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A= \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
a, rút gọn A
b, Tìm x để \(A\le0\)
hmm rút gọn nè :)))
\(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(A=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)\(A=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)\(A=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(ĐKXĐ:x\ge0\)
Ta có: \(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}=\frac{x-1}{x+2\sqrt{x}+1}\)
\(\Rightarrow A^2=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}=\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}\)
\(\Rightarrow A^2+A=\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}+\frac{x-1}{x+2\sqrt{x}+1}\)
\(=\frac{2x-2\sqrt{x}}{x+2\sqrt{x}+1}=\frac{2\left(x-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)^2}\)
\(A\le0\Leftrightarrow\orbr{\begin{cases}A=0\\A< 0\end{cases}}\)
+) A = 0\(\Leftrightarrow2\left(x-\sqrt{x}\right)=0\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\left(tm\right)\)
+) A < 0 \(\Leftrightarrow2\left(x-\sqrt{x}\right)< 0\)(vì \(\left(\sqrt{x}+1\right)^2>0\forall x\ge0\)
\(\Leftrightarrow x-\sqrt{x}< 0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}\\\sqrt{x}-1\end{cases}}\)trái dấu
Mà \(\sqrt{x}>\sqrt{x}-1\Rightarrow\hept{\begin{cases}\sqrt{x}>0\\\sqrt{x}< 1\end{cases}}\Leftrightarrow0< x< 1\)
Vậy 0 < x < 1 thì \(A^2+A\le0\)
Sửa)):
\(0\le x\le1\)nha. Ghi nhầm dấu ở kết luận
Do 2 th là \(\hept{\begin{cases}x=0;x=1\\0< x< 1\end{cases}}\Rightarrow\)\(0\le x\le1\)
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1}{x+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}\)
b: Để P<=0 thì \(\dfrac{-\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}< =0\)
\(\Leftrightarrow\sqrt{x}-1>=0\)
hay x>1
1) +) ta có : \(C-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=\dfrac{3\sqrt{x}-x+\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(x-4\sqrt{x}+4\right)+3}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-2\right)^2+3}{3\left(x+\sqrt{x}+1\right)}\)
không thể cm được đâu bn --> xem lại đề
2) +) ta có : \(D=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)
--> để \(D\in Z\Leftrightarrow\sqrt{x}+2\) là ước của 3 \(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x=1\) vậy \(x=1\)
3) +) tương tự 2)
4) a) +) điều kiện xác định : \(x>0;x\ne4\)
ta có : \(A=\left(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-2}{x+3\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
b) ta có : \(A=3\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=3\Leftrightarrow\sqrt{x}-3=3\sqrt{x}-6\)
\(\Leftrightarrow2\sqrt{x}=3\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\) vậy \(x=\dfrac{9}{4}\)
c) ta có : \(B=A.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{x-9}{x-4}=1-\dfrac{5}{x-4}\)
tương tự 2 )
\(\)
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
-\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\left|x\right|-6x+9\)
\(x< 0\)
\(--->x+3-x-6x+9\)
\(=\left(x-x\right)-6x+3+9\)
\(=-6x+\left(3+9\right)=-6x+12\)
\(x>0\)
\(--->3+x+x-6x+9\)
\(=\left(x+x-6x\right)+\left(3+9\right)\)
\(=\left(2x-6x\right)+12\)
\(=4x+12\)
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
a/ ĐKXĐ: x>= 0 ; x khác 1
b/ \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\dfrac{4\sqrt{x}-8}{1-x}\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}-\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}-\dfrac{8\sqrt{x}}{x-1}\right):\dfrac{8-4\sqrt{x}}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{2}-1-8\sqrt{x}}{x-1}\cdot\dfrac{x-1}{8-4\sqrt{x}}\)
\(=\dfrac{-4\sqrt{x}}{x-1}\cdot\dfrac{x-1}{4\left(2-\sqrt{x}\right)}=\dfrac{-4\sqrt{x}}{4\left(2-\sqrt{x}\right)}=-\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
Làm nốt bài 1 ::v
\(\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\dfrac{3+6\sqrt{3}}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}=\dfrac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}=6-\dfrac{13}{\sqrt{3}+4}=\dfrac{11+6\sqrt{3}}{\sqrt{3}+4}\)
\(A=\left(\dfrac{x-2}{\sqrt{x}-1}-\sqrt{x}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{4+\sqrt{x}}{1-x}\right)\)
\(A_1=\left(\dfrac{x-2}{\sqrt{x}-1}-\sqrt{x}\right)=\dfrac{x-2-x+\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(A_2=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{4+\sqrt{x}}{1-x}\right)=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+4\right)}{x-1}=\dfrac{x-4}{x-1}\)
\(\dfrac{A_1}{A_2}=\dfrac{\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)}.\dfrac{x-1}{x-4}=\dfrac{\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\left\{{}\begin{matrix}x\ne1;4\\A=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=1-\dfrac{1}{\sqrt{x}+2}\end{matrix}\right.\)
\(A=\dfrac{3}{4}=1-\dfrac{1}{4}\Rightarrow\sqrt{x}+2=4;x=4\)
\(P=A.\dfrac{x+21}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}.\dfrac{x+21}{\sqrt{x}+1}=\dfrac{x+21}{\sqrt{x}+2}\)
\(P=\dfrac{6\left(\sqrt{x}+2\right)-6\left(\sqrt{x}-12\right)+x+21}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-3\right)^2}{x+2}+6\ge6\)đẳng thức x =3 thỏa mãn nhận
\(P_{min}=6\)
A\(A\le0< =>\dfrac{\sqrt{x}-1}{\sqrt{x}+4}\le0\)
\(< =>\sqrt{x}-1\le0\left(do\sqrt{x}+4\ge0\right)\)
\(< =>\sqrt{x}\le1< =>x\le1\)
Ta có: \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+4}\) (ĐKXĐ: \(x\ge0\))
Để \(A\le0\Rightarrow A=\dfrac{\sqrt{x}-1}{\sqrt{x}+4}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x\le1\)
Vậy \(x\le1\) thì \(A\le0\)