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Khách
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AP
1
K
14 tháng 3 2023
`x(x+3) - (2x-1) . (x+3) = 0`
`<=>(x+3)(x-2x+1)=0`
`<=>(x+3)(-x+1)=0`
`** x+3=0`
`<=>x=-3`
`** -x+1=0`
`<=>x=1`
`x(x-3) - 5 (x-3) = 0`
`<=>(x-3)(x-5)=0`
`** x-3=0`
`<=>x=3`
`** x-5=0`
`<=>x=5`
`3x + 12 = 0`
`<=>3x=-12`
`<=> x=-4`
`2x (x-2) + 5 (x-2) = 0`
`<=>(x-2)(2x+5)=0`
`** x-2=0`
`<=>x=2`
`** 2x+5=0`
`<=> x= -5/2`
13 tháng 3 2016
bai 1
1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0
<=>(2x)^2-5^2=0
<=>(2x+5)*(2x-5)=0
<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự
DD
8 tháng 7 2017
\(1.\left(x-2\right)\left(x-1\right)=x\left(2x+1\right)+2\)
\(\Leftrightarrow x^2-3x+2=2x^2+x+2\)
\(\Leftrightarrow x^2-2x^2-3x-x=-2+2\)
\(\Leftrightarrow-x^2-4x=0\)
\(\Leftrightarrow x\left(-x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\-x-4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)Vậy S={-4;0}
\(2.\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)=8x\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2-8x=0\)
\(\Leftrightarrow x^2+4x+4-\left(x^2-4x+4\right)-8x=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4-8x=0\)
\(\Leftrightarrow0=0\)(luôn đúng vs mọi giá trị của x)
\(3.\left(2x-1\right)\left(x^3-x+1\right)=2x^3-3x^2+16=0\)
\(\Leftrightarrow2x^4-2x^2+2x-x^3+x-1=2x^3-3x^2+16=0\)
\(\Leftrightarrow2x^4-x^3-2x^2+3x-1=2x^3-3x^2+16=0\)
\(\Leftrightarrow2x^4-x^3-2x^3-2x^2+3x^2+3x-1-16=0\)
\(\Leftrightarrow2x^4-3x^3+x^2+3x-17=0\)
Cái này là phương trình bậc 4 lận, Giải hơi mất thời gian
15 tháng 3 2020
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
15 tháng 3 2020
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
3 tháng 8 2020
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
3 tháng 8 2020
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
10 tháng 2 2020
1) Ta có: 3x-12=5x(x-4)
\(\Leftrightarrow3x-12-5x\left(x-4\right)=0\)
\(\Leftrightarrow3x-12-5x^2+20x=0\)
\(\Leftrightarrow-5x^2+23x-12=0\)
\(\Leftrightarrow-5x^2+20x+3x-12=0\)
\(\Leftrightarrow\left(-5x^2+20x\right)+\left(3x-12\right)=0\)
\(\Leftrightarrow5x\left(-x+4\right)+3\left(x-4\right)=0\)
\(\Leftrightarrow5x\left(4-x\right)-3\left(4-x\right)=0\)
\(\Leftrightarrow\left(4-x\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\5x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{3}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;\frac{3}{5}\right\}\)
2) Ta có: 3x-15=2x(x-5)
\(\Leftrightarrow3x-15-2x\left(x-5\right)=0\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{3}{2}\right\}\)
3) Ta có: 3x(2x-3)+2(2x-3)=0
\(\Leftrightarrow\left(2x-3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}\)
4) Ta có: (4x-6)(3-3x)=0
\(\Leftrightarrow\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{4}=\frac{3}{2}\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};1\right\}\)
10 tháng 2 2020
4) (4x - 6 ) ( 3 - 3x ) = 0
<=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)
29 tháng 8 2021
\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)
\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
29 tháng 8 2021
a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)
Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)
b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy ...
d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
29 tháng 9 2018
\(2x^3-50x=0\)
<=> \(2x\left(x^2-25\right)=0\)
<=> \(2x\left(x-5\right)\left(x+5\right)=0\)
đến đây
bạn tự giải nhé
hk tốt
(4x - 3)2 - (2x + 1)2 = 0
\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0
\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
3x - 12 - 5x(x - 4) = 0
\(\Leftrightarrow\) 3x - 12 - 5x2 + 20x = 0
\(\Leftrightarrow\) -5x2 + 23x - 12 = 0
\(\Leftrightarrow\) 5x2 - 23x + 12 = 0
\(\Leftrightarrow\) 5x2 - 20x - 3x + 12 = 0
\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0
\(\Leftrightarrow\) (x - 4)(5x - 3) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy ...
(8x + 2)(x2 + 5)(x2 - 4) = 0
\(\Leftrightarrow\) (8x + 2)(x2 + 5)(x - 2)(x + 2) = 0
Vì x2 \(\ge\) 0 \(\forall\) x nên x2 + 5 > 0 \(\forall\) x
\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
Chúc bn học tốt!
a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)
b) Ta có: \(3x-12-5x\left(x-4\right)=0\)
\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)
c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)
mà \(2>0\)
và \(x^2+5>0\forall x\)
nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)