\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)

ĐKXĐ x \(\ne\pm5\)

Suy ra : \(\left(x+5\right)^2-\left(x-5\right)^2=20\)

\(\Leftrightarrow\left(x+5+x-5\right)\left(x+5-x+5\right)=20\)

\(\Leftrightarrow2x.10=20\Leftrightarrow20x=20\Leftrightarrow x=1\left(TM\right)\)

Vậy phương trình có tập nghiệm là \(S=\left\{-1\right\}\)

\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\left(ĐK:x\ne\pm5\right)\)

\(\Leftrightarrow\frac{\left(x+5\right)^2-\left(x-5\right)^2}{\left(x+5\right)\left(x-5\right)}=\frac{20}{\left(x+5\right)\left(x-5\right)}\)

\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)

\(\Leftrightarrow20x=20\)

\(\Leftrightarrow x=1\)(thỏa đk nên nhận)

Vậy ...

bạn ơi tập nghiệm là S = {1} chứ ko phải -1 nha mình gõ nhầm

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 \(S=\left\{2018\right\}\)

\(ĐKXĐ:x\ne\pm5\)

\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{x\left(x+25\right)}{x^2-25}\)

\(\Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}=\frac{x^2+25x}{\left(x-5\right)\left(x+5\right)}\)

\(\Rightarrow x^2+10x+25-x^2+10x-25=x^2+25x\)

\(\Leftrightarrow x^2+25x=20x\)

\(\Leftrightarrow x^2+5x=0\)

\(\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\left(ktm\right)\end{cases}}\)

ktm là gì v bn

không thỏa mãn đkxđ

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)$x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}=\frac{5}{24}$

\(a.\frac{7x-3}{x-1}=\frac{3}{2}\)

\(\Leftrightarrow\frac{7x-3}{x-1}-\frac{3}{2}=0\)

\(\Leftrightarrow\frac{2\left(7x-3\right)}{2.\left(x-1\right)}-\frac{3\left(x-1\right)}{2\left(x-1\right)}=0\)

\(\Leftrightarrow\frac{14x-6-3x+3}{2\left(x-1\right)}=0\)

\(\Leftrightarrow11x-3=0\)

\(\Leftrightarrow x=\frac{3}{11}\)

\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)

\(\Leftrightarrow\frac{6-14x}{1+x}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{2\left(6-14x\right)}{2\left(1+x\right)}-\frac{1+x}{2\left(1+x\right)}=0\)

\(\Leftrightarrow\frac{12-28x-1-x}{2\left(1+x\right)}=0\)

\(\Leftrightarrow11-29x=0\)

\(\Leftrightarrow x=\frac{11}{29}\)

\(c.\frac{1}{x-2}+3=\frac{3-x}{x-2}\)

\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}-\frac{3-x}{x-2}=0\)

\(\Leftrightarrow\frac{1+3x-6-3+x}{x-2}=0\)

\(\Leftrightarrow4x-8=0\)

\(\Leftrightarrow x=2\)

\(d.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)

\(\Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\)

\(\Leftrightarrow\frac{x^2+10x+25-x^2+10x-25-20}{x^2-25}=0\)

\(\Leftrightarrow20x-20=0\)

\(\Leftrightarrow x=10\)

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\left(x\ne\pm5\right)\)

\(\Leftrightarrow\frac{x+5}{x-5}+\frac{x-5}{x+5}-\frac{2\left(x^2+25\right)}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25+x^2-10x+25-2x^2-50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Rightarrow\frac{0}{\left(x-5\right)\left(x+5\right)}=0\)

=> PT đúng với mọi x khác \(\pm5\)

Refund QB nhìn logic :V 

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{\left(x+5\right)\left(x-5\right)}\)

\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)

\(20x=2x^2+50\)

\(20x-2x^2-50=0\)

\(2\left(10x-x^2-25\right)=0\)

\(-x^2+10x+25=0\)

\(x^2-10x+25=0\)

\(x^2-2\left(x\right)\left(5\right)+5^2=0\)

\(\left(x-5\right)^2=0\)

\(x-5=0\Leftrightarrow x=5\)

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