Tìm x giúp em với ạ :
8x3 - 50x = 0
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17 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`(2x - 1)^2 - 25 = 0`
`<=> (2x - 1)^2 = 25`
`<=> (2x - 1)^2 = (+-5)^2`
`<=>`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, `S = {-2; 3}`
`b,`
`8x^3 - 50x = 0`
`<=> x(8x^2 - 50) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\8x^2=50\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=\pm\dfrac{5}{2}\end{matrix}\right.\)
Vậy, `S = {-5/2; 0; 5/2}.`
KV
17 tháng 7 2023
a) (2x - 1)² - 25 = 0
(2x - 1)² - 5² = 0
(2x - 1 - 5)(2x - 1 + 5) = 0
(2x - 6)(2x + 4) = 0
2x - 6 = 0 hoặc 2x + 4 = 0
*) 2x - 6 = 0
2x = 6
x = 3
*) 2x + 4 = 0
2x = -4
x = -2
Vậy x = -2; x = 3
b) 8x³ - 50x = 0
2x(4x² - 25) = 0
2x[(2x)² - 5²] = 0
2x(2x - 5)(2x + 5) = 0
2x = 0 hoặc 2x - 5 = 0 hoặc 2x + 5 = 0
*) 2x = 0
x = 0
*) 2x - 5 = 0
2x = 5
x = 5/2
*) 2x + 5 = 0
2x = -5
x = -5/2
Vậy x = -5/2; x = 0; x = 5/2
10 tháng 10 2021
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
HP
29 tháng 10 2021
d. 8x3 - 50x = 0
<=> 2x(4x - 25) = 0
<=> \(\left[{}\begin{matrix}2x=0\\4x-25=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{25}{4}\end{matrix}\right.\)
e. (4x - 3)2 - 3x(3 - 4x) = 0
<=> (4x - 3)2 + 3x(4x - 3) = 0
<=> (4x - 3)(4x - 3 + 3x) = 0
<=> (4x - 3)(7x - 3) = 0
<=> \(\left[{}\begin{matrix}4x-3=0\\7x-3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
29 tháng 10 2021
d) \(8x^3-50x=0\Rightarrow2x\left(4x^2-25\right)=0\)
\(\Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\2x+5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
e) \(\left(4x-3\right)^2-3x\left(3-4x\right)=0\)
\(\Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\)
\(\Rightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-3=0\\7x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
31 tháng 10 2021
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
31 tháng 10 2021
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
30 tháng 9 2016
a) \(\left(2x-1\right)^2-25=0\)
\(\left(2x-1\right)^2=0+25=25\)
\(\left(2x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)
b) \(8x^3-50x=0\)
\(2x\left(4x^2-25\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=0\\4x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{5}{2}\end{array}\right.\end{array}\right.\)
PT
3
DA
0
8x3-50x=0
x(8x2-50)=0
TH1: x=0 TH2: 8x2-50=0
8x2 = 50
x2 = \(\dfrac{25}{4}\)
x = + - \(\dfrac{5}{2}\)
vậy x\(\in\){0,+-\(\dfrac{5}{2}\)}
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