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Ta có \(\frac{A}{B}=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\right)=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\left[\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right]=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)=\sqrt{x}+4\)
Để \(\frac{A}{B}\ge\frac{x}{4}+5\) thì \(\sqrt{x}+4\ge\frac{x}{4}+5\Leftrightarrow\sqrt{x}\ge\frac{x}{4}+1\Leftrightarrow x-4\sqrt{x}+4\le0\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\)
Mà \(\left(\sqrt{x}-2\right)^2\ge0\)
Suy ra \(\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)(tm)
Vậy x=4 thì \(\frac{A}{B}\ge\frac{x}{4}+5\)
\(B=\frac{1}{\sqrt{x}-1}\) (tự rút gọn nha)
\(\frac{A}{B}\ge\frac{x}{4}+5\\ \sqrt{x}+4\ge\frac{x}{4}+5\\ \frac{x}{4}-\sqrt{x}+1\le0\\ x-4\sqrt{x}+4\le0\\ \left(\sqrt{x}-2\right)^2\le0\\ \Rightarrow\sqrt{x}-2=0\\ \Rightarrow x=4\)
Vậy để \(\frac{A}{B}\ge\frac{x}{4}+5\) thì x=4
Câu 1:
\(\frac{A}{B}\ge\frac{x}{4}+5\Leftrightarrow\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\ge\frac{x}{4}+5\)
\(\Rightarrow\sqrt{x}+4\ge\frac{x}{4}+5\Rightarrow x-4\sqrt{x}+4\le0\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2\le0\Rightarrow\sqrt{x}-2=0\Rightarrow x=4\)
Câu 2:
Bạn coi lại đề, biểu thức B không hợp lý
Mới đc câu a ak, thog cảm nha, trih độ mih thấp lắm:
\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}\)
=\(\frac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}-\frac{2b}{a-b}\)
=\(\frac{a+b-2b}{a-b}=\frac{a-b}{a-b}=1\)
\(A=\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)
\(A=\frac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}-1}=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
và \(B=\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{2+\sqrt{2}}{\sqrt{x}+1}\)
\(B=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(B=\sqrt{3}+2+\frac{1}{\sqrt{3}-\sqrt{2}}+\sqrt{2}\)
\(B=\sqrt{3}+\sqrt{2}+\frac{1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)+1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{3-2+1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{2}{\sqrt{3}-\sqrt{2}}+2\)
để A = B thì \(\sqrt{x}-1\)= \(\frac{2}{\sqrt{3}-\sqrt{2}}+2\)
\(\sqrt{x}=\frac{2}{\sqrt{3}-\sqrt{2}}+3\)
\(\sqrt{x}=\frac{2\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+3\)
\(\sqrt{x}=2\sqrt{3}+2\sqrt{2}+3\)
tới bước này tui bí :(( mong các bạn giỏi khác giúp bạn :D
a/ \(B=\frac{1+x}{1+\sqrt{x}+x}\)
b/ Giải phương trình bậc 2 thì dễ rồi ha
c/ \(\frac{1+x}{1+\sqrt{x}+x}>\frac{2}{3}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)đung vì x khac 1
Phương trình bậc hai là\(x-\sqrt{6x}+1=0\) thì giải làm sao bạn ơi??
\(A-B=\frac{2\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{1-\sqrt{x}}+\frac{3\sqrt{x}-1}{x-1}\)
\(\Leftrightarrow M=\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}+\frac{\left(\sqrt{x}+1\right)^2}{x-1}+\frac{3\sqrt{x}-1}{x-1}\)
\(\Leftrightarrow M=\frac{2x-2\sqrt{x}+x+2\sqrt{x}+1+3\sqrt{x}-1}{x-1}=\frac{3x+3\sqrt{x}}{x-1}=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{3\sqrt{x}}{\sqrt{x}-1}\)
Để \(M< 4\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-1}< 4\)
Nếu x>=1
\(\Rightarrow3\sqrt{x}\le4\sqrt{x}-4\)
\(\Leftrightarrow4\le\sqrt{x}\)
\(\Leftrightarrow x\le16\)
Nếu x<1
\(\Rightarrow3\sqrt{x}>4\sqrt{x}-4\)
\(\Leftrightarrow4>\sqrt{x}\)
\(\Rightarrow16>x\)
Ko có x thỏa mãn
1) Khi x = 49 thì:
\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)
2) Ta có:
\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)
\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)
Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)
Vậy x = 4
\(B=\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\)
\(=\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-1}\)
b) \(\frac{A}{B}=\frac{\sqrt{x}+4}{\sqrt{x-1}}:\frac{1}{\sqrt{x}-1}=\sqrt{x}+4\)
Để \(\frac{A}{B}\ge\frac{x}{4}+5\)
\(\Leftrightarrow\sqrt{x}+4\ge\frac{x}{4}+5\)
\(\Leftrightarrow4\sqrt{x}+16\ge x+20\)
\(\Leftrightarrow x-4\sqrt{x}+4\le0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2\le0\)
Mà \(\left(\sqrt{x}-2\right)^2\ge0;\forall x\ge0\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow x=4\)
Vậy ...