Sửa đề: \(\sqrt{11-6\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{2}+2}+\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)

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

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

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

=3-1=2

A=3³=27
\(B=\dfrac{2^{11}.9^2}{3^5.2^8}=\dfrac{2^{11}.3^4}{3^5.2^8}=\dfrac{2^3}{3}=\dfrac{8}{3}\)

A=3^3=9

B=2^3/3^5=8/243

A=3^18:3^15=3^3

B=(2^11.3^4)/(3^5.2^8)

=2^3/3=8/3

Chúc bạn học tốt!

 -11 + y + 7 = (-11 + 7) + y

= - (11 - 7) + y = -4 + y

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5-2\sqrt{6}}-\sqrt{11-4\sqrt{6}}\right)=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-\sqrt{3}\right)^2}\right)\)

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

\(=\sqrt{6}-3\)

a: \(M=\dfrac{18+5x+15+3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{8x+24}{\left(x+3\right)\left(x-3\right)}=\dfrac{8}{x-3}\)

b: Thay x=11 vào M, ta được:

\(M=\dfrac{8}{11-3}=1\)

a) \(M=\dfrac{18}{x^2-9}+\dfrac{5}{x-3}+\dfrac{3}{x+3}.\left(x\ne\pm3\right).\)

\(M=\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{5}{x-3}+\dfrac{3}{x+3}=\dfrac{18+5\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{18+5x+15+3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{24+8x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{8\left(3+x\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{8}{x-3}.\)

b) Thay \(x=11\left(TM\right)\) vào biểu thức M: 

\(\dfrac{8}{11-3}=\dfrac{8}{8}=1.\)

Biểu thức có thể được rút gọn như sau:

√11 + 4√6 − √5 − 2√6
= √11 + (4 - 2)√6 − √5
= √11 + 2√6 − √5

\(16.17-\dfrac{5}{16}\times16+11\)

\(=272-5+11\)

\(=278\)

\(16\cdot17-\dfrac{5}{16}\cdot16+11\\ =16\cdot\left(17-\dfrac{5}{16}\right)+11\\ =16\cdot\dfrac{267}{16}+11\\ =267+11=278\)

\(\dfrac{\left(-6\right).11}{\left(-11\right).\left(-8\right)}=\dfrac{\left(-3\right).2.11}{11.2.4}=-\dfrac{3}{4}\)

\(\dfrac{21.\left(-5\right)}{25.\left(-7\right)}=\dfrac{\left(-3\right).\left(-7\right).\left(-5\right)}{\left(-5\right).\left(-5\right).\left(-7\right)}=\dfrac{3}{5}\)

\(\dfrac{32.9.11}{12.24.22}=\dfrac{4.4.2.3.3.11}{3.4.4.2.3.11.2}=\dfrac{4^2.2.3^2.11}{3^2.4^2.2^2.11}=\dfrac{1}{2}\)

\(\dfrac{\left(-6\right).11}{\left(-11\right).\left(-8\right)}\)=\(\dfrac{-3}{4}\)

\(\dfrac{21.\left(-5\right)}{25.\left(-7\right)}\)=\(\dfrac{2}{5}\)

\(\dfrac{32.9.11}{12.24.22}\)=\(\dfrac{1}{2}\)

Đặt \(B=\frac{\sqrt{11+\sqrt{5}}+\sqrt{11-\sqrt{5}}}{\sqrt{11+2\sqrt{29}}}\)Ta có B>0

\(B^2=2\Rightarrow B=\sqrt{2}\)

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