Giải các phương trình:
a) \(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}=x;\)
b) \(\dfrac{1}{x^2}+\dfrac{1}{\left(x+2\right)^2}=\dfrac{5}{16};\)
c) \(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+1\right)^2}=\dfrac{1}{12}.\)
K
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Những câu hỏi liên quan
H
18 tháng 4 2023
\(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{2}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : \(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{2}{x\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-1\right)}{x\left(x+1\right)}-\dfrac{x+1}{x\left(x+1\right)}=\dfrac{2}{x\left(x+1\right)}\)
`=> x(x-1) -(x+1)=2`
`<=>x^2 - x -x-1=2`
`<=> x^2 -2x-1+2=0`
`<=> x^2 -2x +1=0`
`<=> (x+1)^2=0`
`<=>x+1=0`
`<=>x=-1(ktm)`
Vậy pt vô nghiệm
16 tháng 2 2021
a) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{3x}+\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{12x}+\dfrac{6}{12x}=\dfrac{3x}{12x}\)
Suy ra: \(3x=10\)
\(\Leftrightarrow x=\dfrac{10}{3}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{10}{3}\right\}\)
b) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)
\(\Leftrightarrow\dfrac{3x}{8x^2}-\dfrac{4x}{8x^2}=\dfrac{8}{8x^2}\)
Suy ra: \(3x-4x=8\)
\(\Leftrightarrow-x=8\)
hay x=-8(thỏa ĐK)
Vậy: S={-8}
c)ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{2x}+\dfrac{3}{4x}=\dfrac{5}{2x^2}\)
\(\Leftrightarrow\dfrac{2x}{4x^2}+\dfrac{3x}{4x^2}=\dfrac{10}{4x^2}\)
Suy ra: 2x+3x=10
\(\Leftrightarrow5x=10\)
hay x=2(thỏa ĐK)
Vậy: S={2}
16 tháng 2 2021
d, \(\dfrac{2a}{x+a}=1\) (x \(\ne\) -a)
\(\Leftrightarrow\) \(\dfrac{2a}{x+a}-\dfrac{x+a}{x+a}=0\)
\(\Leftrightarrow\) \(\dfrac{a-x}{x+a}=0\)
\(\Leftrightarrow\) a - x = 0 (x + a \(\ne\) 0)
\(\Leftrightarrow\) x = a (TM)
Vậy S = {a}
Chúc bn học tốt!
R
5 tháng 2 2022
TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
5 tháng 2 2022
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
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 \(x=7\text{v}à\text{x}=15\)
15 tháng 3 2023
1:
a: =>28x-8=9x+3
=>19x=11
=>x=11/19
b: =>(3x-1)(x-1)=(2x+1)(x+1)
=>3x^2-4x+1=2x^2+3x+1
=>x^2-7x=0
=>x=0 hoặc x=7
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 \(\left(x;y\right)=\left\{\left(\dfrac{1}{2};\dfrac{23}{20}\right);\left(\dfrac{2}{3};\dfrac{8}{7}\right)\right\}\)
10 tháng 4 2022
\(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\)
\(ĐK:x\ne0;-10\)
\(\Leftrightarrow\dfrac{12\left(x+10\right)+12x}{12x\left(x+10\right)}=\dfrac{x\left(x+10\right)}{12x\left(x+10\right)}\)
\(\Leftrightarrow12\left(x+10\right)+12x-x\left(x+10\right)=0\)
\(\Leftrightarrow12x+120+12x-x^2-10x=0\)
\(\Leftrightarrow-x^2+14x+120=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)
TT
10 tháng 4 2022
\(o,\dfrac{x}{2x+6}-\dfrac{x}{2x-2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x+3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x+3\right)-2\left(3x+2\right)}{2\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+x-x^2-3x-6x-4=0\)
\(\Leftrightarrow-8x-4=0\)
\(\Leftrightarrow-4\left(2x+1\right)=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)
28 tháng 8 2021
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
dễ quá b ơi
giải đi