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ĐKXĐ :x\(\ge\)0
a) với x=64 thỏa mãn đk; khi đó: A=\(\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)
b)với đk của x thì B xác định ; ta có
B\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
c)Xét M=A:B =\(\dfrac{2+\sqrt{x}}{\sqrt{2}}:\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Để \(M>\dfrac{3}{2}hay\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\Leftrightarrow2\sqrt{x}+4>3\sqrt{x}+3\left(do:\sqrt{x}+1>0\right)\Leftrightarrow\sqrt{x}< 1\Rightarrow x< 1\)
Kết hợp đk x\(\ge\)0. Vậy 0\(\le\)x<1 thì M=A:B>3/2
\(\frac{3}{\sqrt{7}-1}+\frac{3}{\sqrt{7}+1}=\frac{3\left[\sqrt{7}+1+\sqrt{7}-1\right]}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}=\frac{6\sqrt{7}}{6}=\sqrt{7}\)
\(\frac{3}{\sqrt{X}-1}-\frac{2}{\sqrt{X}+1}+\frac{X-7}{X-1}=\frac{3\left(\sqrt{X}+1\right)-2\left(\sqrt{X}-1\right)+X-7}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{X+\sqrt{X}-2}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{\sqrt{X}+2}{\sqrt{X}+1}\)
TÍNH GIÁ TRỊ BIỂU THỨC:
\(\frac{3}{\sqrt{7}-1}\) + \(\frac{3}{\sqrt{7}+1}\)= \(\frac{3\left(\sqrt{7}+1\right)+3\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}\)= \(\frac{3\sqrt{7}+3+3\sqrt{7}-3}{6}\)=\(\frac{6\sqrt{7}}{6}\)=\(\sqrt{7}\)
RÚT GỌN BIỂU THỨC:
\(\frac{3}{\sqrt{X}-1}\)-\(\frac{2}{\sqrt{X}+1}\)+\(\frac{X-7}{X-1}\)
= \(\frac{3\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)-\(\frac{2\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)+\(\frac{X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{3\sqrt{X}+3-2\sqrt{X}+2+X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{X+\sqrt{X}-2}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\sqrt{X}-2}{\sqrt{X}-1}\)
CHÚC EM HỌC TỐT!
\(A=\frac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4}{x-1}\)
b) \(\frac{4}{x-1}=7\)
\(\Leftrightarrow4=7.\left(x-1\right)\)
\(\Leftrightarrow\frac{4}{7}=x-1\)
\(\Leftrightarrow\frac{4}{7}+1=x\)
\(\Leftrightarrow\frac{11}{7}=x\)
\(\Rightarrow x=\frac{11}{7}\)
\(\left(x-2\right)^2\ge0;\forall x\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2+5\ge5\\\left(x-2\right)^2+9\ge9\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x-2\right)^2+5}+\sqrt{\left(x-2\right)^2+9}\ge\sqrt{5}+\sqrt{9}=\sqrt{5}+3\)
Dấu "=" xảy ra khi và chỉ khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
\(A=\sqrt{x^2-8x+16}+\sqrt{x^2-24x+144}\)
\(A=\sqrt{\left(x-4\right)^2}+\sqrt{\left(12-x\right)^2}\)
\(A=\left|x-4\right|+\left|12-x\right|\ge\left|x-4+12-x\right|=8\)
\(A_{min}=8\) khi \(\left(x-4\right)\left(12-x\right)\ge0\Leftrightarrow4\le x\le12\)
\(f\left(x\right)=\sqrt{3-x}+\sqrt{2+x}\ge\sqrt{3-x+2+x}=\sqrt{5}\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}3-x=0\\2+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
Vậy GTNN của \(f\left(x\right)=\sqrt{5}\) khi và chỉ khi x = 3; x = -2
\(b1:=\sqrt{2}\left(\sqrt{3}+1\right).\sqrt{2-\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\sqrt{4-2\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)\\ =2\\ \\ b2:a,=\sqrt{\dfrac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}-3\right)}{\left(2\sqrt{5}-3\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{2}}{\sqrt{2}}.\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{54-14\sqrt{5}}}{2\sqrt{10}-3\sqrt{2}} .\left(\sqrt{10}-\sqrt{2}\right)\\ \)\(=\dfrac{\sqrt{\left(7-\sqrt{5}\right)^2}}{2\sqrt{10}-3\sqrt{2}}.\left(\sqrt{10}-\sqrt{2}\right)\)\(\\ =\dfrac{8\sqrt{10}-12\sqrt{2}}{2\sqrt{10}-3\sqrt{2}}\\ =4\)
ĐK :\(\hept{\begin{cases}x>=0\\x\ne1\end{cases}}\)
Ta có: \(A=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)+x-1}\right]:\left[\frac{\sqrt{x}+1}{x-1}-\frac{2}{x-1}\right]\)
ĐKXĐ : \(x\ge2\)
Ta có : \(A=\dfrac{x+3\sqrt{x-2}}{x+4\sqrt{x-2}+1}\) . Đặt t = \(\sqrt{x-2}\ge0\) \(\Rightarrow x=t^2+2\)
Khi đó : \(A=\dfrac{t^2+2+3t}{t^2+4t+3}=\dfrac{\left(t+2\right)\left(t+1\right)}{\left(t+3\right)\left(t+1\right)}=\dfrac{t+2}{t+3}=1-\dfrac{1}{t+3}\ge1-\dfrac{1}{3}=\dfrac{2}{3}\)
" = " \(\Leftrightarrow t=0\Leftrightarrow x=2\)
Vậy ...
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