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

\(a\text{)}\:\sqrt{25^2-24^2}=\text{ }\sqrt{\left(25-24\right)\left(25+24\right)}=\sqrt{49}=7\)

\(b\text{)}\:\sqrt{21.8^2-18.2^2}=\text{ }\sqrt{3.7.2^6-2^3.3^2}\\ =\sqrt{3.2^3\left(7.2^3-3\right)}=\sqrt{24.53}=\sqrt{1272}=2\sqrt{318}\)

cho bạn link nà
http://hoahoc.tungtung.vn/#/balancer?_k=h5w8j7

gọi 3 số cần tìm là a,b,c(a,b,c là số tự nhiên)

theo đề bài, ta có:

\(\dfrac{a+b+c}{3}=75\Rightarrow a+b+c=225\\ 10b=a\\ 4b=c\)

\(\Rightarrow a+b+c=10b+b+4b=225\\ 15b=225\\ \Rightarrow b=15\Rightarrow a=150\Rightarrow c=60\)

1/

\(\left(x-6\right)^2-x^2+36=\left(x-6+x\right)\left(x-6-x\right)+36\\ =\left(2x-6\right).\left(-6\right)+\left(-6\right)^2=\left(-6\right)\left(2x-6-6\right)=\left(-6\right)\left(2x-12\right)=12\left(6-x\right)\)

2/

\(\dfrac{3}{2}x=\dfrac{4}{3}y=\dfrac{5}{4}z\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}\\ \Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{2y}{\dfrac{3}{2}}=\dfrac{z}{\dfrac{4}{5}}=\dfrac{x-2y+z}{\dfrac{2}{3}-\dfrac{3}{2}+\dfrac{4}{5}}=-\dfrac{16}{-\dfrac{1}{30}}=480\)

suy ra: \(x=\dfrac{480}{\dfrac{2}{3}}=720\\ 2y=\dfrac{480}{\dfrac{3}{2}}=320\Rightarrow y=160\\ z=\dfrac{480}{\dfrac{4}{5}}=600\)

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\\ 2M=2^{2011}-\left(2^{2010}+2^{2009}+...+2^2+2^1\right)\\ 2M-M=\left(2^{2011}-\left(2^{2010}+2^{2009}+...+2^2+2^1\right)\right)-\left(2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\right)\\ 2M-M=2^{2011}+2^{2011}-1\\ M=2^{2012}-1\)

2/

\(•\left|x-3,5\right|\ge0\\ \Rightarrow0,5-\left|x-3,5\right|\le0,5\)

dấu "=" xảy ra khi x=3,5

vậy MAX A=0,5 tại x=3,5

\(•-\left|1,4-x\right|\le0\\ \Rightarrow-\left|1,4-x\right|-2\le-2\)

dấu "=" xảy ra khi x=1,4

vậy MAX B=-2 tại x=1,4

\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\\ n_{H_2SO_4}=0,168mol\)

\(PTHH:Mg\: ­\: ­\: ­\: ­\: ­\: +\: ­\: ­\: ­\: ­\: ­\: H_2SO_4\rightarrow MgSO_4+H_2\uparrow\\ \: ­\: ­\: ­\: ­\: \: ­\: ­\: ­\: ­\: ­\: ­\: ­\: 0,15mol\: ­\: ­\: \: ­\: ­\: ­\: ­\: ­\: ­\: ­\: ­\: 0,168mol\)

Mg đủ, H₂SO₄ dư

=> dd sau pư là MgSO₄ và H₂SO₄ dư

\(\Rightarrow n_{MgSO_4}=0,15mol\\ n_{H_2SO_4\:dư}=0,168-0,15=0,018mol\)

\(m_{dd\:sau\:pư}=3,6+0,168.98-0,15.2=19,764g\)

\(\left[{}\begin{matrix}C_{M\: dd\: H_2SO_4dư}=\dfrac{0,018}{19,764}\approx9,1M\\C_{M\:dd\:MgSO_4}=\dfrac{0,15}{19,764}\approx7,6M\end{matrix}\right.\)