a) 3.5² + 15.2² - 26 : 2

= 3.25 + 15.4 - 13

= 75 + 60 - 13

= 135 - 13

= 122

b) 2021 + 5.[300 - (17 - 7)²]

= 2021 + 5.[300 - 10²]

= 2021 + 5.[300 - 100]

= 2021 + 5.200

= 2021 + 1000

= 3021

c) 3².5 + 2³.10 - 81

= 9.5 + 8.10 - 81

= 45 + 80 - 81

= 125 - 81

= 44

a) 3.5² + 15.2² - 26 : 2

= 3. 25 + 15 . 4 - 26 : 2

=  75  + 60 - 13

= 135 - 13

= 122

b) 2021 + 5[300 - (17 - 7)² ]

= 2021 + 5[300 - 10² ]

=2021 + 5[300 - 100]

=2021 + 5.200

=2021 + 1000

= 3021

c) 3² . 5 + 2³ . 10 - 81

= 9 . 5 + 8 . 10 - 81

= 45 + 80 - 81

= 125 - 81

= 44

a.117 - 35 + (-17) = 117 - 52 = 65

b. (-4) . 17 . (-25) = 1700

c. 47. 129 - 47. 29 +300

= 47.(129-29)+300

=47.100+300

=4700+300 = 5000

d. 25 - 5 . [ 49 - (6-3)² . 5 ] + 1²⁰²¹

= 25 - 5 . ( 49 - (2)² .5 ) + 1 

= 25-5 . ( 49 - 20 ) + 1

= 25-5 .29 + 1 

= 25 - 145 + 1

=-119

a: M=-2021+2021-68-68+17

=-119

b: B=(-1)+(-1)+...+(-1)

=-1x500

=-500

c: C=(1-2-3+4)+(5-6-7+8)+...+(997-998-999+1000)

=0

Ta có: \(2011+5\left[300-\left(17-7\right)^2\right]\)

\(=2011+5\left[300-10^2\right]\)

\(=2011+1000=3011\)

\(2011+5.\left[300-\left(17-7\right)^2\right]\) 

\(=2011+5.\left[300-\left(10\right)^2\right]\) 

\(=2011+5.\left[300-100\right]\) 

\(=2011+5.200\) 

\(=2011+1000\) 

\(=3011\)

a: \(=2011+5\cdot\left[300-10^2\right]\)

\(=2011+5\cdot200\)

=1011

b: \(=695-\left[200+10^2\right]\)

=695-300

=395

a) =2011+5.200

    =3011

b)=695-300

   =395

c)1=29-5.4

     =109

d)=2010-2000:400

  =2010-5

  =2005

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)

\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)

=0

2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)

\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)

\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)

3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)

\(=\dfrac{52546}{15}\)

4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)

\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)

\(=\dfrac{4}{7}\)

1) Tính biểu thức:

e) 2011 + 5[300 – (17 – 7)² ]
= 2011 + 5[300 - 100]
= 2011 + 5 . 200
= 2011 + 1000
= 3011
 

=2011+5[300-10^2]

=2011+5[300-100]

=2011+5*200

=2011+1000

=3011

e) 2011 + 5[300 – (17 – 7)² ]

= 2011  + 5[300 - 10² ]

= 2011  +5[300 -   100 ]

= 2011  +5.   200

=2011  + 1000

=       3011