K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
ND
8 tháng 2 2018
a. Với a = -3 ta được:
\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}+\dfrac{27-3}{x^2-9}=0\)
\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{24}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9+24=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow x=-2\)
8 tháng 2 2018
Giải phương trình :
\(\dfrac{x-a}{x+a}-\dfrac{x+a}{x-a}+\dfrac{3a^2+a}{x^2-a^2}=0\)
a) Với a = -3
\(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}+\dfrac{27+3}{x^2-3^2}=0\)
ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)
Ta có : \(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}+\dfrac{27+3}{x^2-3^2}\)
\(\Leftrightarrow\) \(\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{27+3}{\left(x+3\right)\left(x-3\right)}=0\)
Khử mẫu ta có : \(\left(x-3\right)^2-\left(x+3\right)^2+27+3=0\)
⇔ \(x^2+6x+9-x^2+6x-9+30=0\)
\(\Leftrightarrow12x+30=0\)
\(\Leftrightarrow12x=-30\)
\(\Leftrightarrow x=-\dfrac{5}{2}\)
Tập nghiệm của pt là: \(S=\left\{-\dfrac{5}{2}\right\}\)
b) Với a = 1
\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{3+3}{x^2-1}=0\)
ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)
Ta có : \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{3+3}{x^2-1}=0\)
\(\Leftrightarrow\) \(\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3+3}{\left(x+1\right)\left(x-1\right)}=0\)
Khử mẫu ta có : \(\left(x-1\right)^2-\left(x+1\right)^2+6=0\)
\(\Leftrightarrow x^2+x-1-x^2+x+1+6=0\)
\(\Leftrightarrow2x+6=0\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
Tập nghiệm của pt là : \(S=\left\{-3\right\}\)
28 tháng 6 2020
a) Ta có: 3(x+1)-(4x-3)=5
\(\Leftrightarrow3x+3-4x+3-5=0\)
\(\Leftrightarrow-x+1=0\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(2x+1\right)\left(x-1\right)=x^2-2x+1\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1-x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: S={1;-2}
c) Ta có: |-5x|=3x-16(*)
Trường hợp 1: \(-5x\ge0\Leftrightarrow x\le0\)
(*)\(\Leftrightarrow-5x=3x-16\)
\(\Leftrightarrow-5x-3x=-16\)
\(\Leftrightarrow-8x=-16\)
hay x=2(loại)
Trường hợp 2: \(-5x< 0\Leftrightarrow x>0\)
(*)\(\Leftrightarrow5x=3x-16\)
\(\Leftrightarrow5x-3x=-16\)
\(\Leftrightarrow2x=-16\)
hay x=-8(loại)
Vậy: \(S=\varnothing\)
d) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\frac{x-2}{x-2}-\frac{2}{x-2}=\frac{5x+2}{4-x^2}\)
\(\Leftrightarrow\frac{x-4}{x-2}=\frac{5x+2}{4-x^2}\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{5x+2}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{x^2-2x-8+5x+2}{\left(x-2\right)\left(x+2\right)}=0\)
Suy ra: \(x^2+3x-6=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{33}{4}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2=\frac{33}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{2}=\frac{\sqrt{33}}{2}\\x+\frac{3}{2}=-\frac{\sqrt{33}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{33}-3}{2}\left(tm\right)\\x=\frac{-\sqrt{33}-3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{\sqrt{33}-3}{2};\frac{-\sqrt{33}-3}{2}\right\}\)
2 tháng 6 2022
a: \(2x^4-3x^3+4x+1⋮x^2-1\)
\(\Leftrightarrow2x^4-2x^2-3x^3+3x+2x^2-2+x+3⋮x^2-1\)
\(\Leftrightarrow x+3⋮x^2-1\)
\(\Leftrightarrow x^2-9⋮x^2-1\)
\(\Leftrightarrow x^2-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
\(\Leftrightarrow x\in\left\{\sqrt{2};-\sqrt{2};0;\sqrt{3};-\sqrt{3};\sqrt{5};-\sqrt{5};3;-3\right\}\)
b: \(x^5+2x^4+3x^2+x-3⋮x^2+1\)
\(\Leftrightarrow x^5+x^3+2x^4+2x^2-x^3-x+x^2+1+2x-4⋮x^2+1\)
\(\Leftrightarrow2x-4⋮x^2+1\)
\(\Leftrightarrow4x^2-16⋮x^2+1\)
\(\Leftrightarrow4x^2+4-20⋮x^2+1\)
\(\Leftrightarrow x^2+1\in\left\{1;2;4;5;10;20\right\}\)
hay \(x\in\left\{0;1;-1;\sqrt{3};-\sqrt{3};2;-2;3;-3;\sqrt{19};-\sqrt{19}\right\}\)
MV
12 tháng 8 2017
2.
\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
Vì \(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
\(\Rightarrow x>0\)
Vậy \(x>0\)
KS
2
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 2 2020
\(\Leftrightarrow x^3+x^2-2x+5x^2+5x-10=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)+5\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+2\right)\left(x-1\right)=0\)
b/ \(\Leftrightarrow x^3+5x^2+6x-x^2-5x-6=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)-\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
NH
10 tháng 2 2020
\(x^3+6x^2+3x-10=0\)
\(\Leftrightarrow x^3-x^2+7x^2-7x+10x-10=0\)
\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+5x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-5\end{matrix}\right.\)
Vậy \(S=\left\{1;-2;-5\right\}\)
\(x^3+4x^2+x-6=0\)
\(\Leftrightarrow x^3-x^2+5x^2-5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+3x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{1;-2;-3\right\}\)
J
4
17 tháng 7 2016
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)
Ta xét 3 trường hợp :
1. Với \(x< 1\) , pt trên trở thành : \(1-x+2-x=3\Leftrightarrow2x=0\Leftrightarrow x=0\)(nhận)
2. Với \(1\le x\le2\), pt trên trở thành : \(x-1+2-x=3\Leftrightarrow1=3\)(vô lý - loại)
3. Với \(x>2\) , pt trên trở thành : \(x-1+x-2=3\Leftrightarrow2x=6\Leftrightarrow x=3\)(nhận)
Vậy tập nghiệm của phương trình : \(S=\left\{0;3\right\}\)
17 tháng 7 2016
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow x-1+x-2=3\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=2\)
TN
2
ND
27 tháng 2 2018
\(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-x^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-x^2-4x^2+4x+4x-4=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy......
27 tháng 2 2018
\(\left(x^3-x^2\right)-ã^2+8x-4=0\)
\(< =>x^3-x^2-4x^2+8x-4\)
\(< =>x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(< =>\left(x-1\right)\left(x^2-4x+4=0\right)\)
\(< =>\left(x-1\right)\left(x-2\right)^2=0< =>\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)