tim x
7x + 72x+3 = 344
a,3.(/x/-4/5)+0,2=0,5
b,(x-1)3 = 1/8
c,72x +72x+3= 344
giúp mình với,cảm ơn mọi người nhiều !
b) \(\left(x-1\right)^3=\dfrac{1}{8}\)
\(\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)
\(x-1=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+1\)
\(x=\dfrac{3}{2}\)
72x+72x+3=344 (5-x)(9x2-4)=0 \(\left|2-2x\right|\)-3,75=(-0,5)2
\(\sqrt{x-1}\)+\(\dfrac{2}{3}\)=1 (\(\dfrac{1}{3}\)-\(\dfrac{3}{2}\)x)2=2\(\dfrac{1}{4}\) giúp mình với!!!! cảm ơn nhìu=))❤
(5 - \(x\))(9\(x^2\) - 4) =0
\(\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-\dfrac{2}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { - \(\dfrac{2}{3}\); \(\dfrac{2}{3}\); \(5\)}
72\(x\) + 72\(x\) + 3 = 344
72\(x\) \(\times\) ( 1 + 73) = 344
72\(x\) \(\times\) (1 + 343) = 344
72\(x\) \(\times\) 344 = 344
72\(x\) = 344 : 344
72\(x\) = 1
72\(x\) = 70
\(2x\) = 0
\(x\) = 0
Kết luận: \(x\) = 0
|2 - 2\(x\)| - 3,75 = (-0,5)2
|2 - 2\(x\)| - 3,75 = 0,25
|2- 2\(x\)| =0,25 + 3,75
|2 - 2\(x\)| = 4
\(\left[{}\begin{matrix}2-2x=-4\left(x>1\right)\\2-2x=4\left(x< 1\right)\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=6\left(x\ge1\right)\\2x=-2\left(x\le1\right)\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Kết luận: \(x\) \(\in\) { -1; 3}
Tìm x:
a) 2x + 2 - 3 . 2x = 32
b) (4x - 3)2 = 4x - 3
c) 72x + 72x + 3 = 344
d) (7x - 3)2012 = (3 - 7x)2010
e) (4x2 - 3)3 + 8 = 0
a: =>2^x*4-2^x*3=32
=>2^x=32
=>x=5
b: =>(4x-3)^2-(4x-3)=0
=>(4x-3)(4x-3-1)=0
=>(4x-3)(4x-4)=0
=>x=3/4 hoặc x=1
c: =>7^2x+7^2x*7^3=344
=>7^2x=1
=>2x=0
=>x=0
d: =>(7x-3)^2012-(7x-3)^2010=0
=>(7x-3)^2010*[(7x-3)^2-1]=0
=>(7x-3)^2010*(7x-4)(7x-2)=0
=>x=2/7; x=4/7; x=3/7
e: =>(4x^2-3)^3=-8
=>4x^2-3=-2
=>4x^2=1
=>x^2=1/4
=>x=1/2 hoặc x=-1/2
a) 2x(22 - 3) = 32
2x.1=25
=> x = 5
b) (4x - 3)2 = 4x -3
=> (4x - 3)2 - (4x - 3) = 0
(4x-3)[(4x - 3) - 1] = 0
(4x-3)(4x - 4)=0
\(\Rightarrow\left[{}\begin{matrix}4x-3=0\\4x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=1\end{matrix}\right.\)
c) 72x + 72x+3 = 344
=> 72x(1 + 73) =344
72x . 344 = 344
=> 2x = 0 => x = 0
d) (7x - 3)2012 = (3 - 7x)2010
=> (7x - 3)2012 - (7x - 3)2010 = 0
(7x - 3)2010 [(7x - 3)2 - 1] = 0
\(\Rightarrow\left[{}\begin{matrix}7x-3=0\\\left(7x-3\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\7x=4\\7x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\x=\dfrac{4}{7}\\x=\dfrac{2}{7}\end{matrix}\right.\)
e) (4x2 - 3)3 + 8 = 0
(4x2 - 3)3 = (-2)3
=> 4x2 - 3 = -2
4x2 = 1
x2 = 1/4
=> \(x=\pm\dfrac{1}{2}\)
tìm x biết (72x^2-18x+1)(12x^2-7x+10=5
Tìm x, biết: ( 72x2 - 18x + 1)( 12x2 - 7x + 1)
Tìm x biết: ( 72x2 -18x + 1)( 12x2 - 7x + 1) = 5
tim x : x - ( 432 + 344 ) =236
x - 776 = 236
x = 236 + 776
x = 1012
x - ( 432 + 344 ) = 236
x - 776 = 236
x = 236 +776
x = 1012
Giải phương trình:
a) \(10x^4-27x^3+105x^2-72x+10=0\)
b) (4x-7)(\(x^2-5x+4\) ) (\(2x^2-7x+3\) ) =0
tim x
7x .( 2+ x ) - 7x .( x+ 3) = 14
7x ( 2 + x - x - 3 ) = 14
-7x = 14
x = -2
\(7x\left(2+x\right)-7x\left(x+3\right)=14\)
\(\Rightarrow7x\left[\left(2+x\right)-\left(x+3\right)\right]=14\)
\(\Rightarrow7x\left[2+x-x-3\right]=14\)
\(\Rightarrow7x.\left(-1\right)=14\)
\(\Rightarrow-7x=14\)
\(\Rightarrow x=14:-7\)
\(\Rightarrow x=-2\)
Vậy x = -2
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