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\(1a.A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\) ( x ≥ 0 ; x # 9 )
\(b.A>\dfrac{1}{3}\) ⇔ \(\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\text{⇔}\dfrac{3-\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\)
⇔ \(3-\sqrt{x}>0\)
⇔ \(x< 9\)
Kết hợp ĐKXĐ , ta có : \(0\text{≤}x< 9\)
\(c.\) Tìm GTLN chứ ?
\(A=\dfrac{2}{\sqrt{x}+3}\text{≤}\dfrac{2}{3}\)
⇒ \(A_{MAX}=\dfrac{2}{3}."="x=0\left(TM\right)\)
\(a.VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9=VP\)Vậy , đẳng thức được chứng minh .
\(b.VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)Vậy , đẳng thức được chứng minh .
\(c.VT=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}=8=VP\)Vậy , đẳng thức được chứng minh .
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
a: \(C=\left(\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\dfrac{1}{x+1}\right)\cdot\dfrac{x+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x+1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: \(C\cdot\sqrt{x}=\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\)
a: \(B=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: Để |B|=B thì B>=0
=>\(\sqrt{x}-2>=0\)
hay x>4
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
= \(\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
= \(\dfrac{\left(x+\sqrt{x}+1\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1\)
= \(\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\dfrac{x+2}{\sqrt{x}-1}\)
a) ta có:
\(\left\{{}\begin{matrix}1=1\\\sqrt{x}+1=\sqrt{x}+1\end{matrix}\right.\Rightarrow MTC:\sqrt{x}+1\)
Đặt \(Q\left(x\right)=\dfrac{x+\sqrt{x}}{\sqrt{x}-1}+1\)
\(Q\left(x\right)=\dfrac{x+\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}+1}\)
\(Q\left(x\right)=\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\sqrt{x}+1}\)
\(Q\left(x\right)=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(Q\left(x\right)=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}=\sqrt{x}+1\)
\(\Rightarrow P\left(x\right)=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}.\left(\sqrt{x}+1\right)\)
\(P\left(x\right)=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}.\left(\sqrt{x}+1\right)\)
\(P\left(x\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=x-1\)
b) Thay P(x)=x-1, ta có:
\(2x^2+\left(x-1\right)=0\)
\(\Leftrightarrow x^2+x+x^2+1=0\)
\(\Leftrightarrow x\left(x+1\right)+\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0+1=1\\x=0-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy 2x2+P(x)=0 ⇔ \(x\in\left\{-1;\dfrac{1}{2}\right\}\)
a) \(B=\)\(\dfrac{\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}}{\dfrac{\sqrt{x}}{x+\sqrt{x}}}\) ĐKXĐ: x>0
=\(\dfrac{\dfrac{\sqrt{x}+1+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}}{\dfrac{\sqrt{x}}{x+\sqrt{x}}}\)
\(=\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}}:\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
=\(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}}\times\dfrac{x+\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b)
Theo câu a ) ta có :
B=\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
Xét : \(x+\sqrt{x}+1=x+2.\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
=\(\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) (với mọi x>0) (1)
Xét:
\(\sqrt{x}>0\) (2)
Từ (1) và (2) =>\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}>0\) (ĐPCM)
c) B=\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\) ( theo câu a)
=\(\dfrac{x}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+1\)
=\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\)
Áp dụng BĐT cô si cho \(\sqrt{x}\)và \(\dfrac{1}{\sqrt{x}}\)
Ta có : \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}\)
=2
Vậy :\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge2+1\)
Hay\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge3\)
Min B= 3 Dấu "=" xảy ra khi x=1
CHÚC BẠN HỌC TỐT
thanks nha