Tìm X biết:
a/ 5x (x - 2000) – x + 2000 = 0
b/ x3 – 13x = 0
c/ 3x2 - 6x =0
d/ x(x- 5) +3(x-5)=0
e/9x2 – 4 =0
giúp mình vs thx mn
thực hiện phép tính
a) 5x(3x2-4x+2)
b)(x2-3x)(3x2-x+4)
2) tìm x biết
a) x3-6x2+12x=0
b)x3+9x2+27x+27=0
giúp mình mn ơi
thực hiện phép tính
a) 5x(3x2-4x+2)
b)(x2-3x)(3x2-x+4)
2) tìm x biết
a) x3-6x2+12x=0
b)x3+9x2+27x+27=0
giúp mình mn ơi
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
tìm x bt
a)x(x-5)-4x+20=0
b)x(x+6)-7x-42=0
c)x^3-5x^2-x+5=0
d)4x^2-25-(2x-5)(3x+7)=0
e)x^3+27+(x+3)(x-9)=0
giúp mk vs ah!!1
a) x(x - 5) - 4x + 20 = 0
\(\Leftrightarrow\) x(x - 5) - (4x + 20)
\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x - 4)
Khi x - 5 = 0 hoặc x - 4 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 4
Vậy S = \(\left\{5;4\right\}\)
b) x(x + 6) - 7x - 42 = 0
\(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0
\(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0
\(\Leftrightarrow\) (x + 6)(x - 7) = 0
Khi x - 6 = 0 hoặc x - 7 = 0
\(\Leftrightarrow\) x = 6 \(\Leftrightarrow\) x = 7
Vậy S = \(\left\{6;7\right\}\)
c) x3 - 5x2 - x + 5 = 0
\(\Leftrightarrow\) (x3 - 5x2) - (x + 5) = 0
\(\Leftrightarrow\) x2 (x - 5) - (x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x2 - 1) = 0
\(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0
Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 1 \(\Leftrightarrow\) x = -1
Vậy S = \(\left\{5;1;-1\right\}\)
d) 4x2 - 25 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x)2 - 52 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0
\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0
Khi 2x - 5 = 0 hoặc -x + 12 = 0
\(\Leftrightarrow\) 2x = 5 \(\Leftrightarrow\) -x = -12
\(\Leftrightarrow\) x = \(\dfrac{5}{2}\) \(\Leftrightarrow\) x = 12
Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)
e) x3 + 27 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) x3 - 33 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0
\(\Leftrightarrow\) (x - 3) ( x2 - 3x + 9 + x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 2x) = 0
\(\Leftrightarrow\) (x - 3)x(x - 2)
Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x = 3 \(\Leftrightarrow\) x = 2
Vậy S = \(\left\{3;0;2\right\}\)
Chúc bạn học tốt
a) Ta có: \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
b) Ta có: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
c) Ta có: \(x^3-5x^2-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
Giải các phương trình sau:
a, x2 - 9x +20 = 0
b, x2 - 3x - 18 = 0
c, 2x2 - 9 x + 9 = 0
d, 3x2 - 8x + 4 = 0
e, 3x3 - 6x2 - 9x = 0
f, x(x - 5) - 2 + x = 0
g, x3 + 32 + 6x +8 = 0
h, 2x(x - 2) - 2 + x = 0
i, 5x(1 - x) + x - 1 = 0
k, 4 - 9(x - 1)2 = 0
l, (x - 2)2 - 36(x + 3)2 = 0
\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)
d: \(\Leftrightarrow3x^2-6x-2x+4=0\)
=>(x-2)(3x-2)=0
=>x=2 hoặc x=2/3
e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)
=>x(x-3)(x+1)=0
hay \(x\in\left\{0;3;-1\right\}\)
f: \(\Leftrightarrow x^2-5x-2+x=0\)
\(\Leftrightarrow x^2-4x-2=0\)
\(\Leftrightarrow\left(x-2\right)^2=6\)
hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)
tìm x, biết:
a) 9x2+36=0
b) 3(x+4)-x2-4x=0
c) x(2x-1)-(x-2)(2x+1)=0
d) (2x-3)2-4x2=00
e)1 phần 3.x2-3x=0
f) x3-x2-x+1=0
ráng giúp mình nha
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
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
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Bài 5: Tìm x (Giải phương trinh)
a)x^3-13x=0
b) 5x(x – 2000) – x + 2000 = 0
c) 2x(x – 2) + 3(x – 2) = 0
d) x + 1 = (x + 1)2
e) x + 5x2 = 0
f) x3 + x = 0
Bài 5: Tìm x (Giải phương trình)
a)x^3-13x=0 b) 5x(x – 2000) – x + 2000 = 0
c) 2x(x – 2) + 3(x – 2) = 0 d) x + 5x2 = 0
d) x + 1 = (x + 1)2 e) x3 + x = 0
b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) Ta có: \(2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)
d) Ta có: \(5x^2+x=0\)
\(\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)
tìm x
a)x^2-x-12=0
b)x^2+3x-18=0
c)8x^2+30x+7=0
d)x^3-11x^2+30x=0
e)x^3-7x^2+15x-25=0
giúp mk vs ah!!!!!
a) Ta có: \(x^2-x-12=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
b) Ta có: \(x^2+3x-18=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)
Tìm x∈Z, biết:
a)x.(x-6)=0
b)(-7-x).(-x+5)=0
c)(x+3).(x-7)=0
d)(x-3).(x2+12)=0
e)(x+1).(2-x) ≥0
f)(x-3).(x-5) ≤0
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)