Tính: (\(\sqrt[3]{\dfrac{1}{9}}+4\sqrt[3]{\dfrac{1}{72}}-\sqrt[3]{4}\) )(\(\sqrt[3]{72}+\sqrt[3]{96}+\sqrt[3]{128}\))
Tính:
\(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=-4\sqrt{5}+15\sqrt{2}\)
b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)
\(=4\sqrt{3}-4+4\sqrt{3}+2\sqrt{2}\)
\(=8\sqrt{3}+2\sqrt{2}-4\)
c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3
=6
d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
Tính:
\(A=\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
\(B=\dfrac{1}{\sqrt{2}-1}+\dfrac{14}{3+\sqrt{2}}\)
\(C=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(D=\sqrt{\left(1-\sqrt{2}\right)^2}-3\sqrt{18}+4\sqrt{\dfrac{1}{2}}\)
\(\sqrt{\dfrac{72.x}{128}}=\dfrac{3}{4}\) Giải phương trình: giúp mik vs
\(\sqrt{3}x^2-\sqrt{1587}x=0\)
\(\sqrt{\dfrac{72x}{128}}=\dfrac{3}{4}\)
\(\Leftrightarrow x\cdot\dfrac{9}{16}=\dfrac{9}{16}\)
hay x=1
\(\left(4\sqrt{8}-\sqrt{72}+5\sqrt{\dfrac{1}{2}}\right)2\sqrt{2}\)
\(\dfrac{5+\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\left(\sqrt{3}+\sqrt{5}\right)\)
a: \(\left(4\sqrt{8}-\sqrt{72}+5\sqrt{\dfrac{1}{2}}\right)\cdot2\sqrt{2}\)
\(=\left(4\cdot2\sqrt{2}-6\sqrt{2}+\dfrac{5}{\sqrt{2}}\right)\cdot2\sqrt{2}\)
\(=\left(2\sqrt{2}+\dfrac{5}{\sqrt{2}}\right)\cdot2\sqrt{2}\)
\(=2\sqrt{2}\cdot2\sqrt{2}+\dfrac{5}{\sqrt{2}}\cdot2\sqrt{2}\)
\(=8+10=18\)
b: Sửa đề:\(\dfrac{5+\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\sqrt{5}+1+\sqrt{3}-\sqrt{3}-\sqrt{5}\)
=1
\(\left(4\sqrt{8}-\sqrt{72}+5\sqrt{\dfrac{1}{2}}\right)2\sqrt{2}\\ =\left(8\sqrt{2}-6\sqrt{2}+5\sqrt{\dfrac{1}{2}}\right)2\sqrt{2}\\ =8.2.\sqrt{2}.\sqrt{2}-6.2.\sqrt{2}.\sqrt{2}+5.2.\sqrt{2}.\sqrt{\dfrac{1}{2}}\\ =32-24+10\\ =18\\ \dfrac{5+\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\left(\sqrt{3}+\sqrt{5}\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}\left(\sqrt{3}+\sqrt{5}\right)\\ =\sqrt{5}+1+\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)\\ =\sqrt{5}+1+3+\sqrt{15}\\ =4+\sqrt{5}+\sqrt{15}\)
A\(=\)\((3\sqrt{8}+2\sqrt{50}-4\sqrt{72})\)\(➗\)\(8\sqrt{2}\)
B\(=\)\((-4\sqrt{20}+5\sqrt{500}-3\sqrt{45})\div5 \)
C\(=(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{\sqrt{3}-1}{\sqrt{3}+1})\div\sqrt{48}\)
c: Ta có: \(C=\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right):\sqrt{48}\)
\(=\dfrac{4+2\sqrt{3}-4+2\sqrt{3}}{2}:4\sqrt{3}\)
\(=\dfrac{1}{2}\)
6) (3\(\sqrt{2}\) -\(\sqrt{3}\))(\(\sqrt{3}\)+3\(\sqrt{2}\))
7) \(\sqrt{72}\)+\(\sqrt{4\dfrac{1}{2}}\) - \(\sqrt{32}\) - \(\sqrt{162}\)
6: Ta có: \(\left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)\)
=18-3
=15
7: Ta có: \(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)
\(=6\sqrt{2}+\dfrac{3}{2}\sqrt{2}-4\sqrt{2}-9\sqrt{2}\)
\(=-\dfrac{11}{2}\sqrt{2}\)
tính
A=\(\left(1-\sqrt{7}\right).\dfrac{\sqrt{7}+7}{2\sqrt{7}}\)
B=\(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
C=\(\sqrt{32}-\sqrt{50}+\sqrt{18}\)
D=\(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)
E=\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
Tính:
\(\dfrac{\sqrt[4]{7\sqrt[3]{54}+15\sqrt[3]{128}}}{\sqrt[3]{\sqrt[4]{32}}+\sqrt[3]{9\sqrt[4]{162}}}\)
\(A=\dfrac{\sqrt[4]{7\sqrt[3]{54}+15\sqrt[3]{128}}}{\sqrt[3]{\sqrt[4]{32}}+\sqrt[3]{9\sqrt[4]{162}}}\)
\(\Leftrightarrow A=\dfrac{\sqrt[4]{7\sqrt[3]{3^3.2}+15\sqrt[3]{4^3.2}}}{\sqrt[3]{\sqrt[4]{2^4.2}}+\sqrt[3]{9\sqrt[4]{3^4.2}}}\)
\(\Leftrightarrow A=\dfrac{\sqrt[4]{7.3\sqrt[3]{2}+15.4\sqrt[3]{2}}}{\sqrt[3]{2\sqrt[4]{2}}+\sqrt[3]{9.3\sqrt[4]{2}}}\)
\(\Leftrightarrow A=\dfrac{\sqrt[4]{21\sqrt[3]{2}+60\sqrt[3]{2}}}{\sqrt[3]{2\sqrt[4]{2}}+\sqrt[3]{3^3\sqrt[4]{2}}}\)
\(\Leftrightarrow A=\dfrac{\sqrt[4]{81\sqrt[3]{2}}}{\sqrt[3]{\sqrt[4]{2}}\left(\sqrt[3]{2}+3\right)}=\dfrac{3\sqrt[4]{\sqrt[3]{2}}}{\sqrt[3]{\sqrt[4]{2}}\left(\sqrt[3]{2}+3\right)}\)
\(\Leftrightarrow A=\dfrac{3}{\sqrt[3]{2}+3}\)
1. Tìm số tự nhiên n sao cho :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{n.\left(n+1\right)}=\dfrac{2999}{3000}\)
2. Tính :
a ) \(S=2018.3+2018.4+2018.5+...+2018.2018\)
b ) \(\dfrac{1}{\sqrt{8}+\sqrt{10}}+\dfrac{1}{\sqrt{10}+\sqrt{12}}+\dfrac{1}{\sqrt{12}+\sqrt{14}}+...+\dfrac{1}{\sqrt{200}+\sqrt{202}}\)
c ) \(S=5.21^2+5.21^3+5.21^4+....+5.21^{2018}\)
d ) \(A=9+99+999+9999+...+9..9\)( 99 chữ số 9)
e ) 72+772+7772+...+77...72( 77 chữ số 7 )
2. Tính tổng :
a ) \(S=\dfrac{1}{3\sqrt{1}+3\sqrt{3}}+\dfrac{1}{3\sqrt{3}+3\sqrt{5}}+...+\dfrac{1}{3\sqrt{2017}+3\sqrt{2019}}\)
b ) S = \(\dfrac{1}{\sqrt{2.2}+\sqrt{2.3}}+\dfrac{1}{\sqrt{2.3}+\sqrt{2.4}}+\dfrac{1}{\sqrt{2.4}+\sqrt{2.5}}+...+\dfrac{1}{\sqrt{2.2018}+\sqrt{2.2019}}\)
Câu 1:
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
=>n+1=3000
hay n=2999