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đkxđ: x≥0; x≠4
\(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
\(=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{2+\sqrt{x}}\)
+) A = 1/4 <=> \(\dfrac{2}{2+\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow2+\sqrt{x}=8\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(tm)
Vậy x = 36
đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(A=\dfrac{2+\sqrt{x}+2-\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{4-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{2}{\sqrt{x}+2}\)
để \(A=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x}+2=8\)
\(\Leftrightarrow x=36\left(tm\right)\)
vậy tại x=36 thì A=1/4
\(a.P=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}=\dfrac{\sqrt{x}-3-5+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-8+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\) ( x ≥ 0 ; x # 9 )
\(b.\) \(P=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{2\left(\sqrt{x}+2\right)-\sqrt{x}}{\sqrt{x}+2}=2-\dfrac{\sqrt{x}}{\sqrt{x}+2}\text{≤}2\)
⇒ \(P_{Max}=2."="\) ⇔ \(x=0\)
1, ĐKXĐ: x\(\ge0\);x\(\ne1\)
Rút gọn P với \(x\ge0;x\ne1\)ta có
P=\(\dfrac{-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\sqrt{x}+0,5}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)}{x-\sqrt{x}+1}\right)\)
=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-x\sqrt{x}+x-\sqrt{x}+0,5x-0,5\sqrt{x}+0,5+x\sqrt{x}-x-0,5x+0,5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\dfrac{-1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
=\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
2, Thay x=7-4\(\sqrt{3}\)thỏa mãn đk vào P ta có:
P\(=\dfrac{7-4\sqrt{3}-\sqrt{7-4\sqrt{3}}+1}{\sqrt{7-4\sqrt{3}}}\)
=\(\dfrac{7-4\sqrt{3}-\sqrt{\left(\sqrt{3}-2\right)^2}+1}{\sqrt{\left(\sqrt{3}-2\right)^2}}\)
=\(\dfrac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}\)
\(=\dfrac{6-3\sqrt{3}}{2-\sqrt{3}}=12+6\sqrt{3}-6\sqrt{3}-9\)=3
a) điều kiện xác định : \(x\ge2;x\ne5\)
b) \(P=\dfrac{x-5}{\sqrt{x-2}-\sqrt{3}}=\dfrac{\left(\sqrt{x-2}-\sqrt{3}\right)\left(\sqrt{x-2}+\sqrt{3}\right)}{\sqrt{x-2}-\sqrt{3}}\)
\(\Leftrightarrow P=\sqrt{x-2}+\sqrt{3}\)
c) ta có : \(P=\sqrt{x-2}+\sqrt{3}\ge\sqrt{3}\) \(\Rightarrow\) GTNN của \(P\) là \(\sqrt{3}\)
dấu "=" xảy ra khi \(x=2\)
Câu 1:
a, \(P=a+b-a.b\\ < =>P=2+\sqrt{3}+2-\sqrt{3}-\left(2+\sqrt{3}\right).\left(2-\sqrt{3}\right)\\ < =>P=4-\left(4-2\sqrt{3}+2\sqrt{3}-3\right)\\ < =>P=4-4+2\sqrt{3}-2\sqrt{3}+3\\ P=3\)
b, Gỉai hptrình:
\(\left\{{}\begin{matrix}3x+5=5\\x-2y=-3\end{matrix}\right.\)
Giaỉ hpt là tìm: x= 1; y=2
Câu 1b/ ta có: \(\left\{{}\begin{matrix}3x+y=5\\x-2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\left(-3+2y\right)+y=5\\x-2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y-9=5\\x-2y=-3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x-2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
câu 2:
ta có:
\(P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\)
\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\dfrac{x-1}{x}\)
câu 5; ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\ge\dfrac{4}{2\sqrt{2}}=\sqrt{2}\)
a) Để A xác định
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-4\ne0\\\sqrt{x}-2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Vậy với \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\) thì A xác định
b) \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
Để \(A=\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{-1}{3}\)
\(\Leftrightarrow3\sqrt{x}=-\sqrt{x}+2\)
\(\Leftrightarrow4\sqrt{x}=2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)