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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
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
\(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{x-1}{\left(x+\sqrt{x}+1\right)^2}\)
a) Ta có: \(P=\left(\frac{2-\sqrt{x}}{1-x}-\frac{\sqrt{x}-2}{x+2\sqrt{x}+1}\right):\frac{2}{x^2-2x+1}\)
\(=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\left(x-1\right)^2}{2}\)
\(=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\right)\cdot\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-\left(x-3\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x+3\sqrt{x}-2}{2}\cdot\frac{\left(\sqrt{x}-1\right)}{1}\)
\(=\frac{2\sqrt{x}-4}{2}\cdot\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-1\right)\)
\(=x-3\sqrt{x}+2\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(P+x\le2\)
\(\Leftrightarrow x-3\sqrt{x}+2+x-2\le0\)
\(\Leftrightarrow2x-3\sqrt{x}\le0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-3\right)\le0\)
Trường hợp 1: \(\sqrt{x}\left(2\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\2\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\frac{9}{4}\left(nhận\right)\end{matrix}\right.\)
Trường hợp 2: \(\sqrt{x}\left(2\sqrt{x}-3\right)< 0\)
mà \(\sqrt{x}\ge0\forall x\) thỏa mãn ĐKXĐ
nên \(2\sqrt{x}-3< 0\)
\(\Leftrightarrow2\sqrt{x}< 3\)
\(\Leftrightarrow\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow x< \frac{9}{4}\)
mà \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}0< x< \frac{9}{4}\\x\ne1\end{matrix}\right.\)
Vây: Để \(P+x\le2\) thì \(\left\{{}\begin{matrix}0\le x\le\frac{9}{4}\\x\ne1\end{matrix}\right.\)
Mẫu thức chung là (√x+1)(√x−4)
Bạn quy đồng lên rồi tính là ra
P/s: mình hơi lười. Bạn thông cảm nhé
\(ĐKXĐ:x\ne\pm1\)
\(a.D=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\left(\dfrac{2}{x^2-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)=\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\dfrac{2-x\left(x+1\right)+x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x+1\right)\left(x-1\right)}{1-x^2}=\dfrac{4x}{1-x^2}\)\(b.x=\sqrt{3+\sqrt{8}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{2}+1\left(TM\right)\)
Khi đó : \(D=\dfrac{4\left(\sqrt{2}+1\right)}{1-3-2\sqrt{2}}=\dfrac{4\left(\sqrt{2}+1\right)}{-2\left(1+\sqrt{2}\right)}=-2\)
\(c.D=\dfrac{8}{3}\Leftrightarrow\dfrac{4x}{1-x^2}=\dfrac{8}{3}\)
\(\Leftrightarrow\dfrac{12x-8\left(1-x^2\right)}{3\left(1-x^2\right)}=0\)
\(\Leftrightarrow8x^2+12x-8=0\)
\(\Leftrightarrow2x^2-x+4x-2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
KL.........
\(a.P=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}=\dfrac{\sqrt{x}-3-5+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-8+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\) ( x ≥ 0 ; x # 9 )
\(b.\) \(P=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{2\left(\sqrt{x}+2\right)-\sqrt{x}}{\sqrt{x}+2}=2-\dfrac{\sqrt{x}}{\sqrt{x}+2}\text{≤}2\)
⇒ \(P_{Max}=2."="\) ⇔ \(x=0\)
\(\text{a) }\dfrac{1}{\sqrt{x-1}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}x-1\ge0\\\sqrt{x-1}\ne0\end{matrix}\right.\\ \Rightarrow x-1>0\\ \Rightarrow x>1\)
\(\text{b) }\dfrac{1}{\sqrt{x-\sqrt{2x-1}}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}x-\sqrt{2x-1}\ge0\\\sqrt{x-\sqrt{2x-1}}\ne0\end{matrix}\right.\\ \Rightarrow x-\sqrt{2x-1}>0\\ \Rightarrow x>\sqrt{2x-1}\\ \Rightarrow x^2>2x-1\\ \Rightarrow x^2-2x+1>0\\ \Rightarrow\left(x-1\right)^2>0\\ \Rightarrow\left|x-1\right|>0\\ \Rightarrow\left[{}\begin{matrix}x-1< 0\\x-1>0\end{matrix}\right.\\ \Rightarrow x-1\ne0\\ \Rightarrow x\ne1\)
\(c\text{) }\sqrt{-\dfrac{1}{x}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{x}\ge0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}< 0\\x\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x< 0\left(\text{Vì }1>0\right)\\x\ne0\end{matrix}\right.\Rightarrow x< 0\)
\(\text{d) }\sqrt{\dfrac{a+1}{a^2}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}\dfrac{a+1}{a^2}\ge0\\a^2\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+1\ge0\left(\text{Vì }a^2>0\right)\\a\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ge-1\\a\ne0\end{matrix}\right.\)
\(=\dfrac{8-x}{2+\sqrt[3]{x}}:\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}+\dfrac{\sqrt[3]{x^2}-2\sqrt[3]{x}+2\sqrt[3]{x}}{\sqrt[3]{x}-2}\cdot\dfrac{\sqrt[3]{x^2}-1}{\sqrt[3]{x}\left(\sqrt[3]{x}+1\right)}\)
\(=2-\sqrt[3]{x}+\dfrac{\sqrt[3]{x}-1}{\sqrt[3]{x}-2}\)
\(=\dfrac{4-4\sqrt[3]{x}+\sqrt[3]{x^2}-\sqrt[3]{x}+1}{2-\sqrt[3]{x}}\)
\(=\dfrac{\sqrt[3]{x^2}-5\sqrt[3]{x}+5}{2-\sqrt[3]{x}}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{3\sqrt{x}-1}{\sqrt{x}+2}\)