tìm x
a. 70-5(x-3)=33.5
b, 2x-1=32016x-2016
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a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)
\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)
b) \(\left(x+3\right)^2-5x-15=0\)
\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+3-5\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
c) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
\(2\left(2x-5\right)^2-\left(2x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=70\)
\(2\left(4x^2-20x-25\right)-\left(4x^2-12x+9\right)-\left(4x^2-1\right)=70\)
\(8x^2-40x-50-4x^2+12x-9-4x^2+1=70\)
\(-28x-128=0\)
\(28x=-128\)
\(x=\frac{32}{7}\)
a, 3 - 2x = 3 . (5 - x) + 4
3 - 2x = 15 - 3x + 4
-2x + 3x = 15 + 4 - 3
x = 16
b, 4 - (7x + 2017) = 6 . (5 - x) - 2017
4 - 7x - 2017 = 30 - 6x - 2017
-7x + 6x = 30 - 2017 - 4 + 2017
-x = 26
x = -26
c, 15 - x . (x + 1) = 4 - x^2 + 2x
15 - x^2 - x = 4 - x^2 + 2x
-x^2 - x + x^2 - 2x = 4 - 15
-3x = -11
x = 11/3
d, -4 . (x - 5) + 2016 = 3 . (8 - x) - (2x - 2016)
-4x + 20 + 2016 = 24 - 3x - 2x + 2016
-4x + 3x +2x = 24 + 2016 - 20 - 2016
x = 4
đúng 100%
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
a) TH1: 2x - 1 > 0 => x > 1 nên ta có :
2x - 1 = x - 5
=> 2x - x = 1 - 5
=>x = -4 ( không thỏa mãn vì x phải > 1 )
TH2 : 2x - 1 < 0 = - ( 2x - 1 ) => x < 1 nên ta có :
- 2x + 1 = x - 5
=> - 2x - x = -1 - 5
=> -3 x = - 6
=> x = 2 ( không thỏa mãn )
Vậy phương trình trên vô nghiệm
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
a: =>x^2-25-x^2-3x=10
=>-3x=35
=>x=-35/3
b: =>4x^2-9-4(x^2+4x+4)=5
=>4x^2-9-4x^2-16x-16-5=0
=>-16x-30=0
=>x=-15/8
c: =>9x^2+45x-9x^2+4=7
=>45x=3
=>x=1/15
d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8
=>8x=7
=>x=7/8
70-5.(x-3)=33.5
70-5.(x-3)=27.5
70-5.(x-3)=135
5.(x-3)=70-135
5.(x-3)=-65
x-3=-65:5
x-3=-13
x=-13+3
x=-10
b)2x-1=32016x-2016
....
70-5.(x-3)=33.5
70-5.(x-3)=27.5
70-5.(x-3)=135
5.(x-3)=70-135
5.(x-3)=-65
x-3=-65:5
x-3=-13
x=-13+3
x=-10
sorry nha câu b mình 0 bít