Câu a : B-A= 1001² +1002² +1004² +10072 -1000²-1003²-1005²-1006²

=(1001²-1000²)+(1002²-1003²) (1004²-1005²)+(1007²-1006²)

HĐT số 3= (1001-1000)*(1001+1000)+(1002-1003)*(1002+1003)+(1004-1005)*(1004+1005)+(1007-1006)*(1007+1006)

=2001 -2005-2009+2013=0

vậy A=B

ủng hộ mk nha mọi người

qua de

bài này hình như là sai đề

\(\frac{40^2+51^2+91^2}{79^2}\)= \(\frac{2\times\left(40^2+51^2+91^2\right)}{2\times79^2}\)

=> dpcm

Vế thứ nhất lớn hơn hoặc bằng chứ.Thay a,b,c vào rồi cm bất đẳng thức là xong

\(\dfrac{51}{53}+\dfrac{55}{57}+\dfrac{61}{63}+\dfrac{69}{71}+\dfrac{79}{81}+\dfrac{91}{93}\) 

\(=\left(\dfrac{52}{53}-\dfrac{1}{53}\right)+\left(\dfrac{56}{57}-\dfrac{1}{57}\right)+\left(\dfrac{62}{63}-\dfrac{1}{63}\right)+\left(\dfrac{70}{71}-\dfrac{1}{71}\right)+\left(\dfrac{80}{81}-\dfrac{1}{81}\right)+\left(\dfrac{92}{93}-\dfrac{1}{93}\right)\)

\(=\left(1-\dfrac{1}{53}-\dfrac{1}{53}\right)+\left(1-\dfrac{1}{57}-\dfrac{1}{57}\right)+\left(1-\dfrac{1}{63}-\dfrac{1}{63}\right)+\left(1-\dfrac{1}{71}-\dfrac{1}{71}\right)+\left(1-\dfrac{1}{81}-\dfrac{1}{81}\right)+\left(1-\dfrac{1}{93}-\dfrac{1}{93}\right)\)

\(=\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)\) 

\(=1+1+1+1+1+1\) 

\(=6\)

3:

28/32=21/24

6/15=18/45

11/13=121/143

2:

7/2>4/79

27/9>2

51/17=3/1

a,=\(\dfrac{\left(2-\dfrac{1}{3}+\dfrac{1}{4}\right).12}{\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right).12}\)+\(\dfrac{\left(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{1}{2}\right).20}{\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{2}{5}\right).20}\)

=\(\dfrac{24-4+3}{24+2-3}\) +\(\dfrac{12-5+10}{10+15-8}\)(nhân từng số hạng với 12;20)

=\(\dfrac{23}{23}\)+\(\dfrac{17}{17}\) =1+1=2

b,=\(\dfrac{5.\left(\dfrac{1}{79}\right)+5.\left(\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}\right)+17.\left(\dfrac{1}{83}\right)+\dfrac{1}{5}}\)=\(\dfrac{5.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{5}}\)