6^x.6^3:6^5+81=11^2-4
(2x+1)^3 -2012^0=63
tim x e N
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a)
6^x.6^3:6^5+81=11^2-4
6^x.6^3:6^5+81=117
6x.63:65=36
6x.63=279936
6x=279936:63
6x=1296
6x=64
x=4
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Các bn ơi giúp mình câu này đc ko mong các bạn giúp đỡ 4(x-1)=2x-6(x-2) các bạn giải hộ mình với
1: =>5(2x+6)=40
=>2x+6=8
=>2x=2
=>x=1
2: =>12-(x+3)=256:64=4
=>(x+3)=8
=>x=5
3: =>2x-1=3 hoặc 2x-1=-3
=>x=2 hoặc x=-1
4: \(\Leftrightarrow3^{x+2017}=3^{2015}\)
=>x+2017=2015
=>x=-2
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
Bài làm
a) x² - 3 = 22
=> x² = 25
=> x = + 5
Vậy x = + 5
b) 2x³ + 5 = -11
2x³ = -16
x³ = -8
x = -2
Vậy x = -2
c) ( x + 2 )² = 81
=> x + 2 = 9
=> x = 7
Vậy x = 7
d) ( 2x + 1 )² = 25
=> 2x + 1 = 5
=> 2x = 4
=> x = 2
Vậy x = 2
e) 5x + 2 = 625
5x = 623 ( vô lí )
g) ( 2x - 3 )² = 36.
=> 2x - 3 = 6
=> 2x = 9
=> x = 4,5
Vậy x = 4,5
h) ( 2x - 1 )³ = -8
=> 2x - 1 = -2
=> 2x = -1
=> x = -1/2
Vậy x = -1/2
i) ( x - 1 )x + 2 = ( x - 1 )x + 6
=> [ (x - 1 )x - ( x - 1 )x ] = 6 - 2
=> 0 = 4 ( vô lí )
Vậy x thuộc rỗng.
k) x² + x = 0
=> x( x + 1 ) = 0
=> x = 0 hoặc x + 1 = 0
=> x = 0 hoặc x = -1
Vậy x = 0 hoặc x = -1
a) (5x+1) ^ 2 = 4^2 : 5^ 2
( 5x+1) ^2 = (4:5) ^2
=> (5x+1) = ( 4 : 5) = 0.8
5x = 0.8 - 1
x = 0.7 : 5
x = 0,14
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