1) Tim x
a) 3x(x-1)+x-1=0
b)2(x+3)-x^2-3x=0
làm tương tự như bài này nè!
x(x-2)+x-2=0
=>x*((x-2)-x-3)=0
=>(x-2)(x+10=0
hoac x-3=0 =>x=3
hoac 5x-1=0 =>x=1 phan 5
vậy x = 1 phần 5 ;x=3
hoac 5x-1=0 => x=1 phan 5
K
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Những câu hỏi liên quan
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}
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=\(\dfrac{19}{24}\)
Bài 1.
20 tháng 4 2020
a)
\(\left(4x-10\right)\cdot\left(24+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{24}{5}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{5}{2};-\frac{24}{5}\right\}\)
b)
\(\left(2x-5\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{5}{2};\frac{2}{3}\right\}\)
c)
\(\left(2x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{1}{2};-\frac{1}{3}\right\}\)
d)
\(x\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{0;\frac{1}{2}\right\}\)
e) \(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\)
Do \(x^2\ge0\) Nên \(x^2+4>0\)
\(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{5}\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{-\frac{3}{5};1\right\}\)
....... Còn lại cứ cho mỗi thừa số = 0 rồi tìm x như bình thường thôi bạn
20 tháng 4 2020
1. (4x - 10)(24 + 5x) = 0
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{5}{2}\); \(\frac{-24}{5}\)}
2. (2x - 5)(3x - 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy S = {\(\frac{5}{2}\); \(\frac{2}{3}\)}
3. (2x - 1)(3x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy S = {\(\frac{1}{2}\); \(\frac{-1}{3}\)}
4. x(x2 - 1) = 0
\(\Leftrightarrow\) x(x - 1)(x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy S = {0; 1; -1}
5. (5x + 3)(x2 + 4)(x - 1) = 0
VÌ x2 + 4 > 0 với mọi x nên
\(\Rightarrow\left[{}\begin{matrix}5x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{5}\\x=1\end{matrix}\right.\)
Vậy S = {\(\frac{-3}{5}\); 1}
6. (x - 1)(x + 2)(x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy S = {1; -2; -3}
7. (x - 1)(x + 5)(-3x + 8) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\\-3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\\x=\frac{8}{3}\end{matrix}\right.\)
Vậy S = {1; -5; \(\frac{8}{3}\)}
Chúc bn học tốt!!
AT
12 tháng 10 2017
Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
12 tháng 10 2017
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
13 tháng 12 2021
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
a)\(3x\left(x-1\right)+x-1=0\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\Leftrightarrow\hept{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)
\(S=\left\{1;\frac{1}{3}\right\}\)
b)\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\Leftrightarrow\hept{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}}\)
\(S=\left\{2;-3\right\}\)