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2 tháng 1 2021
\(\frac{36}{x}+\frac{36}{x-12}=\frac{9}{2}\)
ĐKXĐ : x ≠ 0 ; x ≠ 12
pt ⇔ \(36\left(\frac{1}{x}+\frac{1}{x-12}\right)=\frac{9}{2}\)
⇔ \(\frac{x-12}{x\left(x-12\right)}+\frac{x}{x\left(x-12\right)}=\frac{1}{8}\)
⇔ \(\frac{x-12+x}{x\left(x-12\right)}=\frac{1}{8}\)
⇔ \(\frac{2x-12}{x\left(x-12\right)}=\frac{1}{8}\)
⇔ ( 2x - 12 ).8 = x( x - 12 )
⇔ 16x - 96 = x2 - 12x
⇔ x2 - 12x - 16x + 96 = 0
⇔ x2 - 28x + 96 = 0 (1)
Δ' = b'2 - ac = ( b/2 )2 - ac = ( -14 )2 - 96 = 100
Δ' > 0 nên (1) có hai nghiệm phân biệt
\(x_1=\frac{-b+\sqrt{\text{Δ}'}}{a}=\frac{14+\sqrt{100}}{1}=24\)(tm)
\(x_2=\frac{-b-\sqrt{\text{Δ}'}}{a}=\frac{14-\sqrt{100}}{1}=4\)(2)
Vậy phương trình có hai nghiệm x1 = 24 ; x2 = 4
2 tháng 1 2021
\(\frac{36}{x}+\frac{36}{x-12}=\frac{9}{2}\)ĐKXĐ : \(x\ne0;12\)
\(\Leftrightarrow\frac{72\left(x-12\right)}{2x\left(x-12\right)}+\frac{72x}{2x\left(x-12\right)}=\frac{9x\left(x-12\right)}{2x\left(x-12\right)}\)
Khử mẫu : \(72\left(x-12\right)+72x=9x\left(x-12\right)\)
\(\Leftrightarrow72x-864+72x=9x^2-108x\)
\(\Leftrightarrow252x-864-9x^2=0\)
\(\Leftrightarrow9\left(x-24\right)\left(x-4\right)=0\Leftrightarrow x=24;4\)
TQ
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1
\(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)
\(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)
\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\) (do \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\))
\(\Leftrightarrow x=100\)
MA
1
15 tháng 3 2020
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
có : \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
15 tháng 3 2020
\(pt\)\(\Leftrightarrow\)\(({x-90\over10}-1)+({x-76\over12}-2)+\)\(+({x-58\over14}-3)+({x-36\over16}-4)+({x-15\over17}-5)=0\)
\(\Leftrightarrow\)\(({x-100\over10})+({x-100\over12})+({x-100\over14})+({x-100\over16})\)
\(+({x-100\over17})=0\)
\(\Leftrightarrow\)\((x-100)({1\over10}+{1\over12}+{1\over14}+{1\over16}+{1\over17})=0\)
\(\Rightarrow\)\(x-100=0\)
\(\Rightarrow\)\(x=100\)
23 tháng 2 2022
\(\Leftrightarrow36\left(x+6\right)+36\left(x-6\right)=\dfrac{9}{2}\left(x^2-36\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{9}{2}-162-72x=0\)
\(\Leftrightarrow9x^2-144x-324=0\)
\(\Leftrightarrow x^2-16x-36=0\)
=>(x-18)(x+2)=0
=>x=18 hoặc x=-2
23 tháng 2 2022
ĐKXĐ:\(x\ne\pm6\)
\(\dfrac{36}{x-6}+\dfrac{36}{x+6}=4,5\\ \Leftrightarrow36\left(\dfrac{1}{x-6}+\dfrac{1}{x+6}\right)=4,5\\ \Leftrightarrow\dfrac{x+6}{\left(x-6\right)\left(x+6\right)}+\dfrac{x-6}{\left(x-6\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{x+6+x-6}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{2x}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow x^2-36=16x\\ \Leftrightarrow x^2-16x-36=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(18x+36\right)=0\\ \Leftrightarrow x\left(x+2\right)-18\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-18\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=18\left(tm\right)\end{matrix}\right.\)
PT
1
30 tháng 4 2021
\(\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5\)
\(\Leftrightarrow36\left(x-6\right)+36\left(x+6\right)=4,5\left(x^2-36\right)\)
\(\Leftrightarrow36x-216+36x+216=4,5x^2-162\)
\(\Leftrightarrow-4,5x^2+72x+162=0\)
\(\Leftrightarrow\left(x-18\right)\left(-4,5x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-2\end{matrix}\right.\)
1 tháng 5 2021
bạn làm rõ hơn ở chỗ này đc ko, mk ko hiểu
⇔−4,5x2+72x+162=0⇔−4,5x2+72x+162=0
⇔(x−18)(−4,5x−9)=0