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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: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
sửa đề:
\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
ĐKXĐ: \(x\ge0\);\(x\ne\pm1\)
\(A=\dfrac{4x\sqrt{x}+3x+9+x-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}:\dfrac{x+2\sqrt{x}-4\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4x\sqrt{x}+4x}{x-2\sqrt{x}-3}=\dfrac{4x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{4x}{\sqrt{x}-3}\)
ĐKXĐ: \(x>0,x\ne4\)
B= \(\left(x-\sqrt{x}-2\right).\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{4-\sqrt{x}}{x-2\sqrt{x}}\right)\)
= \(\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right).\dfrac{3\sqrt{x}-4+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{4\left(x-1\right)}{\sqrt{x}}\)
ĐKXĐ: \(x\ge0\)
\(K=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
\(K=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right]:\left[\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right]-1\)
\(K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-1\)
\(K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-2\sqrt{x}+1}-1\\K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}-1\\ K=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1 \)
\(K=\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{x+2}{\sqrt{x}-1}\)
a,Điều kiện:x\(\ge\)0;x\(\ne\)1
=\(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)\(\times\)\(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
=\(\dfrac{\sqrt{x}-1_{ }}{\sqrt{x}}\)
b,<=>\(\dfrac{\sqrt{x}_{ }-1}{\sqrt{x}}\)=\(\dfrac{1}{3}\)
<=>3\(\sqrt{x}\)-3=\(\sqrt{x}\)
<=>2\(\sqrt{x}\)=3
<=>x=9/4
\(1a.A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\) ( 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 .
ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ge0\\x-2>0\\x+2>0\\x\ge0\end{matrix}\right.\) và \(4-x\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x>2\\x>-2\\x\ge0\end{matrix}\right.\) và \(x\ne4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)