\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
Rút gọn biểu thức:
\(A=\left|\frac{\left|y-x\right|}{\left|xy\right|}\right|+\left|\frac{y+x}{xy}-\frac{2}{z}\right|+\frac{\left|y-x\right|}{\left|xy\right|}+\frac{y+x}{xy}+\frac{2}{z}\)
với \(x>5\); \(y=\frac{x^2-25}{x+\frac{10x+25}{x}}\); \(z=\frac{x^2-25}{x+\frac{15x+25}{x-5}}\)
\(\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
ĐKXĐ:...
\(\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}{\left(9-x\right)}\)
\(=\frac{5\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{5}{\sqrt{x}+3}\)
Giải PT
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{x\left(x+25\right)}{x^2-25}\)
\(ĐKXĐ:x\ne\pm5\)
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{x\left(x+25\right)}{x^2-25}\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}=\frac{x^2+25x}{\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow x^2+10x+25-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow x^2+25x=20x\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\left(ktm\right)\end{cases}}\)
\(\left(\frac{x}{^{x^2-25}}-\frac{x-5}{x^2+5x}\right)\):\(\left(\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\right)\)
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
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
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{\left(x+5\right)\left(x-5\right)}\)
\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)
\(20x=2x^2+50\)
\(20x-2x^2-50=0\)
\(2\left(10x-x^2-25\right)=0\)
\(-x^2+10x+25=0\)
\(x^2-10x+25=0\)
\(x^2-2\left(x\right)\left(5\right)+5^2=0\)
\(\left(x-5\right)^2=0\)
\(x-5=0\Leftrightarrow x=5\)
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m.n giúp mình với ạ
1, với x khác 5, -5 chứng minh biểu thức sau ko phụ thuộc vào x
\(\left(\frac{3}{25-x^2}+\frac{1}{x^2-10x+25}\right)\times\frac{5-x^2}{2}+\frac{3x}{x+5}\left(x\ne-+5\right)\)
\(\frac{-2}{\left(x+5\right)\left(x-5\right)}\)
Rút gọn
\(\left(\frac{x-5\sqrt{x}}{25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
ĐKXĐ :\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\\sqrt{x}+5\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\\\sqrt{x}\ne-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
- Ta có : \(\left(\frac{x-5\sqrt{x}}{25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{x-9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{-x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\frac{x-5\sqrt{x}-25}{25}\right)\left(\frac{\sqrt{x}+5}{-\sqrt{x}-3}\right)\)
\(=\frac{\left(x-5\sqrt{x}-25\right)\left(\sqrt{x}+5\right)}{-25\left(\sqrt{x}+3\right)}=\frac{x\sqrt{x}+5x-5x-25\sqrt{x}-25\sqrt{x}-125}{-25\left(\sqrt{x}+3\right)}\)
\(=\frac{x\sqrt{x}-125-50\sqrt{x}}{-25\left(\sqrt{x}+3\right)}\)
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
b) \(\left(\frac{x}{3}-5\frac{1}{4}\right)^2-\frac{-2}{5}=1\frac{1}{25}\)
c) \(\frac{17-x}{12}=\frac{3}{17-x}\)
d) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
Cho biểu thức \(P=\frac{\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right)}{\left(\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\right)}\)
a)ĐKXĐ (câu này làm được)
b)rút gọn
c)tìm x để P nguyên
a: ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
b: \(P=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\left(\dfrac{10x-25}{x\left(x+5\right)}-\dfrac{x}{x-5}\right)\)
\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}:\dfrac{\left(10x-25\right)\left(x-5\right)-x^2\left(x+5\right)}{x\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{10x-25}{10x^2-50x-25x+125-x^3-5x^2}\)
\(=\dfrac{10x-25}{-x^3+5x^2-75x+125}\)