Ta có \(x+1=2022\)

\(P\left(x\right)=x^{101}-\left(x+1\right)x^{100}+...+\left(x+1\right)x-1\)

\(=x^{101}-x^{101}-x^{100}+...+x^2+x-1=x-1\)

-> P(x) = 2020 

\(=2022\times\left(1+1+1+1+1+5-7\right)\)

= 2022 x 3

= 6066

2022 + 2022 + 2022 + 2022 + 2022 + 2022 x 5 - 2022 x 7

= 2022 x 1 + 2022 x 1 + 2022 x 1 + 2022 x 1 + 2022 x 1 + 2022 x 5 - 2022 x 7

= 2022 x ( 1 + 1 + 1 + 1 + 1 + 5 - 7 )

= 2022 x 3 

= 6066

chờ tí

6 x 2022/9 + 2022/9 x 4 - 2022/9

=6 x 2022/9 + 2022/9 x 4 - 2022/9x1

=2022/9x(6+4-1)

=2022/9x9=2022

\(50\%+\dfrac{7}{12}-\dfrac{1}{2}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\)

\(=\dfrac{7}{12}\)

_______________

\(2022\cdot67+2022\cdot43-2022\cdot10\)

\(=2022\cdot\left(67+43-10\right)\)

\(=2022\cdot100\)

\(=202200\)

_____________________

\(10,3+6,9+8,7+13,1\)

\(=\left(13,1+6,9\right)+\left(10,3+8,7\right)\)

\(=20+19\)

\(=39\)

___________________

\(17,58\times43+57\times17,58\)

\(=17,58\times\left(43+57\right)\)

\(=17,58\times100\)

\(=1758\)

tớ ghi sai nên cập nhật lại câu hỏi, phần 3 các bạn trả lời sớm mà bị sai thì bỏ qua cho tớ nhé

\(50\%+\dfrac{7}{12}-\dfrac{1}{2}=\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}=\dfrac{7}{12}\)

\(2022\cdot67+2022\cdot43-2022\cdot10=2022\cdot\left(67+43-10\right)=2022\cdot100=202200\)

\(11,3+6,9+8,7+13,1=\left(11,3+8,7\right)+\left(6,9+13,1\right)=20+20=40\)

\(17,58\cdot43+57\cdot17,58=17,58\cdot\left(43+57\right)=17,58\cdot100=1758\)

\(a,50\%+\dfrac{7}{12}-\dfrac{1}{2}\\ =\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\\ =\dfrac{7}{12}\\ b,2022\times67+2022\times43-2022\times10\\ =2022\times\left(67+43-10\right)\\ =2022\times100\\ =202200.\\ c,125-25:3\times12\)

\(=25\times5-25:3\times12\\ =25\times\left(5-\dfrac{1}{3}\right)\times12\\ =25\times\dfrac{14}{3}\times12\\ =1400\)

a,50%+127​−21​=21​+127​−21​=(21​−21​)+127​=127​b,2022×67+2022×43−2022×10=2022×(67+43−10)=2022×100=202200.c,125−25:3×12

=25×5−25:3×12=25×(5−13)×12=25×143×12=1400=25×5−25:3×12=25×(5−31​)×12=25×314​×12=1400

a) 7/12

b) 202200

c) 1400

d) mình ko bt

Đáp số là 2021/2022 em nhé

\(\dfrac{2021}{2022}.\dfrac{7}{16}+\dfrac{9}{16}.\dfrac{2021}{2022}=\dfrac{2021}{2022}\left(\dfrac{7}{16}+\dfrac{9}{16}\right)=\dfrac{2021}{2022}.1=\dfrac{2021}{2022}\)

`2022 \times 123 \times 12 - 2022 \times 11`

`= 2022 \times (123 \times 12 - 11)`

`= 2022 \times (1476 - 11)`

`= 2022 \times 1465`

`= 2962230`

2019 x 45 + 54 x 2019 + 2019/2019 x 2022 - 2018 x 201

=2019*(45+54+1)-2018

=2019*100-2018

=201900-2018

=199882

2019 x 45 + 54 x 2019 + 2019/2019 x 2022 - 2018 x 201

=2019*(45+54+1)-2018

=2019*100-2018

=201900-2018

=199882

2019 x 45 + 54 x 2019 + 2019/2019 x 2022 - 2018 x 201

=2019x(45+54+1)-2018

=2019x100-2018

=201900-2018

=199882

Học Tốt nhóe ☘

=2022(123*12-11)

=2962230

2022 × 123 × 12 - 2022 × 11

= 2022 × (123 × 12 - 11)

= 2022 × (1476 - 11)

= 2022 × 1465

= 2962230

a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)

\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)

98^88+1>98^99+1

=>A<B

b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)

\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

2022^2023>2022^2021

=>2022^2023+2022^2>2022^2021+2022^2

=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)

=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

=>C>D