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1a) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2+\sqrt{2}}}\right)}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{8-4\sqrt{2}-\sqrt{16}+2\sqrt{8}}\)
\(=\sqrt{8-4\sqrt{2}-4+4\sqrt{2}}\)
\(=\sqrt{4}=2\)
1b) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+4\sqrt{3}+3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
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: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
a: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
a) \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\) \(=\left|3+\sqrt{2}\right|-\left|1-\sqrt{2}\right|\)
\(=\left(3+\sqrt{2}\right)-\left(\sqrt{2}-1\right)\) \(=3+\sqrt{2}-\sqrt{2}+1=4\)
b) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\) \(=\left|\sqrt{5}-\sqrt{2}\right|+\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)\) \(=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}=2\sqrt{5}\)
c) \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\) \(=\left|\sqrt{2}+1\right|-\left|\sqrt{2}-5\right|\)
\(=\left(\sqrt{2}+1\right)-\left(5-\sqrt{2}\right)\) \(=\sqrt{2}+1-5+\sqrt{2}=2\sqrt{2}-4\)
thanks bạn nhiều, bạn có thể vào trang mình giúp mình giải bài còn lại được ko
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/ \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\) \(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\).
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) \(\sqrt{\left|x\right|-1}\) biểu thức sau có nghĩa \(\Leftrightarrow\) \(\left|x\right|-1\ge0\)
\(\Leftrightarrow\left|x\right|\ge1\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\hoac\\x\le-1\end{matrix}\right.\)
b) \(\sqrt{\left|x-1\right|-3}\) biểu thức sau có nghĩa \(\Leftrightarrow\left|x-1\right|-3\ge0\)
\(\Leftrightarrow\left|x-1\right|\ge3\) \(\left\{{}\begin{matrix}x-1\ge3\\hoac\\x-1\le-3\end{matrix}\right.\)
c) \(\sqrt{4-\left|x\right|}\) biểu thức sau có nghĩa \(\Leftrightarrow4-\left|x\right|\ge0\)
\(\Leftrightarrow4\ge\left|x\right|\) \(\Leftrightarrow-4\le x\le4\)
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. \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
= \(3-\sqrt{6} +2\sqrt{6}-3\) = \(\sqrt{6}\)
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. \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)^2\right)\)\(2.\left(3+2\sqrt{2}+2-1+3-2\sqrt{2}\right)=2.7=14\)