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18 tháng 11 2019
Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10
DH
22 tháng 6 2017
a, \(6x\left(x-5\right)-\left(2x+7\right)\left(3x-2\right)=0\)
\(\Rightarrow6x^2-30x-\left(6x^2-4x+21x-14\right)=0\)
\(\Rightarrow6x^2-30x-6x^2+4x-21x+14=0\)
\(\Rightarrow-47x=-14\Rightarrow x=\dfrac{14}{47}\)
b, \(\left(8x+3\right)x-\left(4x+1\right)\left(2x-7\right)=5\)
\(\Rightarrow8x^2+3x-\left(8x^2-28x+2x-7\right)=5\)
\(\Rightarrow8x^2+3x-8x^2+28x-2x+7=5\)
\(\Rightarrow29x=5-7\Rightarrow29x=-2\)
\(\Rightarrow x=\dfrac{-2}{29}\)
c, \(\left(2x-3\right)\left(2x+3\right)-x.\left(4x+1\right)=1\)
\(\Rightarrow\left(2x\right)^2-9-4x^2-x=1\)
\(\Rightarrow4x^2-4x^2-x=1+9\)
\(\Rightarrow-x=10\Rightarrow x=-10\)
Chúc bạn học tốt!!!
22 tháng 6 2017
a ) \(6x\left(x-5\right)-\left(2x+7\right)\left(3x-2\right)=0\)
\(\Leftrightarrow6x^2-30x-\left(6x^2-4x+21x-14\right)=0\)
\(\Leftrightarrow6x^2-30x-6x^2+4x-21x+14=0\)
\(\Leftrightarrow-47x+14=0\)
\(\Leftrightarrow x=\dfrac{14}{47}.\)
Vậy ..........................................
b ) \(\left(8x+3\right).x-\left(4x+1\right)\left(2x-7\right)=5\)
\(\Leftrightarrow8x^2+3x-\left(8x^2-28x+2x-7\right)=5\)
\(\Leftrightarrow8x^2+3x-8x^2+28x-2x+7=5\)
\(\Leftrightarrow29x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{29}.\)
Vậy.............
c ) \(\left(2x-3\right)\left(2x+3\right)-x\left(4x+1\right)=1\)
\(\Leftrightarrow4x^2-9-4x^2-x=1\)
\(\Leftrightarrow-x=10\)
\(\Leftrightarrow x=10\)
Vậy..........
KN
28 tháng 7 2015
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
3 tháng 8 2019
\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)
\(x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\)
\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
3 tháng 8 2019
a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)
b) \(x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
c) \(x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)
e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)
g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
12 tháng 8 2016
a) x2 - 5x - 6 = 0
=> x2 - 2x - 3x - 6 = 0
=> (x2 - 2x) + (-3x - 6) = 0
=> x(x - 2) - 3 (x - 2) = 0
=> (x - 2) (x - 3) = 0
=> x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546
BA
19 tháng 6 2019
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
BA
19 tháng 6 2019
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
\(\Leftrightarrow4x^2+12x-5x-15-\left(14x+4x^2-21-6x\right)=0\)
\(\Leftrightarrow4x^2+7x-15-\left(4x^2+8x-21\right)=0\)
\(\Leftrightarrow-x+6=0\)
\(\Leftrightarrow x=6\)