\(\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(2^4-4^2\right)\)

\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(16-16\right)\)

\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).0\)

\(=0\)

-------------------------------------------------

\(\left(7^{1997}-7^{1995}\right):\left(7^{1994}.7\right)\)

\(=\left[7^{1995}\left(7^2-1\right)\right]:7^{1995}\)

\(=7^{1995}.48:7^{1995}\)

\(=48\)

-------------------------------------------------

\(\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-81^2\right)\)

\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(6561-6561\right)\)

\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).0\)

\(=0\)

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\(\left(2^8+8^3\right):\left(2^5.2^3\right)\)

\(=\left[2^8+\left(2^3\right)^3\right]:2^8\)

\(=\left(2^8+2^9\right):2^8\)

\(=2^8.\left(1+2\right):2^8\)

\(=2^8.3:2^8\)

\(=3\)

sos

\((2^{17}+17^{2}).(9^{15}-3^{15}).(2^{4}-4^{2})\)

\(=(2^{17}+17^{2}).(9^{15}-3^{15}).(16-16)\)

\(=(2^{17}+17^{2}).(9^{15}-3^{15}).0\)

\(=0\)

\((7^{1997}-7^{1995}):(7^{1994}.7)\)

\(=(7^{1997}-7^{1995}):7^{1995}\)

\(=7^{1997}:7^{1995}-7^{1995}:7^{1995}\)

\(=7^{2}-1=49-1=48\)

\(1^{2}+2^{3}+3^{4}+4^{5}).(1^{3}+2^{3}+3^{3}+4^{3}).(3^{8}-81^{2})\)

\(=(1^{2}+2^{3}+3^{4}+4^{5}).(1^{3}+2^{3}+3^{3}+4^{3}).(6561-6561)\)

\(=(1^{2}+2^{3}+3^{4}+4^{5}).(1^{3}+2^{3}+3^{3}+4^{3}).0\)

\(=0\)

\((2^{8}+8^{3}):(2^{5}.2^{3})\)

\(=[2^{8}+(2^{3})^{3}]:2^{8}\)

\(=(2^{8}+2^{9}):2^{8}\)

\(=2^{8}:2^{8}+2^{9}:2^{8}\)

\(=1+2=3\)

Bài 2: 

Ta có: \(x\left(x-4\right)-x^2+8=0\)

\(\Leftrightarrow x^2-4x-x^2+8=0\)

\(\Leftrightarrow-4x=-8\)

hay x=2

1)=x²-49-x²

=-49

2)=>x²-4x-x²+8=0

=>-4x+8=0

=>-4x=-8

=>x=2

`a)sqrt{(sqrt7-4)^2}+sqrt7`

`=|sqrt7-4|+sqrt7`

`=4-sqrt7+sqrt7=4`

`b)\sqrt{81a}-sqrt{144a}+sqrt{36a}(a>=0)`

`=9sqrta-12sqrta+6sqrta=3sqrta`

a) Ta có: \(\sqrt{\left(\sqrt{7}-4\right)^2}+\sqrt{7}\)

\(=4-\sqrt{7}+\sqrt{7}\)

=4

b) Ta có: \(\sqrt{81a}-\sqrt{144a}+\sqrt{36a}\)

\(=9\sqrt{a}-12\sqrt{a}+6\sqrt{a}\)

\(=3\sqrt{a}\)

a) \(A=\dfrac{2^6\cdot9^2}{6^4\cdot8}\)

\(=\dfrac{2^6\cdot\left(3^2\right)^2}{3^4\cdot2^4\cdot2^3}\)

\(=\dfrac{2^6\cdot3^4}{3^4\cdot2^7}\)

\(=\dfrac{1}{2}\)

b) \(B=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot9^2}\)

\(=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot\left(3^2\right)^2}\)

\(=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^6}\)

\(=\dfrac{3}{2^2}\)

\(=\dfrac{3}{4}\)

\(A=\dfrac{2^6.9^2}{6^4.8}=\dfrac{2^6.3^4}{2^7.3^4}=\dfrac{1}{2}\)

\(B=\dfrac{2^{13}.3^7}{2^{15}.3^2.9^2}=\dfrac{3^7}{2^2.3^6}=\dfrac{3}{4}\)

\(a,=-x^3+3x^2+2x-6\)

\(\left(x^2-2\right)\left(-x+3\right)\)

\(=-x^3+3x^2+2x-6\)

\(A=\dfrac{1}{2}sin4x-\dfrac{1}{2}sin2x+\dfrac{1}{2}sin12x+\dfrac{1}{2}sin2x\)

\(=\dfrac{1}{2}sin4x+\dfrac{1}{2}sin12x=sin8x.cos4x\)

\(B=\dfrac{1}{2}cos5x+\dfrac{1}{2}cos3x+\dfrac{1}{2}cosx-\dfrac{1}{2}cos5x\)

\(=\dfrac{1}{2}cos3x+\dfrac{1}{2}cosx=cos2x.cosx\)

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+............+2015.2016.(2017-2014)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.............+2015.2016.2017-2014.2015.2016

3A=2015.2016.2017

A=2015.2016.2017:3

A=2015.672.2017

\(A=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{3}\)

\(A=3+3^2+3^3+...+3^{2004}\)

\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2004}\right)\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2005}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2005}\right)-\left(3+3^2+3^3+3^4+...+3^{2004}\right)\)

\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+\left(3^4-3^4\right)+...+\left(3^{2004}-3^{2004}\right)+\left(3^{2005}-3\right)\)

\(\Rightarrow2A=3^{2005}-3\)

\(\Rightarrow A=\dfrac{3^{2005}+3}{2}\)

\(A=\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)

\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}=\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}=\dfrac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\sqrt{\left(\sqrt{3}+1\right)^2}=\dfrac{\sqrt{3}+1}{\sqrt{2}}=\dfrac{\sqrt{2}+\sqrt{6}}{2}\)

\(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}\\ =\sqrt{3+\sqrt{5-1+2\sqrt{3}}}\\ =\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{3+\sqrt{3}-1}\\ =\sqrt{2+\sqrt{3}}\)

Ta có: \(A=\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=\sqrt{3+\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{3+\sqrt{3}-1}\)

\(=\sqrt{2-\sqrt{3}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)