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RK
21 tháng 6 2021
Pt <=> \(\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{1}{550}\)
<=> \(\dfrac{\left(x+5\right)-x}{x\left(x+5\right)}=\dfrac{1}{550}\)
<=> \(\dfrac{5}{x\left(x+5\right)}=\dfrac{1}{550}\)
<=> \(x^2+5x=2750\)
<=> \(x^2+5x-2750=0\)
<=> \(\left(x^2+5x+2,5^2\right)-52,5^2=0\) (bước này hơi tắt xíu nha :<)
<=> \(\left(x+2,5\right)^2-52,5^2=0\)
<=> \(\left(x+55\right)\left(x-50\right)=0\)
<=> \(\left[{}\begin{matrix}x+55=0\\x-50=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-55\\x=50\end{matrix}\right.\)
Vậy nghiệm của phương trình là x \(\in\left\{-55;50\right\}\)
DM
21 tháng 6 2021
\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\)
\(\dfrac{1100\left(x+5\right)-1100x}{x\left(x+5\right)}=\dfrac{2x\left(x+5\right)}{x\left(x+5\right)}\)
\(1100x+5500-1100x=2x^2+10x\)
\(2x^2+10x-5500=0\)
Δ' \(=5^2-2\left(-5500\right)\)
Δ'\(=11025\)
\(\left[{}\begin{matrix}x=50\\x=-55\end{matrix}\right.\)
21 tháng 6 2021
\(\left\{{}\begin{matrix}xy=1100\\y-\dfrac{1100}{x+5}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1100}{x}\left(x\ne0\right)\left(1\right)\\\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\left(2\right)\end{matrix}\right.\)
* giải pt(2)\(=>\dfrac{1100x+5500-1100x}{x^2+5x}=2\)
\(=>5500=2x^2+10x=>2x^2+10x-5500=0\)
\(=>\Delta=10^2-4\left(-5500\right)2=44100>0\)
\(=>\left[{}\begin{matrix}x1=\dfrac{-10+\sqrt{44100}}{2.2}=50\left(TM\right)\left(3\right)\\x2=\dfrac{-10-\sqrt{44100}}{2.2}=-55\left(TM\right)\left(4\right)\end{matrix}\right.\)
thế(3)(4) vào(1)\(=>\left[{}\begin{matrix}y=\dfrac{1100}{50}=22\\y=\dfrac{1100}{-55}=-20\end{matrix}\right.\)
vậy...
G
10 tháng 8 2023
a) Đk: \(x>0;x\ne9;x\ne25\)
Đặt \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)\(:\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}+x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+5}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\)
\(=\dfrac{x}{\sqrt{x}-5}\)
G
10 tháng 8 2023
b) Đk: \(x\ge0;x\ne1;x\ne25\)
Biểu thức
\(=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{x+9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-x-9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)
\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)\(=\dfrac{\left(1-\sqrt{x}\right)^2}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)
23 tháng 7 2021
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
PN
7 tháng 10 2023
\(a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}+35}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
PN
7 tháng 10 2023
\(b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x-5\sqrt{x}-2}{x-9}\)
10 tháng 7 2021
a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)
\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)
=-1
10 tháng 7 2021
Bài 1:
a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:
\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)
b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)
18 tháng 9 2021
a) \(\dfrac{12}{1+\sqrt{5}}+\dfrac{15}{\sqrt{5}}-\dfrac{\sqrt{20}-5}{2-\sqrt{5}}\)
=\(\dfrac{12\left(1-\sqrt{5}\right)}{-4}+\dfrac{15\sqrt{5}}{5}-\dfrac{\left(\sqrt{20}-5\right)\left(2+\sqrt{5}\right)}{-1}\)
=\(-3+3\sqrt{5}-\sqrt{5}+3\sqrt{5}+4\sqrt{5}+10-10-5\sqrt{5}\)
=\(5\sqrt{5}-3\)
b)\(\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{3x}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}}\)
=\(\dfrac{2x-3x+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
=\(\dfrac{-x+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
đk : x khác 0 ; -5
\(1100x+5500-1100x=2x\left(x+5\right)\)
\(\Leftrightarrow2x^2+10x-5500=0\Leftrightarrow x=50;x=-55\)(tm)
= - 55 tm