Bài 1: 

a: \(\left(3x+2\right)^2-4=3x\left(3x+4\right)\)

Bài 2: 

a: \(4x^2+4x+1\)

b: \(9x^2+9x+\dfrac{9}{4}\)

\(37^2+2\times37+13^2.\)

\(=37\times37+2\times37+13^2\)

\(=37\times\left(37+2\right)+13^2\)

\(=37\times39+13^2\)

\(=37\times39+169\)

\(=1443+169=1612\)

\(37^2+2.37.13+13^2=\left(37+13\right)^2=2500\)

\(37^2+2.37.13+13^2\) ( áp dụng Hằng đẳng thứ thứ nhất )

\(=\left(37+13\right)^2=50^2=2500\)

P/s xong rồi easy

áp dụng hằng đẳng thức là đc nha .

37^2 + 2.37.13 + 13^2

= ( 37 + 13 ) ^2

= 50^2

= 2500

chúc bn hk tốt

a: =16,71*(91+97)-188(6,2+0,67)

=16,71*188-188*(6,2+0,67)

=188(16,71-6,2-0,67)=1849,92

b: 56^2+31^2-13^2+56*62

=(56+31)^2-13^2

=87^2-13^2

=74*100=7400

a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)

b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)

c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

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

a) \(1+2+3+...+2021=\dfrac{\left(2021+1\right)\left(\dfrac{2021-1}{1}+1\right)}{2}=2043231\)

b) Giống a

c) \(11+13+15+...131=\dfrac{\left(131+11\right)\left(\dfrac{131-11}{2}+1\right)}{2}=4331\)

a. 1 + 2 + 3 + ... + 2021

= 1 + 2 + 3 + ... + 2020) + 2021

Ta có số cặp từ 1 đên 2020 là: 2020 : 2 = 1010(cặp)

Ta có tổng số đầu và số cuối từ 1 đến 2020 là: 2020 + 1 = 2021

=> Tổng là: 2021 . 1010 + 2021 = 2043231

\(a.\left(35-815\right)-\left(795-65\right)\\ =35-815-795+65\\ =\left(35+65\right)-\left(815+795\right)\\ =100-1610=-1510\)

Bài 2: 

a: Ta có: \(\left(35-815\right)-\left(795-65\right)\)

\(=35-815-795+65\)

\(=100-1600\)

=-1500

b: Ta có: \(5-197-2015-2015\)

\(=-2010-2212\)

=-4222

c: Ta có: \(1259-1409-12+1259+1409\)

\(=2518-12\)

=2506

Các anh các cj giúp em nhé 
Em cảm ơn trước

Giải:

a) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)\) 

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\) 

\(=\dfrac{1.2.3.4}{2.3.4.5}\) 

\(=\dfrac{1}{5}\) 

b) \(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\) 

\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\) 

\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\) 

\(=\dfrac{1}{100}\) 

Chúc bạn học tốt!

A)

\(\left(1-\dfrac{1}{2}\right)x\left(1-\dfrac{1}{3}\right)x\left(1-\dfrac{1}{4}\right)x\left(1-\dfrac{1}{5}\right)\)

\(=\dfrac{1}{2}x\dfrac{1}{3}x\dfrac{1}{4}x\dfrac{1}{5}\)

\(=\dfrac{1x2x3x4}{2x3x4x5}\)

\(=\dfrac{1}{5}\)

B)

\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right)......\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)

\(\dfrac{1}{4}.\dfrac{1}{7}.....\dfrac{94}{97}.\dfrac{97}{100}\)

\(\dfrac{1}{100}\)

học tốt nhé!!!