Tìm số nguyên x
a) x3 - 25x=0
b) (x2-81).(18-3x)=0
c)|2x-5|=|x+3|
Bài 5; Tìm x
a) x2-4=0
b) 2x(x+5)-3(5+x)=0
c) x3-6x2+11x-6=0
a) x² - 4 = 0
x² = 4
x = 2 hoặc x = -2
b) 2x(x + 5) - 3(5 + x) = 0
(x + 5)(2x - 3) = 0
X + 5 = 0 hoặc 2x - 3 = 0
*) x + 5 = 0
x = -5
*) 2x - 3 = 0
2x = 3
x = 3/2
c) x³ - 6x² + 11x - 6 = 0
x³ - x² - 5x² + 5x + 6x - 6 = 0
(x³ - x²) - (5x² - 5x) + (6x - 6) = 0
x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x² - 2x - 3x + 6) = 0
(x - 1)[(x² - 2x) - (3x - 6)] = 0
(x - 1)[x(x - 2) - 3(x - 2)] = 0
(x - 1)(x - 2)(x - 3) = 0
x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0
*) x - 1 = 0
x = 1
*) x - 2 = 0
x = 2
*) x - 3 = 0
x = 3
Vậy x = 1; x = 2; x = 3
Tìm x
a) x2 + 3x - ( 2x+ 6) = 0
b) 5x+ 20- x2- 4x =0
c) 3x2- 3x+ 2x3-2x2= 0
d) x3+ 27= -x2+9
a: ta có: \(x^2+3x-\left(2x+6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
b: Ta có: \(5x+20-x^2-4x=0\)
\(\Leftrightarrow\left(x+4\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\)
Bài 2: tìm x
a) 3x2-x=0
b) x2-25x=0
c) 7x2+2x=0
d) 4x(x-2007)-x+2007
\(a,\Leftrightarrow x\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x-25\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\\ c,\Leftrightarrow x\left(7x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{7}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-2007\right)\left(4x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2007\\x=\dfrac{1}{4}\end{matrix}\right.\)
Tìm x
a, 3/4x*(x2-9)=0
b, x3-16x=0
c, (x-1)(x+2)-x-2=0
d, 3x3-27x=0
e, x2(x+1)+2x(x+1)=0
f, x(2x-3)-2(3-2x)=0
c: =>(x-1)(x+1)=0
hay \(x\in\left\{1;-1\right\}\)
a,
\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0
\(\left\{{}\begin{matrix}\dfrac{3}{4x}=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{3,-3\right\}\)
b,
\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)
\(\left\{{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
=>\(x=\left\{-4,0,4\right\}\)
d,
\(3x^3-27x=0\\ 3x\left(x^2-9\right)=0\\ 3x\left(x-3\right)\left(x+3\right)=0\)
\(\left\{{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{-3,0,3\right\}\)
e,
\(x^2+\left(x+1\right)+2x\left(x+1\right)=0\\ x\left(x+1\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(x=\left\{-2,-1,0\right\}\)
f,
\(x\left(2x-3\right)-2\left(3-2x\right)=0\\ \left(2x-3\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
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 các số thực x, biết:
a) (2x-3)2-49=0
b) 2x(x-5)-7(5-x)=0
c) x2-3x-10=0
a: \(\left(2x-3\right)^2-49=0\)
\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a. (2x - 3)2 - 49 = 0
<=> (2x - 3)2 - 72 = 0
<=> (2x - 3 + 7)(2x - 3 - 7) = 0
<=> (2x + 4)(2x - 10) = 0
<=> \(\left[{}\begin{matrix}2x+4=0\\2x-10=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
b. 2x(x - 5) - 7(5 - x) = 0
<=> 2x(x - 5) + 7(x - 5) = 0
<=> (2x + 7)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
c. x2 - 3x - 10 = 0
<=> x2 - 5x + 2x - 10 = 0
<=> x(x - 5) + 2(x - 5) = 0
<=> (x + 2)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a, (2x - 3)2 - 49 = 0
(2x - 3)2 - 72 = 0
(2x - 3 + 7)( 2x - 3 - 7) = 0
(2x + 4)( 2x - 10) = 0
=> 2x + 4 = 0 => 2x - 10 = 0
2x = - 4 2x = 10
x = - 2 x = 5
Tìm các số nguyên x, biết:
a, (22 + 5)(x2 + 25) = 0
b, (x2 + 7)(x2 - 49) < 0
c, (x2 - 7)(x2 - 49) < 0
d, (x2 - 36)(x2 - 81) ≤ 0
Tìm x:
a)(3x+5).(7-2x)+6x.(x+4)=0
b)x3-25x=0
a) \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)
\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)
\(\Leftrightarrow35x=-35\Leftrightarrow x=-1\)
b) \(x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
a: Ta có: \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)
\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)
\(\Leftrightarrow x=1\)
b: Ta có: \(x^3-25x=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
a. (3x + 5)(7 - 2x) + 6x(x + 4) = 0
<=> 21x - 6x2 + 35 - 10x + 6x2 + 24x = 0
<=> -6x2 + 6x2 + 21x - 10x + 24x = -35
<=> 35x = -35
<=> x = \(\dfrac{-35}{35}=-1\)
b. x3 - 25x = 0
<=> x(x2 - 52)
<=> x(x + 5)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x=0\\x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right.\)
Tìm x,biết:
a)(2x-3)2-49=0
b)2x.(x-5)-7.(5-x)=0
c)x2-3x-10=0
a) \(\Rightarrow\left(2x-3\right)^2=49\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, ⇒ (2x - 3)2 = 49
⇒ (2x - 3)2 = \(\left(\pm7\right)^2\)
⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0
⇒ (x - 5).(2x + 7) = 0
⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c, ⇒ x2 - 5x + 2x - 10 = 0
⇒ (x2 - 5x) + (2x - 10) = 0
⇒ x.(x - 5) +2.(x - 5) = 0
⇒ (x - 5).(x + 2)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Bài 5. Tìm x, biết:
a) x (2x - 7) + 4x -14 = 0
b) x3 - 9x = 0
c) 4x2 -1 - 2(2x -1)2 = 0
d) (x3 - x2 ) - 4x2 + 8x - 4 = 0
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)