3/2x + 2/3 = 1/2x + 5/2
Đề bài tim x. Giúp mình với ạ !!!!
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11 tháng 10 2021
a) \(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\)
\(\Rightarrow32\left(2x+3\right)=24\left(3x-1\right)\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow8x=120\Rightarrow x=15\)
b) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)
\(\Rightarrow17\left(13x-2\right)=76\left(2x+5\right)\)
\(\Rightarrow221x-34=152x+380\)
\(\Rightarrow69x=414\Rightarrow x=6\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 9 2023
Để olm giúp em em nhé!
a, \(\dfrac{x+2}{7x+42}\) = \(\dfrac{x+2}{7.\left(x+6\right)}\) = \(\dfrac{\left(x+2\right)\left(x-6\right)}{7\left(x-6\right)\left(x+6\right)}\) (đk \(x\ne\) \(\mp\) 6)
\(\dfrac{-13x}{x^2-36}\) = \(\dfrac{-13x}{\left(x-6\right)\left(x+6\right)}\) = \(\dfrac{-7.13.x}{7.\left(x-6\right).\left(x+6\right)}\) = \(\dfrac{-91x}{7.\left(x-6\right)\left(x+6\right)}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 9 2023
b, \(\dfrac{7}{4x+16}\) = \(\dfrac{7\left(x-4\right)}{4.\left(x+4\right).\left(x-4\right)}\) (đk \(x\ne\) \(\pm\) 4)
\(\dfrac{15}{x^2-16}\) = \(\dfrac{15.4}{\left(x-4\right)\left(x+4\right).4}\) = \(\dfrac{60}{4.\left(x-4\right).\left(x+4\right)}\)
KN
11 tháng 9 2019
\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)
\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)
\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)
KN
11 tháng 9 2019
\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)
\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)
\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)
\(\Leftrightarrow4x^2+6x-51=0\)
\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)
24 tháng 8 2023
1: =>(x+2018)(6x-3)=0
=>x+2018=0 hoặc 6x-3=0
=>x=1/2 hoặc x=-2018
2: x(x-11)+3(11-x)=0
=>(x-11)(x-3)=0
=>x=11 hoặc x=3
4: =>(x+5)(2x-4)=0
=>2x-4=0 hoặc x+5=0
=>x=2 hoặc x=-5
3: =>(x-3)(x+2)=0
=>x=3 hoặc x=-2
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
Bài 1:
\(6x\left(x+2018\right)-3\left(x+2018\right)=0\)
\(\Leftrightarrow\left(x+2018\right)\left(6x-3\right)=0\)
\(\Leftrightarrow3\left(x+2018\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
\(x\left(x-11\right)+3\left(11-x\right)=0\)
\(\Leftrightarrow x\left(x-11\right)-3\left(x-11\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)
Câu 3:
\(x\left(x-3\right)-2\left(3-x\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Câu 4:
\(2x\left(x+5\right)-4\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
6 tháng 3 2022
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
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