Giải phương trình sau:
\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x+1}{x^2+x}\)
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Những câu hỏi liên quan
19 tháng 8 2021
1: Ta có: \(\dfrac{3}{x+2}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)
Suy ra: \(3x-6-x+1=2x+4\)
\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)
2: Ta có: \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)
Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)
\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)
\(\Leftrightarrow5x=16\)
hay \(x=\dfrac{16}{5}\)
BC
2
25 tháng 2 2021
`1+(x-2)/(1-x)+(2x^2-5)/(x^3-1)=4/(x^2+x+1)(x ne 1)`
`<=>(x^3-1)/(x^3-1)-((x-2)(x^2+x+1))/(x^3-1)+(2x^2-5)/(x^3-1)=(4(x-1))/(x^3-1)`
`<=>x^3-1-(x-2)(x^2+x+1)+2x^2-5=4(x-1)`
`<=>x^3-1-(x^3-x^2-x-2)+2x^2-5=4x-4`
`<=>x^3-1-x^3+x^2+x+2+2x^2-5-4x+4=0`
`<=>3x^2-3x+2=0`
`<=>x^2-2/3 x+2/3=0`
`<=>x^2-2.x. 1/3+1/9+5/9=0`
`<=>(x-1/3)^2=-5/9` vô lý
Vậy phương trình vô nghiệm.
25 tháng 2 2021
ĐKXĐ: \(x\ne1\)
Ta có: \(1+\dfrac{x-2}{1-x}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\)
\(\Leftrightarrow\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
Suy ra: \(x^3-1-\left(x^3+x^2+x-2x^2-2x-2\right)+2x^2-5=4x-4\)
\(\Leftrightarrow x^3-1-x^3+x^2+x+2+2x^2-5-4x+4=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
mà 3>0
nên x(x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
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 \(x=7\text{v}à\text{x}=15\)
I
2
NH
22 tháng 2 2023
\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\left(x\ne-2;x\ne3\right)\\ < =>\dfrac{9x-2}{x^2-3x+2x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{x\left(x-3\right)+2\left(x-3\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{\left(x-3\right)\left(x+2\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)
suy ra: \(9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(< =>9x-2+2x^2-6x-\left(x^2+2x-x-2\right)=x^2+2x-3x-6\)
\(< =>9x-2+2x^2-6x-x^2-2x+x+2=x^2-x-6\)
\(< =>2x^2-x^2-x^2+9x-6x-2x+x+x=6+2-2\)
\(< =>3x=6\\ < =>x=2\left(tm\right)\)
22 tháng 2 2023
ĐKXĐ: \(x\ne\left\{-2;3\right\}\)
\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)
\(\Leftrightarrow\dfrac{9x-2}{\left(x+2\right)\left(x-3\right)}+\dfrac{2x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)
\(\Leftrightarrow9x-2+2x^2-6x-x^2-x+2=x^2-x-6\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-2\left(loại\right)\)
Vậy: PT vô nghiệm.
19 tháng 8 2021
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2021
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
28 tháng 8 2021
a: Ta có: \(6-4x=5(x+3)+3\)
\(\Leftrightarrow6-4x-5x-12-3=0\)
\(\Leftrightarrow-9x=9\)
hay x=-1
b: Ta có: \(\dfrac{x+3}{2}-1=\dfrac{x-1}{3}+\dfrac{x+5}{6}\)
\(\Leftrightarrow15x+45-30=10x-30+5x+25\)
\(\Leftrightarrow15=-5\left(loại\right)\)
c: Ta có: \(\left(x-2\right)\left(2x+1\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
29 tháng 8 2021
d: Ta có: \(\dfrac{2}{x^2-2x}+\dfrac{1}{x}=\dfrac{x+2}{x-2}\)
\(\Leftrightarrow2+x-2=x^2+2x\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
HP
16 tháng 3 2021
1.
ĐK: \(x\ne7;x\ne-1;x\ne3\)
\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)
TH1: \(x< -1\)
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)
\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)
\(\Leftrightarrow x^2-5x+22\ge0\)
\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)
TH2: \(-1< x< 3\)
\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)
\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)
\(\Leftrightarrow-x^2+5x-22\ge0\)
\(\Rightarrow\) vô nghiệm
TH3: \(3< x< 7\)
Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)
\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)
TH4: \(x>7\)
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)
\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)
\(\Leftrightarrow x^2-5x+22\le0\)
\(\Rightarrow\) vô nghiệm
Vậy ...
Các bài kia tương tự, chứ giải ra mệt lắm.
Y
17 tháng 1 2023
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
ĐKXĐ:\(x\ne-1,x\ne0\)
\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x+1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}-\dfrac{2x+1}{x\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-1+x-2x-1}{x\left(x+1\right)}=0\\ \Rightarrow x^2-x-2=0\\ \Leftrightarrow x^2-2x+x-2=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm `S={2}`
\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x+1}{x^2+x}\left(đk:x\ne0,-1\right)\)
\(\Leftrightarrow\dfrac{x-1}{x}+\dfrac{1}{x+1}-\dfrac{2x+1}{x\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)+x-2x-1}{x\left(x+1\right)}=0\)
\(\Leftrightarrow x^2+x-x-1+x-2x-1=0\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Delta=b^2-4ac=\left(-1\right)^2-4.\left(-2\right)=9>0\Rightarrow\sqrt{\Delta}=3\)
\(\Rightarrow\)PT có 2 nghiệm \(x_1,x_2\)
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{1+3}{2}=2\left(n\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{1-3}{2}=-1\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)