tim x biết x(x+1)(x-6)-x3 = 5x
bài 1: tim x, biết
a,x.(x - 2) + x - 2 = 0
b,x3 + x + x + 1 = 0
c,5x.(x - 4) = 2x + 8
d,(5x - 4)2 - 49x2 = 0
a,x(x-2)+x-2=0
⇔ (x-2)(x+1)=0
⇔ x=2;x=-1
b,x3+x2+x+1=0
⇔ x2(x+1)+x+1=0
⇔ (x+1)(x2+1)=0
⇔ x=-1
Tim X
3) -12 + (2x – 9) + x= 0
4) 11 + (15 - x) = 1
5) 4 - (27 - 3) = x - (13 - 4)
6) 8 - (x - 10) = 23 - (- 4 +12)
7) 105 – 5(10 – 5x) = -20
8) (x -1)(8-2x)(3x+123) = 0
9) (x2 - 25)(x+ 10) = 0
10) x(x2+5) =
3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)
4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)
5)
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
\(3.-12+\left(2x-9\right)+x=0.\)
\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)
Vậy \(x=7.\)
\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)
Vậy \(x=25.\)
\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)
Vậy \(x=-11.\)
\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)
Vậy \(x=3.\)
\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)
Vậy \(x=-3.\)
\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)
Vậy \(x\in\left\{1;4;-41\right\}.\)
\(9.\left(x^2-25\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)
Vậy \(x\in\left\{5;-5;-10\right\}.\)
\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)
Tìm x , biết :
a) ( 3x -1 ) (2x+7) - ( x +1) (6x-5 ) = 16
b) ( 10x +9 )x - ( 5x -1 ) (2x+3 )= 8
c) ( 3x - 5 ) ( 7- 5x ) + ( 5x +2 )( 3x-2 ) -2 = 0
d) x(x + 1) ( x+6 ) - x3 = 5x
a/ pt đãcho tương đương với
6x\(^2\)+ 21x -2x-7-6x+5x-6x+5= 16
<=>18x=18
=> x=1
b/ pt đã cho tương đương với
10x\(^2\)+9x-10x\(^2\)-15x+2x+3= 8
<=> -4x=5
<=.> x=-\(\frac{5}{4}\)
c/ pt đã cho tương đương với
21x-15x\(^2\)-35+25x+15x\(^2\)-10x+6x-4-2=0
<=>42x=41
<=> x= \(\frac{41}{42}\)
d/ pt đã cho tương đương với
( x\(^2\)+x )(x+6)-x\(^3\)=5x
<=> x\(^3\)+6x\(^2\)+x\(^2\)+6x-x\(^3\)=5x
<=> 8x\(^2\)+6x-5x=0
<=>8x\(^2\)+16x-10x-5x=0
<=> (x+2)2x-5(x+2)=0
<=> (x+2)(2x-5)=0
<=>x+2=0 hoặc 2x+5=0
=> x=-2 hoặc x= -\(\frac{5}{2}\)
x(x + 1)(x+ 6) _ x3 = 5x
\(x\left(x+1\right)\left(x+6\right)-x^3=5x\\ \Rightarrow x\left(x^2+7x+6\right)-x^3=5x\\ \Rightarrow x^3+7x^2+6-x^3-5x=0\\ \Rightarrow7x^2-5x+6=0\\ \)
đến đây nghiệm xấu lắm bạn xem lại đề đúng không
Ta có: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)
\(\Leftrightarrow x\left(x^2+7x+6\right)-x^3-5x=0\)
\(\Leftrightarrow x^3+7x^2+6x-x^3-5x=0\)
\(\Leftrightarrow7x^2+x=0\)
\(\Leftrightarrow x\left(7x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{7}\end{matrix}\right.\)
Tìm x, biết:
a) 2(5x-8)-3(4x-5) = 4(3x-4) + 11;
b) 2 x ( 6 x - 2 x 2 ) + 3 x 2 ( x - 4 ) = 8;
c) 2 ( x 3 - 1 ) - 2 x 2 ( x + 2 x 4 ) + ( 4 x 5 + 4 ) x = 6;
d)(2x)2(4x-2)-(x3 -8x2) = 15.
a) x = 2 7 b) x = 2.
c) x = 2 d) x = 1.
Tìm x biết:
a) 5x(x – 2) + 3x – 6 = 0
b) x 3 – 9 x = 0
Tìm x biết:
a) (x+5).(2x+1)=0
b) x.(x+2)-3.(x+2)=0
c) 2x.(x-5)-x.(3+2x)=26
d) x2-10x-8x+16=0
e) x2-10x=25
f) 5x.(x-1)=x-1
g) 2.(x+5)-x2-5x=0
h) x2+5x-6=0
i) (2x-3)2-4.(x+1).(x-1)=49
j) x3+x2+x+1=0
k) x3-x2=4x2-8x+4
Mn ơi giúp em vs ạ,em cảm ơn trc ạ
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
tìm x biết:
a)x2 + 3x = 0 b) x3 – 4x = 0
c) 5x(x-1) = x-1 d) 2(x+5) - x2-5x = 0
e) 2x(x-5)-x(3+2x)=26 f) 5x.(x – 2012) – x + 2012 = 0
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)