22-(-5/7)0+(1/3)4.36
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25 tháng 10 2023
a: \(\sqrt{4\cdot36}=\sqrt{144}=12\)
b: \(\left(\sqrt{8}-3\sqrt{2}\right)\cdot\sqrt{2}\)
\(=\left(2\sqrt{2}-3\sqrt{2}\right)\cdot\sqrt{2}\)
\(=-\sqrt{2}\cdot\sqrt{2}=-2\)
c: \(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}=\dfrac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}=-\sqrt{7}\)
d: \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{\sqrt{5}-2}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)+2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4+2\sqrt{5}+4=4\sqrt{5}\)
ML
2
NT
0
24 tháng 9 2021
\(1,\\ a,ĐK:x-2\ge0\Leftrightarrow x\ge2\\ b,ĐK:2-3x\ge0\Leftrightarrow x\le\dfrac{2}{3}\\ 2,\\ a,=\sqrt{16}-3\sqrt{4}=4-6=-2\\ b,=\dfrac{-\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{7}\\ c,=\sqrt{4}\cdot\sqrt{36}=2\cdot6=12\\ d,=\sqrt{\dfrac{25}{81}}\cdot\sqrt{\dfrac{16}{49}}=\dfrac{5}{9}\cdot\dfrac{4}{7}=\dfrac{20}{63}\\ 3,\\ a,=\sqrt{19+2\sqrt{34}}-\sqrt{19-2\sqrt{34}}\\ =\sqrt{\left(\sqrt{17}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{17}-\sqrt{2}\right)^2}=\sqrt{17}+\sqrt{2}-\sqrt{17}+\sqrt{2}=2\sqrt{2}\\ b,=3-4+2\cdot5=9\)
\(4,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow x+5=4\Leftrightarrow x=-1\left(tm\right)\\ 5,\\ a,B=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ b,B=\dfrac{5}{2}\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{5}{2}\\ \Leftrightarrow2\sqrt{x}+4=5\sqrt{x}\\ \Leftrightarrow3\sqrt{x}=4\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\)
NT
0
6 tháng 4 2020
1/ \(-12;-7;-1;0;4;10;\left|-18\right|\)
2/ \(-13;-5;0;+3;\left|+4\right|;12;\left|-15\right|\)
3/ \(\left|-26\right|;19;-\left(-12\right);0;-4;-\left|35\right|\)
4/ \(-\left(-73\right);\left|+7\right|;4;\left|-1\right|;0;+\left(-5\right);-\left(+22\right)\)
6 tháng 4 2020
Bài 20:
*Tăng dần
1/ -12; -7; -1; 0; +4; 10; |-18|
2/ -13; -5; 0; +3; |+4|; 12; |-15|
*Giảm dần
3/ |-26|; +19; -(-12); 0; -4; -|-35|
4/ -(-73); | +7|; 4; |-1|; 0; +(-5); -(+22)
Chúc bạn học tốt !!!
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)