giải phương trình sau:
(2x+5)^2=(x+2)^2
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16 tháng 2 2022
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
DT
16 tháng 2 2022
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
Y
12 tháng 4 2022
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
Y
12 tháng 4 2022
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
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
ND
0
AH
2
20 tháng 2 2022
a, ĐKXĐ:\(x\ne-5\)
\(\dfrac{2x-5}{x+5}=3\\ \Rightarrow2x-5=3\left(x+5\right)\\ \Leftrightarrow3x+15-2x+5=0\\ \Leftrightarrow x+20=0\\ \Leftrightarrow x=-20\)
b, ĐKXĐ:\(x\ne3\)
\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\\ \Rightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x^2-x-6=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\\ \Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\\ \Leftrightarrow x\left(\dfrac{x+1}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x-3}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4}{2\left(x+1\right)\left(x-3\right)}\right)=0\\ \Leftrightarrow x.\dfrac{x+1+x-3-4}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(2x-6\right)}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x}{x+1}=0\\ \Rightarrow x=0\left(tm\right)\)
21 tháng 3 2021
\(\dfrac{1}{x+2}\)+\(\dfrac{5}{x-2}\)=\(\dfrac{2x-12}{x^2-4}\)
(đkxđ: x≠2, x≠-2)
⇔ \(\dfrac{x-2}{x^2-4}\)+\(\dfrac{5\left(x+2\right)}{x^2-4}\)= \(\dfrac{2x-12}{x^2-4}\)
⇔ x-2+5(x+2)=2x-12
⇔ x-2+5x+10=2x-12
⇔ 4x=-20
⇔ x=-5(tm)
LA
24 tháng 3 2017
a/ 4x + 20 = 0
⇔4x = -20
⇔x = -5
Vậy phương trình có tập nghiệm S = {-5}
b/ 2x – 3 = 3(x – 1) + x + 2
⇔ 2x-3 = 3x -3+x+2
⇔2x – 3x = -3+2+3
⇔-2x = 2
⇔x = -1
Vậy phương trình có tập nghiệm S = {-1}
LA
24 tháng 3 2017
câu tiếp theo
a/ (3x – 2)(4x + 5) = 0
3x – 2 = 0 hoặc 4x + 5 = 0
- 3x – 2 = 0 => x = 3/2
- 4x + 5 = 0 => x = – 5/4
Vậy phương trình có tập nghiệm S= {-5/4,3/2}
b/ 2x(x – 3) – 5(x – 3) = 0
=> (x – 3)(2x -5) = 0
=> x – 3 = 0 hoặc 2x – 5 = 0
* x – 3 = 0 => x = 3
* 2x – 5 = 0 => x = 5/2
Vậy phương trình có tập nghiệm S = {0, 5/2}
29 tháng 8 2021
a: Ta có: \(3x-5\ge2\left(x-6\right)-12\)
\(\Leftrightarrow3x-5\ge2x-24\)
hay \(x\ge-19\)
b: Ta có: \(2\left(5-2x\right)\ge3-x\)
\(\Leftrightarrow10-4x-3+x\ge0\)
\(\Leftrightarrow-3x\ge-7\)
hay \(x\le\dfrac{7}{3}\)
PT
0
\(\left(2x+5\right)^2=\left(x+2\right)^2\)
<=>\(2x+5=x+2\)
<=>\(2x-x=2-5\)
<=>\(x=-3\)