Giải phương trình sau:
\(\dfrac{x}{18}\) - \(\dfrac{x}{24}\) = \(\dfrac{3}{2}\)
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23 tháng 1 2024
a: ĐKXĐ: \(x\notin\left\{3;-5\right\}\)
\(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)
=>\(\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
=>\(\dfrac{5x+25-3x+9}{15}=\dfrac{5x+25-3x+9}{\left(x-3\right)\left(x+5\right)}\)
=>(x-3)(x+5)=15
=>\(x^2+2x-15-15=0\)
=>\(x^2+2x-30=0\)
=>\(\left(x+1\right)^2=31\)
=>\(\left[{}\begin{matrix}x+1=\sqrt{31}\\x+1=-\sqrt{31}\end{matrix}\right.\Leftrightarrow x=-1\pm\sqrt{31}\left(nhận\right)\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2+x+1}=3-x\)
=>\(\left\{{}\begin{matrix}x^2+x+1=\left(3-x\right)^2\\x< =3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =3\\x^2-6x+9=x^2+x+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =3\\-7x=-8\end{matrix}\right.\Leftrightarrow x=\dfrac{8}{7}\left(nhận\right)\)
c:
ĐKXĐ: \(x\in R\)
\(x^2-x+\sqrt{x^2-x+24}=18\)
=>\(x^2-x+24+\sqrt{x^2-x+24}=42\)
=>\(\left(\sqrt{x^2-x+24}\right)^2+\left(\sqrt{x^2-x+24}\right)-42=0\)
=>\(\left(\sqrt{x^2-x+24}+7\right)\left(\sqrt{x^2-x+24}-6\right)=0\)
=>\(\sqrt{x^2-x+24}-6=0\)
=>\(x^2-x+24=36\)
=>\(x^2-x-12=0\)
=>(x-4)(x+3)=0
=>\(\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
21 tháng 9 2021
Đặt \(\dfrac{1}{y-1}=a\), hpt tở thành
\(\left\{{}\begin{matrix}\dfrac{5}{x+1}+a=10\\\dfrac{1}{x-2}+3a=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15}{x+1}+3a=30\left(1\right)\\\dfrac{1}{x-1}+3a=18\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\), ta được:
\(\dfrac{15}{x+1}-\dfrac{1}{x-1}=12\\ \Leftrightarrow\dfrac{15x-15-x-1}{\left(x-1\right)\left(x+1\right)}=12\\ \Leftrightarrow12x^2-12=14x-16\\ \Leftrightarrow12x^2-14x+4=0\\ \Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Với \(x=\dfrac{1}{2}\Leftrightarrow\dfrac{10}{3}+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{10y-7}{3\left(y-1\right)}=10\)
\(\Leftrightarrow30y-30=10y-7\Leftrightarrow y=\dfrac{23}{20}\)
Với \(x=\dfrac{2}{3}\Leftrightarrow3+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{1}{y-1}=7\Leftrightarrow7y-7=1\Leftrightarrow y=\dfrac{8}{7}\)
Vậy \(\left(x;y\right)=\left\{\left(\dfrac{1}{2};\dfrac{23}{20}\right);\left(\dfrac{2}{3};\dfrac{8}{7}\right)\right\}\)
28 tháng 4 2023
\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow x\left(x+8\right)=105\)
\(\Leftrightarrow x^2+8x-105=0\)
\(\Leftrightarrow x^2-7x+15x-105=0\)
\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)
Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)
28 tháng 1 2022
\(\dfrac{x^2-26}{10}+\dfrac{x^2-25}{11}\ge\dfrac{x^2-24}{12}+\dfrac{x^2-23}{13}\)
\(\Leftrightarrow\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)
\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)
\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)
\(\Leftrightarrow\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\Rightarrow x^2-36\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-6\\x\ge6\end{matrix}\right.\)
28 tháng 1 2022
Bất phương trình đó tương đương với:
\(\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)
⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)
⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)
⇔ \(\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)
+)Vì \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}\) nên \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\)
⇔ \(x^2-36\ge0\)
⇔ \(x^2\ge36\)
⇔ \(\sqrt{x^2}\ge6\)
⇔ \(\left|x\right|\ge6\)
⇔ \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)
➤ Vậy \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)
H
26 tháng 3 2021
Bài 4 :
24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ
Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0
Suy ra quãng đường AB là 36x(km)
Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)
Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)
Ta có phương trình:
\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)
Vậy quãng đường AB dài 36.2 = 72(km)
9 tháng 2 2021
Sửa đề: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)Ta có: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
mà \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}>0\)
nên 100-x=0
hay x=100
Vậy: S={100}
9 tháng 2 2021
Ta có : \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}+5=0\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
Thấy : \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\ne0\)
\(\Rightarrow100-x=0\)
\(\Leftrightarrow x=100\)
Vậy ...
15 tháng 5 2023
a: =>\(\dfrac{5x-15+4x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)
=>\(\dfrac{9x-23}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)
=>9x^2-23x=x^2-5x+6
=>8x^2-18x-6=0
=>\(x=\dfrac{9\pm\sqrt{129}}{8}\)
b: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>216x+18=275x-100
=>-59x=-118
=>x=2
25 tháng 5 2021
Đk:\(x\ge0\)
Pt \(\Leftrightarrow2\sqrt{x}+5=36+3\left(\sqrt{x}-3\right)\)
\(\Leftrightarrow-\sqrt{x}=22\) (vô nghiệm)
Vậy phương trình vô nghiệm
NA
19 tháng 3 2022
a,\(\left(3x-2\right)\left(x+3\right)=9x^2-4\\ \Leftrightarrow\left(3x-2\right)\left(x+3\right)-\left(3x-2\right)\left(3x+2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x+3-3x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(-2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b, ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{x-4}{x+2}-\dfrac{x+1}{x-2}=\dfrac{24}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{24}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{x^2-6x+8-x^2-3x-2-24}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow-9x-18=0\\ \Leftrightarrow x=-2\left(ktm\right)\)
\(\Leftrightarrow\dfrac{4x-3x}{72}=\dfrac{3}{2}\\ \Leftrightarrow\dfrac{x}{72}=\dfrac{3}{2}\\ \Leftrightarrow x=108\)
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