Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b) ĐKXĐ : \(x\ne\pm1\)
\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1+2x-\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{2}{\sqrt{x}}\)
a) ĐKXĐ : \(x\ge0;x\ne16\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)
Ta có :
\(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+1}-\sqrt{x+2}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}=-\sqrt{x+1}+\sqrt{x+2}\)
Tương tự :
\(\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}=-\sqrt{x+2}+\sqrt{x+3}\)
\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)
....
\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)
Từ những ý trên , pt trở thành :
\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)
\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)
\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)
\(\Leftrightarrow121x-900480=0\)
\(\Leftrightarrow x=\dfrac{900480}{121}\)
a: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-\left(2\sqrt{x}+1\right)+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1\)
\(=x-\sqrt{x}\)
b: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)
\(=\dfrac{x-4+5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))
\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(x+2\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x\sqrt{x}+2x+\sqrt{x}}\)
c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
mình nhầm mẫu nhé :v mình làm lại
\(=\left(\dfrac{x-\sqrt{x}-2x+4\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\right):\dfrac{2-\sqrt{x}}{x-1}\)
\(=\dfrac{-x+3\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{2-\sqrt{x}}=\dfrac{\left(2-\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2-\sqrt{x}\right)\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
a: ĐKXĐ: x>0
\(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
b: ĐKXĐ: x>=0; x<>16
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}\)
\(=\dfrac{x+16}{x+16}\cdot\dfrac{\sqrt{x}+2}{x-16}=\dfrac{\sqrt{x}+2}{x-16}\)
c: ĐKXĐ: x>=0; x<>25
\(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
\(=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-10\sqrt{x}+25}{x-25}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
d: \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{-3\sqrt{x}-9}{x-9}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{-3}{\sqrt{x}-3}\)
a.
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ge-1\)
\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
c.
ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)
\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)
\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)
Dấu "=" xảy ra khi và chỉ khi \(x=-1\)
a) \(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\) (ĐK: \(x\ne4,x\ge0\))
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
\(Q=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(Q=\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(Q=\dfrac{6-3\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(Q=\dfrac{3\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(Q=\dfrac{3}{2+\sqrt{x}}\)
b) \(Q=\dfrac{6}{5}\) khi:
\(\dfrac{3}{2+\sqrt{x}}=\dfrac{6}{5}\)
\(\Leftrightarrow15=12+6\sqrt{x}\)
\(\Leftrightarrow6\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)