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2 13 + 2 5 trên 2 10 + 2 2
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TU
0
20 tháng 6 2023
a: \(=\dfrac{2^{12}\cdot3^{14}}{3^{12}\cdot2^{12}}=3^2=9\)
b: \(=\dfrac{7^3\cdot2\cdot5^3}{5^2\cdot7^2\cdot6}=7\cdot5\cdot\dfrac{1}{3}=\dfrac{35}{3}\)
d: =2^5(2^8+1)/2^2(2^8+1)=2^3=8
c: \(=\dfrac{5^3\cdot3^6\cdot2^8\cdot5^4\cdot3^4\cdot2^2}{2^{10}\cdot3^{10}\cdot5^5}=5^2\)
TH
1
TH
0
6 tháng 1 2022
a: =>(x+10)(x-1)=0
=>x=-10 hoặc x=1
b: \(A=x^3-1-\left(x+5\right)\left(x^2-3\right)-5x^2-10x-5\)
\(=x^3-5x^2-10x-6-x^3+3x-5x^2+15\)
=-7x+9
=110/13
24 tháng 6 2021
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
NM
2
13 tháng 3 2018
=\(\frac{7.\left(2^2\right)^5.3^{10}.3+2^{10}.2^3.\left(3^2\right)^5}{2^{10}.3^{10}+2^{10}.3^{10}.2^2}\)
=\(\frac{7.2^{10}.3^{10}.3+2^{10}.2^3.3^{10}}{2^{10}.3^{10}+2^{10}.3^{10}.2^2}\)
=\(\frac{2^{10}.3^{10}\left(7.3+2^3\right)}{2^{10}.3^{10}\left(1+2^2\right)}\)
=\(\frac{7.3+2^3}{1+2^2}\)
13 tháng 3 2018
\(\frac{7.4^5.3^{11}+2^{13}.9^5}{6^{10}+2^{12}.3^{10}}=\frac{7.\left(2^2\right)^5.3^{11}+2^{13}.\left(3^2\right)^5}{\left(2.3\right)^{10}+2^{12}.3^{10}}=\frac{7.2^{10}.3^{11}+2^{13}.3^{10}}{2^{10}.3^{10}+2^{12}.3^{10}}\)
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