\Rightarrow x-2=-3\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
b) \(\left(2x+1\right)^4=81\)
\(\Rightarrow\left(2x+1\right)^4=3^4=\left(-3\right)^4\)
\(\left\{{}\begin{matrix}\left(2x+1\right)^4=3^4\Rightarrow2x+1=3\Rightarrow x=1\\\left(2x+1\right)^4=\left(-3\right)^4\Rightarrow2x+1=-3\Rightarrow x=-2\end{matrix}\right.\)
Vậy \(x=1;x=-2\)
c) Bạn xem lại đề bài nhé!
d) \(\left(5x-2\right)^{10}=\left(5x-2\right)^{100}\)
\(\Rightarrow\left(5x-2\right)^{10}-\left(5x-2\right)^{100}=0\)
\(\Rightarrow\left(5x-2\right)^{10}.\left[1-\left(5x-2\right)^{90}\right]=0\)
+) TH1: \(\left(5x-2\right)^{10}=0\)
\(\Rightarrow5x-2=0\)
\(\Rightarrow x=\dfrac{2}{5}\)
+) TH2: \(1-\left(5x-2\right)^{90}=0\)
\(\Rightarrow\left(5x-2\right)^{90}=1\)
\(\Rightarrow\left(5x-2\right)^{90}=1^{90}=\left(-1\right)^{90}\)
\(\Rightarrow\left\{{}\begin{matrix}\left(5x-2\right)^{90}=1^{90}\Rightarrow5x-2=1\Rightarrow x=\dfrac{3}{5}\\\left(5x-2\right)^{90}=\left(-1\right)^{90}\Rightarrow5x-2=-1\Rightarrow x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{5};\dfrac{2}{5};\dfrac{3}{5}\right\}\)
16 tháng 11 2017
a) \(\left(x-3\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=1^2\\\left(x-3\right)^2=-1^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
b) \(\left(x-\dfrac{1}{7}\right)^2=0\)
\(\Rightarrow x-\dfrac{1}{7}=0\)
\(\Rightarrow x=0+\dfrac{1}{7}\)
\(\Rightarrow x=\dfrac{1}{7}\)
c) \(\left(2x+3\right)^3=-27\)
\(\Rightarrow\left(2x+3\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x+3=-3\)
\(\Rightarrow2x=-6\)
\(\Rightarrow x=-3\)
d) \(-\left(5+35x\right)^2=36\)
\(\Rightarrow\left[{}\begin{matrix}\left(-5-35x\right)^2=6^2\\\left(-5-35x\right)^2=\left(-6\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-5-35x=6\\-5-35x=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}35x=-11\\35x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{35}\\x=\dfrac{1}{35}\end{matrix}\right.\)
16 tháng 11 2017
a) (x-3)mũ 2 = 1
Vậy x-3 = 1( vì 1 mũ 2 sẽ bằng 1)
=> x = 1+3 = 4
b) (x - 1/7) mũ 2 = 0
Vậy x - 1/7 = 0 ( vì 0 mũ 2 sẽ bằng 0)
=> x = 0 + 1/7 = 1/7
c) (2x + 3 ) mũ 3 = -27
vậy 2x + 3 = -3 ( vì -3 mũ 3 sẽ bằng -27)
=> 2x = -3-3 = -6
=> x = -6/2 = -3
KL
0
TT
23 tháng 9 2017
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
MN
25 tháng 4 2017
a. P(x)+Q(x)=(3x4 + x3- x2- \(\dfrac{1}{4}\)x)+(3x4- 4x3+x2-\(\dfrac{1}{4}\))=6x4-3x3+\(\dfrac{1}{2}\)
Tương tự làm P(x)-Q(X) nhé !!!
b. Thay x = 0 vào đa thức P(x) ta có :
.....................................................
thay x = 0 vào đa thức Q(x) ta có:
......................................................
=> đpcm
11 tháng 1 2018
bài 1:
|x| = \(\dfrac{1}{3}\) => x = \(\pm\)\(\dfrac{1}{3}\) |y| = 1 => y = \(\pm\)1
a
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\) +5 = 2.\(\dfrac{1}{9}\) - 1 + 5
= \(\dfrac{2}{9}\) - 1 + 5 = \(\dfrac{2-9+45}{9}\) = \(\dfrac{38}{9}\)
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{-1}{3}\right)^2\) - 3\(\left(\dfrac{-1}{3}\right)\) + 5
= 2.\(\dfrac{1}{9}\) - (-1) + 5 = \(\dfrac{2}{9}\) + 1 +5
= \(\dfrac{2+9+45}{9}\) = \(\dfrac{56}{9}\)
b) +) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\).1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - 1 + 1 = \(\dfrac{2}{9}\) - 1 + 1
= \(\dfrac{2-9+9}{9}\) = \(\dfrac{2}{9}\)
+) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{-1}{3}\right)\)\(^2\) - 3\(\left(\dfrac{-1}{3}\right)\). 1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - (-1) + 1 = \(\dfrac{2}{9}\) + 1 + 1
= \(\dfrac{2+9+9}{9}\) = \(\dfrac{20}{9}\)
11 tháng 1 2018
bài 3
x.y.z = 2 và x + y + z = 0
A = ( x + y )( y +z )( z + x )
= x + y . y + z . z + x = ( x + y + z ) + ( x . y . z )
= 0 + 2 = 2
bài 4
a) | 2x - \(\dfrac{1}{3}\) | - \(\dfrac{1}{3}\) = 0 => | 2x - \(\dfrac{1}{3}\) | = \(\dfrac{1}{3}\)
=> 2x - \(\dfrac{1}{3}\) = \(\pm\) \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\)= \(\dfrac{1}{3}\)
=> 2x = \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
x = \(\dfrac{2}{3}\) : 2 = \(\dfrac{2}{3}\) . \(\dfrac{1}{2}\) = \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\) = \(\dfrac{-1}{3}\)
2x = \(\dfrac{-1}{3}\) + \(\dfrac{1}{3}\) = 0
x = 0 : 2 = 2
a, \(4.3^{2x}-2.9^x-54=0\)
\(\Rightarrow3^{2x}\left(4-2\right)=54\)
\(\Rightarrow3^{2x}=27=3^3\)
Vì \(3\ne\pm1;3\ne0\) nên \(2x=3\Rightarrow x=\dfrac{3}{2}\)
b, \(\dfrac{1}{2}.2^x+4.2^x-288=0\)
\(\Rightarrow2^x\left(\dfrac{1}{2}+4\right)=288\)
\(\Rightarrow2^x=64=2^6\)
Vì \(2\ne\pm1;2\ne0\) nên \(x=6\)