Tìm x, biết
a) (x:3-3) (x:6-6)=0
b) 4x + 5x -x =128
K
Khách
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K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
19 tháng 10 2021
\(a,x\left(x+9\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Rightarrow x\left(x^2+4x+4\right)=0\\ \Rightarrow x\left(x+2\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(x-5-4\right)\left(x-5+4\right)=0\\ \Rightarrow\left(x-9\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\\ d,\Rightarrow3\left(x+2\right)-x\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ e,\Rightarrow x^3+6x^2+12x+8-x^3-6x^2=4\\ \Rightarrow12x=-4\Rightarrow x=-\dfrac{1}{3}\\ g,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
N
1
9 tháng 2 2022
a: \(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x-1\right)=0\)
=>x=-1 hoặc x=1
b: \(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
hay \(x\in\left\{-1;2;-2\right\}\)
c: \(x^3+x^2+4=0\)
\(\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\)
\(\Leftrightarrow\left(x+2\right)\cdot\left(x^2-x+2\right)=0\)
=>x+2=0
hay x=-2
e: \(\Leftrightarrow x^4-2x^3-3x^3+6x^2-x^2+2x+3x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-3x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)=0\)
hay \(x\in\left\{2;3;-1;1\right\}\)
30 tháng 7 2021
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
TC
30 tháng 7 2021
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
N
giải các Phương trình sau
a) (5x+3)(x2+1)(x-1)=0
b) (4x-1)(x-3)-(x-3)(5x+2)=0
c) (x+6)(3x-1)+x2-36 =0
1
23 tháng 2 2023
a: =>(5x+3)(x-1)=0
=>x=1 hoặc x=-3/5
b: =>(x-3)(4x-1-5x-2)=0
=>(x-3)(-x-3)=0
=>x=-3 hoặc x=3
c: =>(x+6)(3x-1+x-6)=0
=>(x+6)(4x-7)=0
=>x=7/4 hoặc x=-6
13 tháng 7 2021
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
13 tháng 7 2021
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: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
20 tháng 10 2021
\(a,\Leftrightarrow\left(x-4\right)\left(x^2+5\right)>0\\ \Leftrightarrow x-4>0\left(x^2+5\ge5>0\right)\\ \Leftrightarrow x>4\\ b,\Leftrightarrow\left(x-y\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=y\left(vô.lí.do.x\ne y\right)\\x=\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow S=x^2-x=\dfrac{25}{9}-\dfrac{5}{3}=\dfrac{10}{9}\)
S
22 tháng 3 2021
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
22 tháng 3 2021
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
a. (x : 3 - 3)(x : 6 - 6) = 0
<=> \(\orbr{\begin{cases}x:3-3=0\\x:6-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=9\\x=36\end{cases}}\)
Vậy ...
b. 4x + 5x - x = 128
<=> x(4 + 5 - 1) = 128
<=> 8x = 128
<=> x = 16
a) (x:3-3) (x:6-6)=0
Vì (x:3-3) (x:6-6)=0
=> (x:3- 3) = 0 hoặc (x:6-6)=0
x : 3 = 3 x : 6 =6
x = 1 x = 1
Vậy x = 1
b) 4x + 5x -x =128
=> 8x = 128
=> x = 16
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