Cho \(B=\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{2}\left(x\ge0;x\ne4\right)\)
1.Rút gọn \(B\)
2.Tìm x để \(B< \frac{2}{3}\)
3.Tìm Min \(B\)
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a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
\(A=\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)
\(A=\frac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}-1}=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
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=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)
\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{2x}{\sqrt{x}-2}\)
B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
:V
Câu đầu cho x > 0 thì dễ hơn ......
Sử dụng BĐT AM - GM ta dễ có:\(D=\sqrt{x}+\frac{9}{\sqrt{x}+2}=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-2\ge2\sqrt{\left(\sqrt{x}+2\right)\cdot\frac{9}{\sqrt{x}+2}}-2=4\)
Đẳng thức xảy ra tại x=1
\(E=\frac{x+1}{\sqrt{x}}\ge\frac{2\sqrt{x}}{\sqrt{x}}=2\) Đẳng thức xảy ra tại x=1
Làm 2 cái thôi còn lại tương tự bạn nhé :)
+ Ta có: \(D=\sqrt{x}+\frac{9}{\sqrt{x}+2}\)
\(D=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-2\)
Áp dụng bất đẳng thức Cô-si cho phương trình \(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\) ta có:
\(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge\sqrt{\left(\sqrt{x}+2\right).\left(\frac{9}{\sqrt{x}+2}\right)}=\sqrt{9}=3\)
\(\Rightarrow\)\(D\ge3-2=1\)
Dấu bằng xảy ra khi và chỉ khi: \(\sqrt{x+2}=\frac{9}{\sqrt{x}+2}\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)^2=9\)
\(\Leftrightarrow\sqrt{x}+2=\pm3\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+2=-3\\\sqrt{x}+2=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-5\left(L\right)\\\sqrt{x}=1\end{cases}}\)
\(\Leftrightarrow x=\pm1\)
Vậy \(S=\left\{\pm1\right\}\)
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
1. \(B=\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{2}=\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}-2}{2}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
b,
\(B=\frac{\sqrt{x}+1}{\sqrt{x}+2}< \frac{2}{3}=>3\sqrt{x}+3< 2\sqrt{x}+4=>\sqrt{x}< 1=>0\le x< 1\)
Vậy ...