Khi x=2 thì A=4+2+12=18

\(\sqrt{6-4\sqrt{2}}\)\(+\sqrt{22-12\sqrt{2}}\)

\(=\sqrt{4-4\sqrt{2}+2}\)\(+\sqrt{18-12\sqrt{2}+4}\)

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

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

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

\(=0+2\sqrt{2}\)\(=2\sqrt{2}\)

\(\sqrt{17-12\sqrt{2}}\)\(+\sqrt{9+4\sqrt{2}}\)

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

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

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

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

\(=4+0=4\)

=5x^2+5x-2x-2-(5x^2+x-15x-3)-17x-51

=5x^2-14x-53-5x^2+14x+3

=-50

a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)

b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)

c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)

d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

4 + 2 2 = 4 + 2 = 4 + 2

a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012

2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013

3M=2^0+2^2013

M=(2^0+2^2013)÷3

Vậy.......

b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012

3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013

4N=3-3^2013

N=(3-3^2013)÷4

Vậy........

K tao nhé ko lên lớp tao đánh m😈😈😈

Bt dễ thế mà ko làm dc😂😂😂😂😂

phần c đâu

Do x>=17 nên |x-17|>=0

Do đó |x-17|=x-17

=>A=x-17+18-x

=>A=1

Chúc em học tốt^^

Khi   \(x\ge17\)thì   \(x-17\ge0\)hay  \(\left|x-17\right|=x-17\)

\(\Rightarrow A=x-17+18-x=1\)

tk nka !!11

Do x>=17 nên |x-17|>=0

Do đó |x-17|=x-17

=>A=x-17+18-x

=>A=1

Chúc em học tốt^^