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a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
=>16x-7=13x+2
=>3x=9
hay x=3
b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)
=>x+2017=0
hay x=-2017
e: \(\left(2x-3\right)^2=144\)
=>2x-3=12 hoặc 2x-3=-12
=>2x=15 hoặc 2x=-9
=>x=15/2 hoặc x=-9/2
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.
a.
| x | = 5,6
=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)
Vậy \(x\in\left\{-5,6;5,6\right\}\)
b, \(\left|x-3,5\right|=5\)
=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy \(x\in\left\{-1,5;8,5\right\}\)
c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)
d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)
=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)
=> \(\left|4x\right|=13,75\)
=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)
Vậy \(x\in\left\{-3,4375;3,4375\right\}\)
e, ( x - 1 ) 3 = 27
=> x - 1 = 3
=> x = 4
Vậy x = 4
f, ( 2x - 3)2 = 36
=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)
Vậy x\(\in\left\{-1,5;4,5\right\}\)
g, \(5^{x+2}=625\)
=> \(5^{x+2}=5^4\)
=> x + 2 = 4
=> x = 2
Vậy x = 2
h, ( 2x - 1)3 = -8
=> 2x - 1 = -2
=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)
=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)
=> \(\dfrac{1}{32.2^{31}}=2^x\)
=> \(\dfrac{1}{2^{36}}=2^x\)
=> x = -36
Vậy x = -36
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) \(-\left(5+35x\right)^2=36\)
\(\Rightarrow\left[{}\begin{matrix}\left(-5-35x\right)^2=6^2\\\left(-5-35x\right)^2=\left(-6\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-5-35x=6\\-5-35x=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}35x=-11\\35x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{35}\\x=\dfrac{1}{35}\end{matrix}\right.\)
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
1) Tính
a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625
\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9
2) Tìm x thuộc Q, biết:
a) 3x + 2 = 27
=> 3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)
\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)
\(\dfrac{1}{2}x-3=3^{ }\)
\(\dfrac{1}{2}x=3+3\)
\(\dfrac{1}{2}x=9\)
\(x=9:\dfrac{1}{2}\)
\(x=18\)
c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)
\(x-\dfrac{1}{2}=-3\)
\(x=-3+\dfrac{1}{2}\)
\(x=\dfrac{-5}{2}\)
d) 5 . 5x + 1 = 125
5x + 1 = 125 : 5
5x + 1 = 25
5x + 1 = 52
x + 1 = 2
x = 2 - 1
x = 1.
1/vì (1,782x-2-1,78x):1,78x=0
nên 1,78x2-2-1,78x=0
=>1,782x-2=1,78x
=>2x-2=x
2x=x+2
=>x=2
2/vì cơ số bằng nhau nên ta có
x-2=1;-1;0
ta có: x-2=1 => x=3
x-2=-1 => x=1
x-2=0 => x=2
3/ta có
(x+2)3=33 =>x+2=3 =>x=1
mik mệt rồi bạn cứ gải tiếp đi
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 \(x\in\left\{\dfrac{1}{5};\dfrac{2}{5};\dfrac{3}{5}\right\}\)
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