a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1

Bài 1: Thực hiện các phép tính sau:

\(a)\)Chưa rỏ đề

\(b)\)\(5025\div5-25\div5\)

\(=\)\(1005-5\)

\(=\)\(1000\)

\(c)\)\(218-180\div2\div9\)

\(=\)\(218-10\)

\(=\)\(208\)

\(d)\)\(\left(328-8\right)\div32\)

\(=\)\(320\div32\)

\(=\)\(10\)

Bài 1:

a) ( Tôi không nhìn rõ đầu bài )

b) 5025  :  5  -  25  :  5

=  ( 5025  -  25 )  :  5

=        5000   :   5

=       1000

c)   218  -  180  :  2  :  9

=     218   -    180  :  (  2  .  9 )

=      218    -    180   :  18

=       218    -    10

=       208

d)   (  328  -  8  )  :  32

=      320   :   32

=     10 

Bài 2: Thực hiện phép tính:

\(a)\)\(27.75+25.27-150\)

\(=\)\(27.\left(75+25\right)-150\)

\(=\)\(27.100-150\)

\(=\)\(2700-150\)

\(=\)\(2550\)

\(b)\)\(12\div\left\{400\div\left[500-\left(125+25.7\right)\right]\right\}\)

\(=\)\(12\div\left\{400\div\left[500-\left(125+175\right)\right]\right\}\)

\(=\)\(12\div\left\{400\div\left[500-300\right]\right\}\)

\(=\)\(12\div\left\{400\div200\right\}\)

\(=\)\(12\div2\)

\(=\)\(6\)

\(c)\)\(13.17-256\div16+14\div7-1\)

\(=\)\(221-16+2-1\)

\(=\)\(206\)

\(d)\)\(18\div3+182+3.\left(51\div17\right)\)

\(=\)\(6+182+3.3\)

\(=\)\(6+182+9\)

\(=\)\(197\)

a, = 3.25 + 15.4 - 16 : 2 = 75 + 60 - 8 = 135 - 8 = 127

a: =75+60-8=127

a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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

bài 3:

\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)

\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)

\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)

Bài 1:

a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)

\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)

\(=\dfrac{30}{30}-1\)

=1-1

=0

b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)

\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)

\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)

=3

c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)

\(=9-12\cdot\dfrac{9-8}{12}\)

=9-1

=8

a: =1/6+14/6-3/6=12/6=2

b: =-13/8+5/4:(-5/4)

=-13/8-1=-21/8

c: =-3/8(2/5+14/5)

=-3/8*16/5

=-6/5

d: =5/34(1/4+11/9-2/9+29/4)

=5/34*(15/2+1)

=5/34*17/2
=5/4

b: =-45-27+8=-72+8=-64

c: =12:6=2

d: =-24:8=-3

e: =-36:9=-4

f: =14:7=2

Bài 3: 

a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)

b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)

c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)