a] 23/16

c)7/45

d)1/2

b)83/48

Đúng nhé mình làm tong toán nâng cao lớp 5 rồi

a: =371-271-531+731=100+200=300

b: =-321-28-327+27

=-350-300

=650

c: =-424-371+424+29=-342

d: =-249+248=-1

a)371+(-271)+(-531)+731=300

b)-312+(-327+(-28)+27=650

c)-424+(-371)-(-424)-(-29)=-342

d)-249+197+248+(-197)=-1

e)3+(-5)+(-9)+11+(-13)=-13

g)-8+10+(-12)+14+(-16)+18=6

các bạn giải giúp mình nhé

\(=\dfrac{1}{11}-\dfrac{1}{11}+\dfrac{1}{41152}=\dfrac{1}{41152}\)

623/12

1/15

5/12

2/7

4/15

1/4

=1/4

a: \(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\)

\(=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}\)

\(=\dfrac{4}{6}=\dfrac{2}{3}\)

b: \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\)

\(=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}\)

\(=\dfrac{31}{24}\)

c: \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)=\dfrac{3}{4}:\dfrac{7}{20}=\dfrac{3}{4}\cdot\dfrac{20}{7}=\dfrac{15}{7}\)

d: \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\)

\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=\dfrac{7}{8}\)

a)-1-2-3-4-5-6-....-80

=(-1)+(-2)+(-3)+(-4)+(-5)+(-6)+...+(-80)

Khoảng cách giữa các số:(-1)-(-2)=1

Tổng các số hạng:(-1)-(-80)+1=80 số

Tổng:[(-1)+(-80)].80:2= -3240

=>-1-2-3-4-5-6+......-80=-3240

b,1-2+3-4+5-6+......+2021-2022

=(1-2)+(3-4)+(5-6)+...+(2021-2022)

=(-1)+(-1)+(-1)+...+(-1)

Tổng số cặp là:

(2022-1+1):2=1011 cặp

-1.1011=-1011

=>1-2+3-4+5-6+......+2021-2022= -1011

c, Đề bài sai

d,-4-8-12-16-.......-2020

=-4+(-8)+(-12)+(-16)+...+(-2020)

Khoảng cách giữa các số:-4-(-8)=4

Tổng các số hạng:[-4-(-2020]:4+1=505 số

Tổng:[-4+(-2020)].505:2=-511060

=>-4-8-12-16-.......-2020=-511060

Mọi người giúp mình với ạ! Thank you everyone!

\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{225}\right)\)

\(=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{224}{225}\)

\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot...\cdot\frac{14\cdot16}{15\cdot15}\)

\(=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot14\cdot16}{2\cdot2\cdot3\cdot3\cdot...\cdot25\cdot25}\)

\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot14\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot16\right)}{\left(2\cdot3\cdot4\cdot...\cdot15\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot15\right)}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot14}{2\cdot3\cdot4\cdot...\cdot15}\cdot\frac{3\cdot4\cdot5\cdot...\cdot16}{2\cdot3\cdot4\cdot...\cdot15}\)

\(=\frac{1}{15}\cdot\frac{16}{2}\)

\(=\frac{1}{15}\cdot8\)

\(=\frac{8}{15}\)

( 1- 1/4 ) . ( 1 - 1/9 ) . ( 1 - 1/16 ) ...( 1 - 1/225 ) 

= 3/4 . 8/9 . 15/16 ... 224/225

= 3 . 8 . 15 ... 224 / 4 . 9 . ..225 

= 3 . 2 . 4 . 3 . 5 ... 14 . 16 / 2 . 2 . 3 . 3 ... 15 . 15 

= ( 3 . 4 . 5 ... 16 ) . ( 2 . 3 . 4 ... 14 ) /  ( 2 . 3 ... 15 ) . ( 2 . 3 ... 15 ) 

=          16/ 2 . 15

=           `8/15

Bài 1 không có cơ sở để tính biểu thức.

Bài 2:

a. 

$(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$

$=[(6x+1)-(6x-1)]^2=2^2=4$

b.

$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^8-1)(2^8+1)(2^{16}+1)$
$=(2^{16}-1)(2^{16}+1)=2^{32}-1$

c.

$2C=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^{16}+1)$

$=(5^4-1)(5^4+1)(5^8+1)(5^{16}+1)$

$=(5^8-1)(5^8+1)(5^{16}+1)$
$=(5^{16}-1)(5^{16}+1)=5^{32}-1$

$\Rightarrow C=\frac{5^{32}-1}{2}$

a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)

b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)

c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)