\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
Những câu hỏi liên quan
\(\frac{x+5}{x-5}\)--\(\frac{x-5}{x+5}\)=\(\frac{20}{x^2-25}\)
Xét vế trái
x+5/x-5 - x-5/x+5
Quy đồng mẫu số
(x+5)²/(x-5)(x+5) - (x-5)²/(x+5)(x-5)
=x²+10x+25-x²+10x-25 /(x-5)(x+5)
=20x/x²-25
Đúng 0
Bình luận (0)
Giải các phương trình:
a)\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2}-25\)
b) x2+ 6x+9= 144
c) \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)
( mink đag cần gấp)
b) \(x^2+6x+9=144\)
\(\Leftrightarrow\left(x+3\right)^2=12^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)
b, Ta có : \(x^2+6x+9=144\)
=> \(\left(x+3\right)^2=12^2\)
=> \(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{9,-15\right\}\)
c, Ta có : \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)
=> \(\frac{2-x}{2016}-1=\frac{1-x}{2017}+\frac{-x}{2018}\)
=> \(\frac{2-x}{2016}+1=\frac{1-x}{2017}+1+\frac{-x}{2018}+1\)
=> \(\frac{2-x}{2016}+\frac{2016}{2016}=\frac{1-x}{2017}+\frac{2017}{2017}+\frac{-x}{2018}+\frac{2018}{2018}\)
=> \(\frac{2018-x}{2016}=\frac{2018-x}{2017}+\frac{2018-x}{2018}\)
=> \(\frac{2018-x}{2016}-\frac{2018-x}{2017}-\frac{2018-x}{2018}=0\)
=> \(\left(2018-x\right)\left(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
=> \(2018-x=0\)
=> \(x=2018\)
Vậy phương trình có tập nghiệm là \(S=\left\{2018\right\}\)
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
3) frac{1-x}{x+1}-frac{3+2x}{x+1}0
13) frac{x+2}{x}-frac{x^2+5x+4}{xleft(x+2right)}frac{x}{x+2}
14) frac{1}{x+1}-frac{5}{x-2}frac{20}{left(x+1right)left(2-xright)}
16) frac{x+5}{x-5}-frac{x-5}{x+5}frac{20}{x^2-25}
17) frac{3x+2}{3x-2}-frac{6}{2+3x}frac{9x^2}{9x^2-4}
18) frac{x-1}{x}+frac{1}{x+1}frac{2x-1}{2x^2+2}
19) frac{2}{x+1}-frac{3x+1}{left(x+1right)}frac{1}{left(x+1right)left(x-2right)}
20) frac{x+5}{3x-6}-frac{1}{2}frac{2x-3}{2x-4}
Đọc tiếp
3) \(\frac{1-x}{x+1}-\frac{3+2x}{x+1}=0\)
13) \(\frac{x+2}{x}-\frac{x^2+5x+4}{x\left(x+2\right)}=\frac{x}{x+2}\)
14) \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{20}{\left(x+1\right)\left(2-x\right)}\)
16) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
17) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
18) \(\frac{x-1}{x}+\frac{1}{x+1}=\frac{2x-1}{2x^2+2}\)
19) \(\frac{2}{x+1}-\frac{3x+1}{\left(x+1\right)}=\frac{1}{\left(x+1\right)\left(x-2\right)}\)
20) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)
Giải PT
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{x\left(x+25\right)}{x^2-25}\)
\(ĐKXĐ:x\ne\pm5\)
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{x\left(x+25\right)}{x^2-25}\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}=\frac{x^2+25x}{\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow x^2+10x+25-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow x^2+25x=20x\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\left(ktm\right)\end{cases}}\)
Bài 2: Giải các phương trình sau:
a \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\)
b \(\left(x+1\right)\left(x+2\right)=\left(2-x\right)\left(x+2\right)\)
c \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
d\(x^2+6x+9=114\)
Bài 1:Tìm x,biết
a) x-frac{20}{11.13}-frac{20}{13.15}-frac{20}{15.17}-...-frac{20}{53.55}frac{3}{11}
b)x+frac{15}{90.94}+frac{15}{94.98}+frac{15}{98.102}+...+frac{15}{146.150}frac{2}{3}
c)x-frac{1}{2}-frac{1}{6}-frac{1}{12}-frac{1}{20}-frac{1}{30}-frac{1}{42}-frac{1}{56}frac{5}{24}
d)8-frac{8-frac{8}{5}+frac{8}{25}-frac{8}{125}}{9-frac{9}{5}+frac{9}{25}-frac{9}{125}}:frac{161616}{151515}frac{4+frac{4}{73}-frac{4}{115}}{5+frac{5}{73}-frac{1}{23}}
Đọc tiếp
Bài 1:Tìm x,biết
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
b)\(x+\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}=\frac{2}{3}\)
c)\(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}=\frac{5}{24}\)
d)\(8-\frac{8-\frac{8}{5}+\frac{8}{25}-\frac{8}{125}}{9-\frac{9}{5}+\frac{9}{25}-\frac{9}{125}}:\frac{161616}{151515}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{1}{23}}\)
a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)
Đúng 0
Bình luận (0)
1, giải các pt sau
a, \(\frac{7x-3}{x-1}=\frac{3}{2}\)
b, \(\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
c, \(\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
d, \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\) \(\frac{20}{x^2-25}\)
giúp mik v mn ơi
\(a.\frac{7x-3}{x-1}=\frac{3}{2}\)
\(\Leftrightarrow\frac{7x-3}{x-1}-\frac{3}{2}=0\)
\(\Leftrightarrow\frac{2\left(7x-3\right)}{2.\left(x-1\right)}-\frac{3\left(x-1\right)}{2\left(x-1\right)}=0\)
\(\Leftrightarrow\frac{14x-6-3x+3}{2\left(x-1\right)}=0\)
\(\Leftrightarrow11x-3=0\)
\(\Leftrightarrow x=\frac{3}{11}\)
\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{6-14x}{1+x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{2\left(6-14x\right)}{2\left(1+x\right)}-\frac{1+x}{2\left(1+x\right)}=0\)
\(\Leftrightarrow\frac{12-28x-1-x}{2\left(1+x\right)}=0\)
\(\Leftrightarrow11-29x=0\)
\(\Leftrightarrow x=\frac{11}{29}\)
\(c.\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}-\frac{3-x}{x-2}=0\)
\(\Leftrightarrow\frac{1+3x-6-3+x}{x-2}=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow x=2\)
\(d.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\)
\(\Leftrightarrow\frac{x^2+10x+25-x^2+10x-25-20}{x^2-25}=0\)
\(\Leftrightarrow20x-20=0\)
\(\Leftrightarrow x=10\)
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
mình cần gấp mong các bạn giúp đỡ
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\left(x\ne\pm5\right)\)
\(\Leftrightarrow\frac{x+5}{x-5}+\frac{x-5}{x+5}-\frac{2\left(x^2+25\right)}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{x^2+10x+25+x^2-10x+25-2x^2-50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Rightarrow\frac{0}{\left(x-5\right)\left(x+5\right)}=0\)
=> PT đúng với mọi x khác \(\pm5\)
Refund QB nhìn logic :V
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{\left(x+5\right)\left(x-5\right)}\)
\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)
\(20x=2x^2+50\)
\(20x-2x^2-50=0\)
\(2\left(10x-x^2-25\right)=0\)
\(-x^2+10x+25=0\)
\(x^2-10x+25=0\)
\(x^2-2\left(x\right)\left(5\right)+5^2=0\)
\(\left(x-5\right)^2=0\)
\(x-5=0\Leftrightarrow x=5\)
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Xem thêm câu trả lời