Tìm x biết:
a. 5x(x - 4) - x2 + 16 = 0
b. x + 6x2 + 9x3 = 0
c. x2 - 4x + 3 = 0
bài 1 giải các bất phương trình sau
a, -x2 +5x-6 ≥ 0
b, x2-12x +36≤0
c, -2x2 +4x-2≤0
d, x2 -2|x-3| +3x ≥ 0
e, x-|x+3| -10 ≤0
bài 2 xét dấu các biểu thức sau
a,<-x2+x-1> <6x2 -5x+1>
b, x2-x-2/ -x2+3x+4
c, x2-5x +2
d, x-< x2-x+6 /-x2 +3x+4 >
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
Bài 1: Giải các pt sau: 1) x2 + 5x + 6 = 0 2)
x2 - x - 6 = 0
3) (x2 + 1) (x2 + 4x + 4) = 0
4) x3 + x2 + x + 1 = 0
5) x2 - 7x + 6 = 0
6) 2x2 - 3x - 5 = 0
7) x2 + x - 12 = 0
8) 2x3 + 6x2 = x2 + 3x
9) (3x - 1) (x2 + 2) = (3x - 1)(7x - 10)
Bài 2: Cho biểu thức A = (5x - 3y + 1) (7x + 2y -2) a) Tìm x sao cho với y = 2 thì A = 0 b) Tìm y sao cho với x = -2 thì A = 0
Bài 1: Giải các pt sau: 1) x2 + 5x + 6 = 0
2) x2 - x - 6 = 0
3) (x2 + 1) (x2 + 4x + 4) = 0
4) x3 + x2 + x + 1 = 0
5) x2 - 7x + 6 = 0
6) 2x2 - 3x - 5 = 0
7) x2 + x - 12 = 0
8) 2x3 + 6x2 = x2 + 3x
9) (3x - 1) (x2 + 2) = (3x - 1)(7x - 10)
Bài 2: Cho biểu thức A = (5x - 3y + 1) (7x + 2y -2) a) Tìm x sao cho với y = 2 thì A = 0 b) Tìm y sao cho với x = -2 thì A = 0
Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
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) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
Tìm x biết:
a/ 5x( x- 3) = x – 3 b/ x3 - x = 0 c/ x2 – 7x + 6 = 0
d/ x2 – 4 + ( x – 2)2 = 0 e/ x2 – 16 –( x +4) = 0 f/ x2 + x – 2 = 0
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Rút gọn biểu thức:
a) M = 1 x + 2 + 2 x − 2 − 2 x x 2 − 4 , với x ≠ ± 2 ;
b) N = x 2 + 5 x + 6 x 2 + x − 12 : x 2 + 4 x + 4 x 2 − 3 x , với x ≠ 0 ; − 4 ; 2 ; 3 .
a) MTC = (x -2)(x + 2). Ta rút gọn được M = 1 x − 2
b) Gợi ý: x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) ; x 2 + x − 12 = ( x − 3 ) ( x + 4 )
Ta có N = ( x + 2 ) ( x + 3 ) ( x − 3 ) ( x + 4 ) : ( x + 2 ) 2 x ( x − 3 ) = x ( x + 3 ) ( x + 2 ) ( x + 4 )
Tìm x:
a) 36x3-4x=0
b) 3x(x-2)-2+x=0
c) (x3-x2)-4x2+8x-4=0
d) x2-6x-16=0
e) x4-6x2-7=0
a) (x – 2)(x2 + 2x + 4) – x( x2 +2) = 12 b) (x – 3)2 – (x+2)(x–2) = 16
c) x3 – 9x = 0 d) x3 – 6x2 + 9x – 54 = 0
giúp e vs ạ
\(a,\Leftrightarrow x^3-8-x^3-2x=12\Leftrightarrow-2x=20\Leftrightarrow x=-10\\ b,\Leftrightarrow x^2-6x+9-x^2+4=16\Leftrightarrow=-6x=3\Leftrightarrow x=-\dfrac{1}{2}\\ c,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-6\right)+9\left(x-6\right)=0\\ \Leftrightarrow\left(x^2+9\right)\left(x-6\right)=0\\ \Leftrightarrow x=6\left(x^2+9>0\right)\)
Hãy giải các phương trình sau đây :
1, x2 - 4x + 4 = 0
2, 2x - y = 5
3, x + 5y = - 3
4, x2 - 2x - 8 = 0
5, 6x2 - 5x - 6 = 0
6,( x2 - 2x )2 - 6 (x2 - 2x ) + 5 = 0
7, x2 - 20x + 96 = 0
8, 2x - y = 3
9, 3x + 2y = 8
10, 2x2 + 5x - 3 = 0
11, 3x - 6 = 0
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
Tìm x:
a) 36x3-4x=0
b) 3x(x-2)-2+x=0
c) (x3-x2)-4x2+8x-4=0
d) x2-6x-16=0
e) x4-6x2-7=0
(Mình cần gấp ạ)
a) Ta có: \(36x^3-4x=0\)
\(\Leftrightarrow4x\left(9x^2-1\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
b) Ta có: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
d) Ta có: \(x^2-6x-16=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
e) Ta có: \(x^4-6x^2-7=0\)
\(\Leftrightarrow\left(x^2-7\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)