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a: \(\left(3+\sqrt{5}\right)^2=14+6\sqrt{5}\)
\(\left(2\sqrt{2}+\sqrt{6}\right)^2=14+4\sqrt{12}\)
mà \(6\sqrt{5}< 4\sqrt{12}\)
nên \(3+\sqrt{5}< 2\sqrt{2}+\sqrt{6}\)
c: \(\sqrt{14}-\sqrt{13}=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)
\(\sqrt{12}-\sqrt{11}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)
mà \(\dfrac{1}{\sqrt{14}+\sqrt{13}}< \dfrac{1}{\sqrt{12}+\sqrt{11}}\)
nên \(\sqrt{14}-\sqrt{13}< \sqrt{12}-\sqrt{11}\)
\(P=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(P=\dfrac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)
\(P=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}\)
\(P=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
\(P=-1+\sqrt{100}=-1+10=9\)
1, đk: \(x>0\) và \(x\ne4\)
Ta có: A=\(\dfrac{1}{2\sqrt{x}-x}=\dfrac{1}{-\left(x-2\sqrt{x}+1\right)+1}=\dfrac{1}{-\left(\sqrt{x}-1\right)^2+1}\)
Ta luôn có: \(-\left(\sqrt{x}-1\right)^2\le0\) với \(x>0\) và \(x\ne4\)
\(\Rightarrow-\left(\sqrt{x}-1\right)^2+1\le1\)
\(\Rightarrow A\ge1\). Dấu "=" xảy ra <=> x=1 (t/m)
Vậy MinA=1 khi x=1
2, đk: \(x\ge0;x\ne1;x\ne9\)
Ta có: B=\(\dfrac{1}{x-4\sqrt{x}+3}=\dfrac{1}{\left(x-4\sqrt{x}+4\right)-1}=\dfrac{1}{\left(\sqrt{x}-2\right)^2-1}\)
Ta luôn có: \(\left(\sqrt{x}-2\right)^2\ge0\) với \(x\ge0;x\ne1;x\ne9\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2-1\ge-1\)
\(\Rightarrow B\le-1\). Dấu "=" xảy ra <=> x=4 (t/m)
Vậy MaxB=-1 khi x=4
3, đk: \(x\ge0;x\ne15+4\sqrt{11}\)
Ta có: C=\(\dfrac{1}{4\sqrt{x}-x+7}=\dfrac{1}{-\left(x-4\sqrt{x}+4\right)+11}=\dfrac{1}{-\left(\sqrt{x}-2\right)^2+11}\)
Ta luôn có: \(-\left(\sqrt{x}-2\right)^2\le0\) với \(x\ge0;x\ne15+4\sqrt{11}\)
\(\Rightarrow-\left(\sqrt{x}-2\right)^2+11\le11\)
\(\Rightarrow C\ge\dfrac{1}{11}\). Dấu "=" xảy ra <=> x=4 (t/m)
Vậy MinC=\(\dfrac{1}{11}\) khi x=4
.Cảm ơn!!
đk : \(x\ne4\) ; \(x\ge0\)
1) a) Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\)
Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{6-3\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\) = \(\dfrac{3\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{3}{2+\sqrt{x}}\)
b) ta có Q = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{6}{4+2\sqrt{x}}\) = \(\dfrac{6}{5}\)
\(\Leftrightarrow\) \(4+2\sqrt{x}=5\) \(\Leftrightarrow\) \(2\sqrt{x}=1\) \(\Leftrightarrow\) \(\sqrt{x}=\dfrac{1}{2}\) \(\Leftrightarrow\) \(x=\dfrac{1}{4}\)
c) điều x nguyên ; x \(\ge\) 0 ; x\(\ne\) 4
ta có Q nguyên \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) nguyên
\(\Rightarrow\) \(2+\sqrt{x}\) là ước của 3 là 3 ; 1 ; -1 ; -3
mà \(2+\sqrt{x}\ge2\) (đk :\(x\ge0\)) vậy còn lại 3
\(\Leftrightarrow\) \(2+\sqrt{x}=3\) \(\Leftrightarrow\) x = 1 (tmđk)
vậy x = 1 nguyên thì Q nguyên
2) a) \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) = \(4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}\)
= \(4\sqrt{a}-5\sqrt{10a}\)
b) \(\left(2\sqrt{3}+5\right)\sqrt{3}-\sqrt{60}\) = \(6+5\sqrt{3}-\sqrt{60}\)
c) \(\left(\sqrt{99}-\sqrt{8}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-2\sqrt{22}-11+3\sqrt{22}\)
= \(22+\sqrt{22}\)
1: \(=3+2\sqrt{2}+\sqrt{5}-2=1+2\sqrt{2}+\sqrt{5}\)
2: \(=\dfrac{-\sqrt{7}-\sqrt{5}}{2}-\dfrac{2\left(\sqrt{7}+1\right)}{6}\)
\(=\dfrac{-3\sqrt{7}-3\sqrt{5}-2\sqrt{7}-2}{6}=\dfrac{-5\sqrt{7}-3\sqrt{5}-2}{6}\)
3: \(=-\sqrt{3}-\sqrt{2}-\dfrac{-2\sqrt{3}+3\sqrt{2}}{2}\)
\(=\dfrac{-2\sqrt{3}-2\sqrt{2}+2\sqrt{3}-3\sqrt{2}}{2}=-\dfrac{5\sqrt{2}}{2}\)
mjk đang cần gấp lắm
a: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2\left(\sqrt{5}+1\right)\)
\(=2\sqrt{5}-2\sqrt{5}-2=-2\)
c: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
1) a) \(\sqrt{27}\) + \(\sqrt{75}\) - \(\sqrt{\dfrac{1}{3}}\) = \(3\sqrt{3}\) + \(5\sqrt{3}\) - \(\dfrac{\sqrt{3}}{3}\) = \(8\sqrt{3}\) - \(\dfrac{\sqrt{3}}{3}\)
= \(\dfrac{23\sqrt{3}}{3}\)
b) \(\sqrt{4+2\sqrt{3}}\) \(-\sqrt{4-2\sqrt{3}}\)
= \(\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\) \(-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}\)
= \(\sqrt{\left(\sqrt{3}+1\right)^2}\) \(-\sqrt{\left(\sqrt{3}-1\right)^2}\)
= \(\left(\sqrt{3}+1\right)\) \(-\left(\sqrt{3}-1\right)\)
= \(\sqrt{3}+1-\sqrt{3}+1\)
= 2
2) \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right)\) : \(\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
= \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\) : \(\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\left(\dfrac{a-1}{\left(\sqrt{a}-1\right)\sqrt{a}}\right)\) : \(\left(\dfrac{\left(\sqrt{a}-1\right)+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}\right)\) : \(\left(\dfrac{\left(\sqrt{a}-1\right)+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\dfrac{\sqrt{a}+1}{\sqrt{a}}\) : \(\dfrac{2}{\sqrt{a}+1}\) = \(\dfrac{\sqrt{a}+1}{\sqrt{a}}\) . \(\dfrac{\sqrt{a}+1}{2}\) = \(\dfrac{\left(\sqrt{a}+1\right)^2}{2\sqrt{a}}\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6\right)^2=\left(1-x\right)^2\\-3< =x< =1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6+x-1\right)\left(2x+6+1-x\right)=0\\-3< =x< =1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+5\right)\left(x+7\right)=0\\-3< =x< =1\end{matrix}\right.\Leftrightarrow x=-\dfrac{5}{3}\)
b: \(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=2x\)
\(\Leftrightarrow4\sqrt{x-3}=2x\)
\(\Leftrightarrow2\sqrt{x-3}=x\)
\(\Leftrightarrow\sqrt{4x-12}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2=4x-12\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
a) \(\sqrt{x-3}\) xác định
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy..
b) \(\sqrt{3-2x}\) xác định
\(\Leftrightarrow3-2x\ge0\)
\(\Leftrightarrow x\le-\dfrac{3}{2}\)
Vậy..
c) \(\sqrt{4x^2-1}\) xác định
\(\Leftrightarrow4x^2-1\ge0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1\ge0\\2x+1\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ge\dfrac{-1}{2}\end{matrix}\right.\)\(\Rightarrow x\ge\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1\le0\\2x+1\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\le\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow x\le\dfrac{-1}{2}\)
Vậy ...
d) \(\sqrt{3x^2+2}\) xác định
\(\Leftrightarrow3x^2+2\ge0\)
mà \(3x^2\ge0\)
\(\Rightarrow3x^2+2>0\)
Vậy...
e) \(\sqrt{2x^2+4x+5}\) xác định
\(\Leftrightarrow2x^2+4x+5\ge0\)
mà \(2x^2+4x\ge0\)
\(2x\left(x+2\right)\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x\ge0\\x+2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-2\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x\le0\\x+2\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x\le-2\end{matrix}\right.\)\(\Rightarrow x\le-2\)
\(\Rightarrow2x^2+4x+5>0\)
Vậy...
( Câu này không chắc lắm nha )
Bài 2: Tách sẵn ra cho bạn luôn nhé, không thì bạn nhấn máy tính ra cũng được :v
a) \(-\dfrac{7}{9}\sqrt{\left(-27\right)^2+6\sqrt{1}}\)
\(=-\dfrac{7}{9}\sqrt{\left(-3\right)^2.\left(-9\right)^2+6}\)
\(=\dfrac{-7}{9}\sqrt{735}\)
\(=\dfrac{-7}{9}\sqrt{49.15}\)
\(=\dfrac{-49\sqrt{15}}{9}\)
b) \(\sqrt{49}\sqrt{12^2}+\sqrt{256}:\sqrt{8^2}\)
\(=84+2=86\)
c)\(\sqrt{\left(\sqrt{3-1}\right)^2-\sqrt{\left(\sqrt{3+1}\right)^2}}\)
\(=\sqrt{2-2}\)
= 0
DAT P = Q:R \(Q=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(3\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(3\sqrt{a}-1\right)}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{3\sqrt{a}+1}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(=\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(R=1-\dfrac{2\sqrt{a}-a+1}{3\sqrt{a}+1}=\dfrac{a+\sqrt{a}}{3\sqrt{a}+1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{3\sqrt{a}+1}\)
\(\Rightarrow P=Q:R=\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\times\dfrac{3\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(P=\dfrac{3}{3\sqrt{a}-1}\)
\(P>\dfrac{3}{\left|1-3\sqrt{5}\right|}\Leftrightarrow\dfrac{3}{3\sqrt{a}-1}>\dfrac{3}{3\sqrt{5-1}}\)
\(3\sqrt{a}-1< 3\sqrt{5}-1\)
\(\Rightarrow0\le\sqrt{a}\le\sqrt{5}\)
\(a=\) 0 ;1 ;2 ;3 ;4
a lớn nhất \(\Rightarrow a\) = 4
Bạn rút gọn được P chưa ?~!