a, x (2x-1 ) (3x -126) =0
b, ( 2x -4) ^3 = 8
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19 tháng 2 2021
a)(3x-1)(4x-8)=0
⇔3x-1=0 hoặc 4x-8=0
1.3x-1=0⇔3x=1⇔x=1/3
2.4x-8=0⇔4x=8⇔x=2
phương trình có 2 nghiệm:x=1/3 và x=2
b)(x-2)(1-3x)=0
⇔x-2=0 hoặc 1-3x=0
1.x-2=0⇔x=2
2.1-3x=0⇔-3x=1⇔x=-1/3
phương trình có 2 nghiệm:x=2 và x=-1/3
c)(x-3)(x+4)-(x-3)(2x-1)=0
⇔(x+4)(2x-1)=0
⇔x+4=0 hoặc 2x-1=0
1.x+4=0⇔x=-4
2.2x-1=0⇔2x=1⇔x=1/2
phương trình có hai nghiệm:x=-4 và x=1/2
d)(x+1)(x+2)=2x(x+2)
⇔(x+1)(x+2)-2x(x+2)=0
⇔2x(x+1)=0
⇔2x=0 hoặc x+1=0
1.2x=0⇔x=0
2.x+1=0⇔x=-1
phương trình có 2 nghiệm:x=0 và x=-1
5 tháng 2 2022
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
4 tháng 2 2021
\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{5;-2\right\}\)
\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)
Câu d xem lại đề
HP
26 tháng 10 2021
a. 3x2 - 2x - 1 = 0
<=> 3x2 - 3x + x - 1 = 0
<=> 3x(x - 1) + (x - 1) = 0
<=> (3x + 1)(x - 1) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
b. \(\dfrac{x+1}{3}+\dfrac{2x+3}{5}=\dfrac{3}{4}\)
<=> \(\dfrac{20\left(x+1\right)}{60}+\dfrac{12\left(2x+3\right)}{60}=\dfrac{45}{60}\)
<=> 20x + 20 + 24x + 36 = 45
<=> 44x = -11
<=> x = \(-\dfrac{1}{4}\)
26 tháng 10 2021
a) \(3x^2-2x-1=0\) \(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) Pt\(\Rightarrow\)\(5\cdot4\left(x+1\right)+3\cdot4\cdot\left(2x+3\right)=3\cdot3\cdot5\)
\(\Leftrightarrow44x=-11\Rightarrow x=-\dfrac{1}{4}\)
10 tháng 12 2021
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
7 tháng 10 2021
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2021
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
19 tháng 4 2023
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
VD
21 tháng 3 2022
\(a,\left(2x-3\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=-2\end{matrix}\right.\\ b,2x-\left(3-5x\right)=4\left(x+3\right)\\ \Leftrightarrow2x-3+5x=4x+12\\ \Leftrightarrow7x-3-4x-12=0\\ \Leftrightarrow3x-15=0\\ \Leftrightarrow x=5\)
\(c,ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\dfrac{1}{x-2}-\dfrac{2}{x+1}=\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}-\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}-\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x+1-x+2-11+3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow3x-8=0\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)
