Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
a/ \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\) \(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\).
b/ \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\Rightarrow A^2=8+2\sqrt{4^2-\left(\sqrt{10+2\sqrt{5}}\right)^2}=8+2\sqrt{6-2\sqrt{5}}\) \(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{5}+1\)
c/ \(B=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\Rightarrow\sqrt{2}B=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\sqrt{5}+2=2\Rightarrow B=\sqrt{2}\)
a. \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
= \(3-\sqrt{6} +2\sqrt{6}-3\) = \(\sqrt{6}\)
b. \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
= \(\sqrt{8\sqrt{3}}-2.5\sqrt{12}+4\sqrt{8\sqrt{3}}\)
= \(5\sqrt{8\sqrt{3}}-5\sqrt{4.\sqrt{12}}=5\sqrt{8\sqrt{3}}-5\sqrt{4.2\sqrt{3}}\)
= \(5\sqrt{8\sqrt{3}}-5\sqrt{8\sqrt{3}}=0\)
c. \(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\) = \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}.\left(\sqrt{3}+1\right)\)
=\(\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
= 3 - 1 = 2
d. \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)}{\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\) = \(\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}\)
= \(\dfrac{2\sqrt{5}}{\sqrt{2}}\)= \(\sqrt{10}\)
e. \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)^2\right)\)\(2.\left(3+2\sqrt{2}+2-1+3-2\sqrt{2}\right)=2.7=14\)
a: \(=\left|x-4\right|-\left|x-2\right|\)
\(=\left|3\sqrt{2}-1-4\right|-\left|3\sqrt{2}-1-2\right|\)
\(=5-3\sqrt{2}-\left(3\sqrt{2}-3\right)=-6\sqrt{2}+8\)
b: \(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\left|\sqrt{7}-1+1\right|+\left|\sqrt{7}-1-1\right|\)
\(=\sqrt{7}+4-\sqrt{7}=4\)
\(P=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(P=\dfrac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)
\(P=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}\)
\(P=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
\(P=-1+\sqrt{100}=-1+10=9\)
Bài 1 :
\(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
\(=\left(150\sqrt{2}-45\sqrt{2}+10\sqrt{2}\right):\sqrt{10}\)
\(=115\sqrt{2}:\sqrt{10}\)
\(=23\sqrt{5}\)
\(\sqrt{1-x-2x^2}=\sqrt{\left(1+x\right)\left(1-2x\right)}\le\dfrac{1+x-2x+1}{2}=\dfrac{-x+2}{2}\)
(AM-GM)
do đó \(A\le\dfrac{x}{2}+\dfrac{-x+2}{2}=1\)
Dấu = xảy ra khi 1+x=1-2x <=> x=0 (tmđk)
a) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|=\left(\sqrt{5}-\sqrt{2}\right)-\left(\sqrt{5}+\sqrt{2}\right)\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}=-2\sqrt{2}\)
c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|=\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
d) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=2\sqrt{6+2\sqrt{5}}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{5}-2=2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)
a) \(\sqrt{36-6\times2\sqrt{5}+5-5}\)
=\(\sqrt{\left(6-\sqrt{5}\right)^2-5}\)=\(\sqrt{\left(6-\sqrt{5}-\sqrt{5}\right)\left(6-\sqrt{5}+\sqrt{5}\right)}\)
=\(\sqrt{\left(5-2\sqrt{5}+1\right)\times6}\)
=\(\sqrt{\left(\sqrt{5}-1\right)^2\times6}\)
=(\(\sqrt{5}-1\))\(\times\)\(\sqrt{6}\)
Câu b muộn rùi nghỉ đây bạn tự nghĩ đi dễ mà
Mk bt lm hết rùi nhưng dù sao cũng cảm ơn nha ><