Giải phương trình:
\(\dfrac{t +3} {t - 2}\) +\(\dfrac{t - 2} { t+3}\) = \(\dfrac{ 5t +15} {t^2 + t -6}\)
MAI HỌC RỒI !! HELP ME /?!/
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
6 tháng 2 2018
a, (3x-1)(x2+2)=(3x-1)(7x-10)
<=>(3x-1)(x2+2)-(3x-1)(7x-10)=0
<=>(3x-1)(x2+2-7x+10)=0
<=>(3x-1)(x2-7x+12)=0
<=>(3x-1)(x2-3x-4x+12)=0
<=>(3x-1)(x-3)(x-4)=0
<=>\(\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
Vậy ft có tập nghiệm S=\(\left\{\dfrac{1}{3},3,4\right\}\)
b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (ĐKXĐ:t\(\ne2;t\ne-3\))
<=>\(\dfrac{\left(t+3\right)^2+\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{t^2-2t+3t-6}\)
<=>\(\dfrac{t^2+6t+9+t^2-4t+4}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)
=>2t2+2t+13=5t+15
<=>2t2+2t-5t+13-15=0
<=>2t2-3t-2=0
<=>2t2-4t+t-2=0
<=>(t-2)(2t+1)=0
<=>\(\left[{}\begin{matrix}t-2=0\\2t+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-1}{2}\left(tmđkxđ\right)\end{matrix}\right.\)
Vậy ft có nghiệm duy nhất x=\(\dfrac{-1}{2}\)
G
6 tháng 2 2018
Giải:
a) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
Chia cả hai vế cho 3x-1, ta được:
\(x^2+2=7x-10\)
\(\Leftrightarrow x^2-7x+10+2=0\)
\(\Leftrightarrow x^2-7x+12=0\)
\(\Leftrightarrow x^2-4x-3x+12=0\)
\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy ...
b) \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (1)
ĐKXĐ: \(t\ne2;t\ne-3\)
\(\left(1\right)\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)
\(\Rightarrow\left(t+3\right)^2+\left(t-2\right)^2=5t+15\)
\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)
\(\Leftrightarrow2t^2+2t+13=5t+15\)
\(\Leftrightarrow2t^2+2t+13-5t-15=0\)
\(\Leftrightarrow2t^2-3t-2=0\)
\(\Leftrightarrow2t^2-4t+t-2=0\)
\(\Leftrightarrow2t\left(t-2\right)+\left(t-2\right)=0\)
\(\Leftrightarrow\left(2t+1\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2t+1=0\\t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{2}\left(tm\right)\\t=2\left(ktm\right)\end{matrix}\right.\)
Vậy ...
AP
4 tháng 3 2018
a,
\(\dfrac{1+x+3-3x-3+x}{1-x}=0\\ \dfrac{1-x}{1-x}=0\\ =>1-x=0\\ =>x=1\\ \)
9 tháng 3 2018
a, \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) ĐKXĐ: t\(\ne\)2,t\(\ne\)-3
\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)
\(\Rightarrow\left(t+3\right)\left(t+3\right)+\left(t-2\right)\left(t-2\right)=5t+15\)
\(\Leftrightarrow t^2+6t+9+t^2-4t+4-5t-15=0\)
\(\Leftrightarrow-3t-2=0\)
\(\Leftrightarrow-3t=2\)
\(\Leftrightarrow t=\dfrac{-2}{3}\) (tđk)
\(\Rightarrow S=\left\{\dfrac{-2}{3}\right\}\)
b, \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)ĐKXĐ: x\(\ne\)\(\dfrac{2}{7}\)
\(\Leftrightarrow\) \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)
\(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)
\(\Leftrightarrow\) \(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}+1=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+8+2-7x=0\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x+10=0\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{5}{2};-8\right\}\)
ND
9 tháng 3 2018
ĐKXĐ: x khác 2 và x khác -3
\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)
\(\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t+3\right)\left(t-2\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t+3\right)\left(t-2\right)}=\dfrac{5t+15}{t^2+t-6}\)
\(\Rightarrow t^2+6t+9+t^2-4=5t+15\)
\(\Leftrightarrow2t^2+t-10=0\)
\(\Leftrightarrow2t^2-4t+5t-10=0\)
\(\Leftrightarrow2t\left(t-2\right)+5\left(t-2\right)=0\)
\(\Leftrightarrow\left(2t+5\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy..................
Q
22 tháng 6 2017
1) \(\dfrac{2}{15}\cdot6\dfrac{5}{11}+\dfrac{5}{11}\cdot\dfrac{-2}{15}-\dfrac{2}{15}\cdot2015^0\)
\(=\dfrac{2}{15}\cdot\dfrac{71}{11}-\dfrac{1}{11}\cdot\dfrac{2}{3}-\dfrac{2}{15}\cdot1\)
\(=\dfrac{142}{165}-\dfrac{2}{33}-\dfrac{2}{15}\)
\(=\dfrac{2}{3}\)
2) \(\dfrac{5}{2\cdot7}+\dfrac{3}{14\cdot11}+\dfrac{4}{11\cdot7}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot16}\)
\(=\dfrac{5}{14}+\dfrac{3}{154}+\dfrac{4}{77}+\dfrac{1}{210}+\dfrac{13}{240}\)
\(=\dfrac{39}{80}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1
Nhận thấy \(t=0\) ko phải nghiệm
Với \(t\ne0\) pt tương đương:
\(\dfrac{3}{t+3+\dfrac{2}{t}}+\dfrac{2}{t+1+\dfrac{2}{t}}=1\)
Đặt \(t+\dfrac{1}{t}+1=x\Rightarrow t+\dfrac{2}{t}+3=x+2\)
Pt trở thành:
\(\dfrac{3}{x+2}+\dfrac{2}{x}=1\)
\(\Rightarrow3x+2\left(x+2\right)=x\left(x+2\right)\)
\(\Leftrightarrow x^2-3x-4=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t+\dfrac{2}{t}+1=-1\\t+\dfrac{2}{t}+1=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t^2+2t+2=0\left(vn\right)\\t^2-3t+2=0\end{matrix}\right.\)
\(\Rightarrow t=\left\{1;2\right\}\)
15 tháng 8 2017
a) \(\left(\dfrac{3}{4}-5x\right)-\left(2x-\dfrac{17}{4}\right)\)=\(\dfrac{44}{7}\)
\(\left(\dfrac{3}{4}-\dfrac{17}{4}\right)-\left(5x-2x\right)\) = \(\dfrac{44}{7}\)
\(\dfrac{-14}{4}-3x\) = \(\dfrac{44}{7}\)
3x= \(\dfrac{-14}{4}-\dfrac{44}{7}\)=\(\dfrac{-98}{28}-\dfrac{176}{28}\)= \(\dfrac{-274}{28}=\dfrac{-137}{14}\)
x= \(\dfrac{-137}{14}:3\)
x= \(\dfrac{-137}{42}\)
7 tháng 3 2017
3.
\(A=1-3+3^2-3^3+...-3^{2009}-3^{2010}\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...-3^{2010}+3^{2011}\)
\(\Rightarrow4A=3-3^2+3^3-3^4+...-3^{2010}+3^{2011}+\left(1-3+3^2-3^3+...-3^{2009}+3^{2010}\right)\)\(\Rightarrow4A=3^{2011}-1\)
\(\Rightarrow4A=3^{2011}\)
\(\Rightarrow\)ĐPCM
LS
7 tháng 3 2017
mik đg cần gấp câu 1 và câu 2 câu 3 mik lm đk r mơn bn nhé nhưng bn có thể giúp mik câu 1 và câu 2 k?
30 tháng 4 2017
Bài 1:
a)=2.( \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{97}-\dfrac{1}{99}\)
=2. (1/3-1/99)
=2. (33/99-1/99)
=2. 32/99
=64/99
b) tương tự như trên.
30 tháng 4 2017
Bài 1 :
a) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(=2\left(\dfrac{33}{99}-\dfrac{1}{99}\right)\)
\(=2.\dfrac{32}{99}\)
\(=\dfrac{2.32}{99}\)
\(=\dfrac{64}{99}\)
b) \(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)
\(=2\left(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\right)\)
\(=3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=3\left(1-\dfrac{1}{51}\right)\)
\(=3.\dfrac{50}{51}\)
\(=\dfrac{3.50}{51}\)
\(=\dfrac{1.50}{17}\)
\(=\dfrac{50}{17}\)
\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)(đkxđ: t khác 2, t khác -3)
<=>\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)
<=>\(\dfrac{\left(t+3\right)^2}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)^2}{\left(t+3\right)\left(t-2\right)}=\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)
=>t^2+6t+9+t^2-4t+4=5t+15
<=>2t^2-2t-5t=15-9-4=0
<=>2t^2-7t=0
<=> t(2t-7)=0
<=>t=0
2t-7=0<=>t=-7/2
vậy.....