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a) 3\(\sqrt{3}\)=\(\sqrt{27}\)>\(\sqrt{12}\)
c) \(\frac{1}{3}\)\(\sqrt{51}\)=\(\sqrt{\frac{51}{9}}\)<\(\frac{1}{5}\)\(\sqrt{150}\)=\(\sqrt{\frac{150}{25}}\)=\(\sqrt{6}\)
b) 3\(\sqrt{5}\)=\(\sqrt{45}\)< 7=\(\sqrt{49}\)
d) \(\frac{1}{2}\sqrt{6}\)=\(\sqrt{\frac{6}{4}}\)=\(\sqrt{\frac{3}{2}}\)< 6\(\sqrt{\frac{1}{2}}\)=\(\sqrt{\frac{36}{2}}\)=\(\sqrt{18}\)
a) Ta có:
Vì nên
Vậy .
b) Ta có:
Vì nên
Vậy .
c) Ta có:
Vì nên
Vậy .
d) Ta có:
Vì nên
Vậy .
a) \(1=\sqrt{1}< \sqrt{2}\)
b) \(2=\sqrt{4}>\sqrt{3}\)
c) \(6=\sqrt{36}< \sqrt{41}\)
d) \(7=\sqrt{49}>\sqrt{47}\)
e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)
f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)
g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)
h) \(\sqrt{3}>0>-\sqrt{12}\)
i) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
B2:
3) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2019}+\sqrt{2020}}\)
\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2020}-\sqrt{2019}}{2020-2019}\)
\(=\sqrt{2}-1+\sqrt{3}-2+...+\sqrt{2020}-\sqrt{2019}\)
\(=\sqrt{2020}-1\)
Câu 4: a) ĐK: \(x^2\ge9\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
b) ĐK: \(x^2-3x+2\ge0\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)
c) Đk: \(-3\le x< 5\)
d) x + 3 và 5 - x đồng dấu. Xét hai trường hợp:
\(\left\{{}\begin{matrix}x+3\ge0\\5-x>0\left(\text{do mẫu phải khác 0}\right)\end{matrix}\right.\Leftrightarrow-3\le x< 5\)
\(\left\{{}\begin{matrix}x+3< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>5\end{matrix}\right.\) do x ko thể đồng thời thỏa mãn cả hai nên loại.
Câu 1:
a) Đặt \(A=x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\)
\(A=x+\left|x+2\right|\cdot\left(x-2\right)\)
+) Với \(x\ge-2\):
\(A=x+\left(x+2\right)\left(x-2\right)=x+x^2-4\)
+) Với \(x< -2\):
\(A=x-\left(x+2\right)\left(x-2\right)=x-x^2+4\)
b) \(B=\sqrt{m^2-6m+9-2m}\)
\(B=\sqrt{m^2-8m+9}\)
Bạn xem lại đề nhé :)
c) \(C=1+\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(C=1+\sqrt{x-1}\)
d) \(D=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(D=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(D=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(D=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
+) Xét \(x\ge8\):
\(D=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
+) Xét \(4< x< 8\):
\(D=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
Vậy....
a) Ta có: \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
\(=\sqrt{3}\left(2+\sqrt{16}-\sqrt{25}-\sqrt{81}\right)\)
\(=\sqrt{3}\left(2+4-5-9\right)\)
\(=-8\sqrt{3}\)
b) Ta có: \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)
\(=\left(\frac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)\)
\(=7-5=2\)
c) Ta có: \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\left(\sqrt{3}+1\right)\cdot\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\cdot\left|\sqrt{3}-1\right|\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
d) Ta có: \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)
\(=\sqrt{2}\cdot\left(5+\sqrt{9}-\sqrt{49}-\sqrt{144}\right)\)
\(=\sqrt{2}\cdot\left(5+3-7-12\right)\)
\(=-11\sqrt{2}\)
e) Ta có: \(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)
\(=\left(\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=3-5=-2\)
g) Ta có: \(\left(\sqrt{3}-1\right)\cdot\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\)
\(=\left(\sqrt{3}-1\right)\cdot\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\cdot\left|\sqrt{3}+1\right|\)
\(=\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1>0\))
\(=3-1=2\)
a) Ta thấy:
\(\left(3+\sqrt{5}\right)^2=\left(\sqrt{9}+\sqrt{5}\right)^2=9+5+2\sqrt{45}=14+2\sqrt{45}\)
\(\left(2\sqrt{2}+\sqrt{6}\right)^2=\left(\sqrt{8}+\sqrt{6}\right)^2=8+6+2\sqrt{48}=14+2\sqrt{48}\)
Vì \(45< 48\)
\(\Rightarrow\sqrt{45}< \sqrt{48}\)
\(\Rightarrow2\sqrt{45}< 2\sqrt{48}\)
\(\Rightarrow14+2\sqrt{45}< 14+2\sqrt{48}\)
\(\Rightarrow\left(3+\sqrt{5}\right)^2< \left(2\sqrt{2}+\sqrt{6}\right)^2\)
Do \(3+\sqrt{5}>0;2\sqrt{2}+\sqrt{6}>0\)
\(\Rightarrow3+\sqrt{5}< 2\sqrt{2}+6\)
b) Ta thấy:
Vì \(26>3\)
\(\Rightarrow\sqrt{26}>\sqrt{3}\)
\(\Rightarrow\sqrt{26}+1>\sqrt{3}\)
\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{27}+\sqrt{3}\)
Mà \(\sqrt{27}+\sqrt{3}=3\sqrt{3}+\sqrt{3}=4\sqrt{3}=\sqrt{48}\)
\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{48}\)