So sánh
\(\sqrt{50+2}va\sqrt{50}+\sqrt{2}\)
\(\sqrt{63-27}va\sqrt{63}-\sqrt{27}\)
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So sánh:
a) 7 và \(\sqrt{24}+\sqrt{3}\)
b) \(\sqrt{50+2}\) và \(\sqrt{50}+\sqrt{2}\)
c) \(\sqrt{63-27}\) và \(\sqrt{63}-\sqrt{27}\)
Help me!!! Mk cần gấp!!!
a, ta có:
\(\sqrt{24}=4,89\\ \sqrt{3}=1,73\)
\(\Rightarrow\sqrt{24}+\sqrt{3}=4,89+1,73=6,62\)
vì 7>6,62 nên 7>\(\sqrt{24}+\sqrt{3}\)
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\(7=5+2=\sqrt{25}+\sqrt{4}>\sqrt{24}+\sqrt{3}\)
\(\left(\sqrt{50+2}\right)^2=\left|50+2\right|=50+2\\ \left(\sqrt{50}+\sqrt{2}\right)^2=50+20+2>50+2=\left(\sqrt{50+2}\right)^2\\ \Rightarrow\sqrt{\left(\sqrt{50+2}\right)^2}< \sqrt{\left(\sqrt{50}+\sqrt{2}\right)^2}\\ \Leftrightarrow\left|\sqrt{50+2}\right|< \left|\sqrt{50}+\sqrt{2}\right|\\ \Leftrightarrow\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)
\(\left(\sqrt{63-27}\right)^2=\left|63-27\right|=63-27\\ \left(\sqrt{63}-\sqrt{27}\right)^2=63-2\sqrt{90}+27>63-20+27=63-7>63-27=\left(\sqrt{63-27}\right)^2\\ \Rightarrow\sqrt{\left(\sqrt{63-27}\right)^2}< \sqrt{\left(\sqrt{63}-\sqrt{27}\right)^2}\\ \Leftrightarrow\left|\sqrt{63-27}\right|< \left|\sqrt{63}-\sqrt{27}\right|\\ \Leftrightarrow\sqrt{63-27}< \sqrt{63}-\sqrt{27}\)
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b, Ta có:
\(\sqrt{50+2}=\sqrt{52}=7,21\\
\sqrt{50}+\sqrt{2}=7,07+1,41=8,48
\)
vì 7,21<8,48 nên \(\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)
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Xem thêm câu trả lời
So sánh
\(\sqrt{63-27}\) và \(\sqrt{63}-\sqrt{27}\)
So sánh
\(\sqrt{63}-\sqrt{27}\) và \(\sqrt{63-27}\)
so sánh \(\sqrt{63}-\sqrt{27}\) và \(\sqrt{63-27}\)
Ta có:
\(\sqrt{63-27}=\sqrt{36}=6\)
\(\sqrt{63}-\sqrt{27}
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1.So sánh: \(\sqrt{63-27}\) và\(\sqrt{63}-\sqrt{27}\)
2.Cho N = \(\frac{9}{\sqrt{x}-5}\) Tìm x thuộc Z để N có giá trị nguyên
So sánh
\(\sqrt{8}_{ }\)_\(\sqrt{5}\)và 1
\(\sqrt{63-27}\) và \(\sqrt{63}\)_\(\sqrt{27}\)
Ta có\(8< 16\Rightarrow\sqrt{8}< \sqrt{16}=4\)
và \(5< 9\Rightarrow\sqrt{5}< \sqrt{9}=3\)
\(\Rightarrow\sqrt{8}-\sqrt{5}< \sqrt{16}-\sqrt{9}=4-3=1\)
Vậy \(\sqrt{8}-\sqrt{5}< 1\)
Ta có \(\sqrt{63-27}=\sqrt{36}=6\)
lại có\(63< 64\Rightarrow\sqrt{63}< \sqrt{64}=8\)và \(27>4\Rightarrow\sqrt{27}>\sqrt{4}=2\)
\(\Rightarrow\sqrt{63}-\sqrt{27}< \sqrt{64}-\sqrt{4}=8-2=6\)
mà\(\sqrt{63-27}=6\Rightarrow\sqrt{63}-\sqrt{27}< \sqrt{63-27}\)
Vậy\(\sqrt{63}-\sqrt{27}< \sqrt{63-27}\)
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Rút Gọn1.3sqrt{3}+4sqrt{12}-5sqrt{27}2.sqrt{32}-sqrt{50}+sqrt{18}3.sqrt{72}+sqrt{4frac{1}{2}}-sqrt{32}-sqrt{162}4.left(sqrt{325}-sqrt{117}+2sqrt{208}right):sqrt{13}5.left(sqrt{12}-sqrt{48}-sqrt{108}-sqrt{192}right):2sqrt{3}6.left(2sqrt{112}-5sqrt{7}+2sqrt{63}-2sqrt{28}right)sqrt{7}7.left(2sqrt{27}-3sqrt{48}+3sqrt{75}-sqrt{192}right)left(1-sqrt{3}right)8.7sqrt{24}-sqrt{150}-5sqrt{54}9.2sqrt{20}-sqrt{50}+3sqrt{80}-sqrt{320}10.sqrt{32}-sqrt{50}+sqrt{98}-sqrt{72}11.3sqrt{2}-4sqrt{18}+2sqrt{32}-sqrt{...
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Rút Gọn
1.\(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
2.\(\sqrt{32}-\sqrt{50}+\sqrt{18}\)
3.\(\sqrt{72}+\sqrt{4\frac{1}{2}}-\sqrt{32}-\sqrt{162}\)
4.\(\left(\sqrt{325}-\sqrt{117}+2\sqrt{208}\right):\sqrt{13}\)
5.\(\left(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\right):2\sqrt{3}\)
6.\(\left(2\sqrt{112}-5\sqrt{7}+2\sqrt{63}-2\sqrt{28}\right)\sqrt{7}\)
7.\(\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)\)
8.\(7\sqrt{24}-\sqrt{150}-5\sqrt{54}\)
9.\(2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
10.\(\sqrt{32}-\sqrt{50}+\sqrt{98}-\sqrt{72}\)
11.\(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)
12.\(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
13.\(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
14.\(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)
15.\(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}\)
16.\(10\sqrt{28}-2\sqrt{275}-3\sqrt{343}-\frac{3}{2}\sqrt{396}\)
Rút gọn các biểu thức:
1. Aleft(sqrt{5}-2right)left(sqrt{5}+2right)
2. B left(sqrt{45}+sqrt{63}right)left(sqrt{7}-sqrt{5}right)
3. C left(sqrt{5}+sqrt{3}right)left(5-sqrt{15}right)
4. D left(sqrt{32}-sqrt{50}+sqrt{27}right)left(sqrt{27}+sqrt{50}-sqrt{32}right)
5. E left(sqrt{3}+1right)^2-2sqrt{3}+4
6. F left(sqrt{15}-2sqrt{3}right)^2+12sqrt{5}
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Rút gọn các biểu thức:
1. A=\(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
2. B= \(\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
3. C= \(\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)\)
4. D= \(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
5. E= \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
6. F= \(\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}\)
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
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So sánh
a) \(\sqrt{8}-\sqrt{5}\) với 1
b) \(\sqrt{63-27}với\sqrt{63}-\sqrt{27}\)
Giúp mk nha, thứ 2 mk phải nộp rồi
Cách giải đầy đủ nha
a, Ta có: 1= \(\sqrt{9}-\sqrt{4}\)
Nx: \(\begin{cases} 0<8<9\\ 5>4>0 \end{cases}\)
=>\(\begin{cases} \sqrt{8}<\sqrt{9}\\ \sqrt{5}>\sqrt{4} \end{cases}\)
=>\(\sqrt{8}-\sqrt{5}<\sqrt{9}-\sqrt{4}\)
=>\(\sqrt{8}-\sqrt{5}<1\)
b,Ta có: \(\sqrt{63-27}=\sqrt{36}=6\)
\(\sqrt{63}-\sqrt{27}<\sqrt{64}-\sqrt{25}\\ =>\sqrt{63}-\sqrt{27}<3\\ =>\sqrt{63}-\sqrt{27}<6(vì 3<6)\\ =>\sqrt{63}-\sqrt{27}<\sqrt{63-27} \)
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