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\(=\sqrt{4x-1-2\sqrt{4x-1}+1}+\sqrt{4x-1+2\sqrt{4x-1}+1}\)
\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)
\(=\left|\sqrt{4x-1}-1\right|+\sqrt{4x-1}+1\)
\(=\left[{}\begin{matrix}2\sqrt{4x-1}\text{ nếu }x\ge\dfrac{1}{2}\\2\text{ nếu }\dfrac{1}{4}\le x< \dfrac{1}{2}\end{matrix}\right.\)
\(A=1-\left(\dfrac{2}{1+2\sqrt{x}}-\dfrac{5\sqrt{x}}{4x-1}-\dfrac{1}{1-2\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(A=1-\dfrac{2\left(2\sqrt{x}-1\right)-5\sqrt{x}+\left(2\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}+1\right)^2}\)
\(A=1-\dfrac{4\sqrt{x}-2-5\sqrt{x}+2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=1-\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=1-\dfrac{2\sqrt{x}+1}{2\sqrt{x}-1}=\dfrac{2\sqrt{x}-1-2\sqrt{x}-1}{2\sqrt{x}-1}=\dfrac{-2}{2\sqrt{x}-1}\)
Tick hộ nha
a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
1) Áp dụng HTL:
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{4^2}+\dfrac{1}{\left(4\sqrt{2}\right)^2}=\dfrac{3}{32}\Rightarrow AH=\dfrac{4\sqrt{6}}{3}\left(cm\right)\)
Áp dụng đ/lý Pytago:
\(BC^2=AB^2+AC^2\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+\left(4\sqrt{2}\right)^2}=4\sqrt{3}\left(cm\right)\)
Bài 2:
a) \(pt\Leftrightarrow\sqrt{\left(2x+1\right)^2}=3\Leftrightarrow\left|2x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b) \(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=2\sqrt{x}.\dfrac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{x}+2\)
\(\sqrt{4x-2\sqrt{4x-1}}+\sqrt{4x+2\sqrt{4x-1}}\)(với \(x\ge\dfrac{1}{4}\))
\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)
\(=\left(\sqrt{4x-1}-1\right)+\left(\sqrt{4x-1}+1\right)\)
\(=2\sqrt{4x-1}\) (với \(x\ge\dfrac{1}{4}\))
1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)
\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)
\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)
\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)
\(=-8\sqrt{3}\)
2) \(A=\sqrt{12-4x}\) có nghĩa khi:
\(12-4x\ge0\)
\(\Leftrightarrow4x\le12\)
\(\Leftrightarrow x\le\dfrac{12}{4}\)
\(\Leftrightarrow x\le3\)
3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
Quy đồng lên ta được:
A=\(\sqrt{1+4X}\) - \(\sqrt{1-4X}\)