\(37.46+37.54-5^3.2^3\\ =37\left(46+54\right)-125.8\\ =37.100-1000\\ =3700-1000\\ =2700\)

\(37\cdot46+37\cdot54-5^3\cdot2^3\)
\(\Rightarrow37\cdot\left(46+54\right)-125\cdot8\)
\(\Rightarrow37\cdot100-1000\)
\(\Rightarrow3700-1000\)
\(\Rightarrow2700\)

hic cíu mng oi

a: ĐKXĐ: \(x\notin\left\{10;-10;\sqrt{10};-\sqrt{10}\right\}\)

b: \(A=\dfrac{5x^3+50x+2x^2+20+5x^3-50x-2x^2+20}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)

\(=\dfrac{10x^3+40}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)

\(-2003+\left(-21+75+2003\right)=-2003+2003-21+75=54\)

\(8,=\left(2x-3\right)\left(2x+3\right)\\ 9,=\left(1-5a^2\right)\left(1+5a^2\right)\)

8) \(-9+4x^2=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

9) \(1-25a^4=1-\left(5a^2\right)^2=\left(1-5a^2\right)\left(1+5a^2\right)\)

=5079275

tính hẳn ra hộ em là đầu tiên phải làm như nào rồi thế nào được không ạ !

- _ -

\(=\left(5x^2+10xy\right)-\left(4x+8y\right)\)

\(=5x\left(x+2y\right)-4\left(x+2y\right)\)

\(=\left(x+2y\right)\left(5x-4\right)\)

5x² + 10xy - 4x - 8y

= ( 5x² +10xy )-( 4x+8y )

= 5x( x + 2y) - 4( x+2y)

=(x+2y)(5x-4)

Chọn câu C

e: \(E=\frac{8}{77}+\frac{12}{187}+\frac{4}{323}+\frac{8}{21}+\frac{10}{456}\)

\(=\frac{8}{7\cdot11}+\frac{12}{11\cdot17}+\frac{4}{17\cdot19}+\frac{10}{19\cdot24}+\frac{8}{21}\)

\(=2\left(\frac{4}{7\cdot11}+\frac{6}{11\cdot17}+\frac{2}{17\cdot19}+\frac{5}{19\cdot24}\right)+\frac{8}{21}\)

\(=2\left(\frac17-\frac{1}{11}+\frac{1}{11}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{24}\right)+\frac{8}{21}\)

\(=2\left(\frac17-\frac{1}{24}\right)+\frac{8}{21}=2\cdot\frac{24-7}{7\cdot24}+\frac{8}{21}=2\cdot\frac{17}{7\cdot24}+\frac{8}{21}\)

\(=\frac{17}{7\cdot12}+\frac{8}{21}=\frac{17}{84}+\frac{32}{84}=\frac{49}{84}=\frac{7}{12}\)

f: \(G=\frac{2}{35}-\frac{3}{28}+\frac{3}{10}+\frac{7}{78}-\frac{9}{52}\)

\(=\frac{2}{5\cdot7}-\frac{3}{4\cdot7}+\frac{3}{2\cdot5}+\frac{7}{78}-\frac{9}{4\cdot13}\)

\(=\frac15-\frac17-\frac14+\frac17+\frac12-\frac15+\frac{7}{78}-\frac14-\frac{1}{13}\)

\(=\left(\frac15-\frac15\right)+\left(-\frac17+\frac17\right)+\left(-\frac14+\frac12-\frac14\right)+\frac{7}{78}-\frac{1}{13}\)

\(=\frac{7}{78}-\frac{1}{13}=\frac{7}{78}-\frac{6}{78}=\frac{1}{78}\)