Tìm x biết :
a) ( 2x - 1 )3 = ( 2x - 1 )4 .
b) ( 2017 )x + 2 = ( 2018 - 53 )x + 2 .
c) 5x + 2 + 5x = 650 .
d) 81x = ( -3 )7
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a )
\(5x\left(x-3\right)+7\left(x-3\right)=0\)
\(\Rightarrow\left(5x+7\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+7=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-7\\x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{7}{5}\\x=3\end{cases}}}\)
Vậy ...
b )
\(x^{2017}=x^{2018}\)
\(\Rightarrow x^{2017}-x^{2018}=0\)
\(\Rightarrow x^{2017}\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^{2017}=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy ...
c )
\(2x^2=x\)
\(\Rightarrow2x^2:x=1\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy ...
e )
\(x^5=x^4\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)( làm tương tự như phần b )
~
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
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, x - 3 : 2 = 5 14 : 5 12
=> x - 3 : 2 = 5 2
=> x - 3 : 2 = 25
=> x – 3 = 25
=> x = 53
b, 30 : x - 7 = 15 19 : 15 18
=> 30 : x - 7 = 15
=> x – 7 = 2
=> x = 9
c, x 70 = x
=> x 70 - x = 0
=> x ( x 69 - 1 ) = 0
=>
d, 2 x + 1 3 = 9 . 81
=> 2 x + 1 3 = 9 3
=> 2x + 1 = 9
=> x = 4
e, 5 x + 5 x + 2 = 650
=> 5 x 1 + 5 2 = 650
=> 5 x . 26 = 650
=> 5 x = 25
=> x = 2
f, 4 x - 1 2 = 25 . 9
=> 4 x - 1 2 = 5 2 . 3 2
=> 4 x - 1 2 = 15 2
=> 4x – 1 = 15
=> x = 4
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a, \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)
\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)
\(\Leftrightarrow-14x-9=5\)
\(\Leftrightarrow-14x=14\)
\(\Leftrightarrow x=-1\)
Vậy....
b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)
\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)
\(\Leftrightarrow-10+17x=-44\)
\(\Leftrightarrow17x=-34\)
\(\Leftrightarrow x=-2\)
Vậy....
c, \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)
\(\Leftrightarrow10x+10=30\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy....
d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)
\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)
\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)
\(\Leftrightarrow14x-3=7\)
\(\Leftrightarrow14x=10\)
\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)
Vậy...
a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0
\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)
\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)
\(-5x-8=0\)
\(x=-\frac{8}{5}\)
Mai mik làm mấy bài kia sau
c) \(5^{x+2}+5^x=650\)
\(\Leftrightarrow 5^x(5^2+1)=650\)
\(\Leftrightarrow 5^x.26=650\)
\(\Rightarrow 5^x=25=5^2\Rightarrow x=2\)
d) \(81^x=(-3)^7\)
Ta thấy \(81^x>0, \forall x\in\mathbb{R}\)
\((-3)^7<0\)
Do đó pt đã cho vô nghiệm.
Lời giải:
a) \((2x-1)^3=(2x-1)^4\)
\(\Leftrightarrow (2x-1)^4-(2x-1)^3=0\)
\(\Leftrightarrow (2x-1)^3[(2x-1)-1]=0\)
\(\Leftrightarrow (2x-1)^3(2x-2)=0\)
\(\Rightarrow \left[\begin{matrix} 2x-1=0\\ 2x-2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=1\end{matrix}\right.\)
b) \(2017^{x+2}=(2018-5^3)^{x+2}\)
\(\Rightarrow \left[\begin{matrix} x+2=0(1)\\ 2017=2018-5^3(2)\end{matrix}\right.\)
(1)\(\Rightarrow x=-2\)
(2): hiển nhiên vô lý
Vậy pt có nghiệm $x=-2$