Bài giải:

14 lần số cây lớp 4A bằng 15 lần số cây lớp 4B
⇒ Số cây lớp 4A và lớp 4B theo tỉ lệ 15 : 14.

Tổng số phần là:

\(15+14=29(\text{ph}\overset{ˋ}{\hat{\text{a}}}\text{n})\)

72 cây ứng với 29 phần.
Một phần là:

\(72 : 29\)

Số cây lớp 4A là:

\(15\times\left(\right.72:29\left.\right)=1080:29=37(\text{c}\hat{\text{a}}\text{y})\)

Số cây lớp 4B là:

\(14\times\left(\right.72:29\left.\right)=1008:2935(\text{c}\hat{\text{a}}\text{y})\)

Đáp số:

• Lớp 4A: khoảng 37 cây
• Lớp 4B: khoảng 35 cây

What

2019/1028>2021/2020

Xin tick nha

Bài 1:

Ta có:

\(N=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)

Do \(\hept{\begin{cases}\frac{2017}{2018+2019}< \frac{2017}{2018}\\\frac{2018}{2018+2019}< \frac{2018}{2019}\end{cases}\Rightarrow\frac{2017}{2018+2019}+\frac{2018}{2018+2019}< \frac{2017}{2018}+\frac{2018}{2019}}\)

                                                     \(\Leftrightarrow N< M\)

Vậy \(M>N.\)

Bài 2:

Ta có:

\(A=\frac{2017}{987653421}+\frac{2018}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}\)

\(B=\frac{2018}{987654321}+\frac{2017}{24681357}=\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)

Do \(\hept{\begin{cases}\frac{2017}{987654321}+\frac{2017}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}\\\frac{1}{24681357}>\frac{1}{987654321}\end{cases}}\)

\(\Rightarrow\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}>\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)

                                                                     \(\Leftrightarrow A>B\)

Vậy \(A>B.\)

Bài 3:

\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{3}{2016}\)

                                                                \(=1+1+1+1-\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}+\frac{3}{2016}\)

                                                                \(=4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)\)

Do \(\hept{\begin{cases}\frac{1}{2017}< \frac{1}{2016}\\\frac{1}{2018}< \frac{1}{2016}\\\frac{1}{2019}< \frac{1}{2016}\end{cases}\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}< \frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}=\frac{3}{2016}}\)

\(\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\)âm

\(\Rightarrow4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)>4\)

Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}>4.\)

Bài 4:

\(\frac{1991.1999}{1995.1995}=\frac{1991.\left(1995+4\right)}{\left(1991+4\right).1995}=\frac{1991.1995+1991.4}{1991.1995+4.1995}\)

Do \(\hept{\begin{cases}1991.1995=1991.1995\\1991.4< 1995.4\end{cases}}\Rightarrow1991.1995+1991.4< 1991.1995+1995.4\)

\(\Rightarrow\frac{1991.1995+1991.4}{1991.1995+4.1995}< \frac{1991.1995+1995.4}{1991.1995+4.1995}=1\)

\(\Rightarrow\frac{1991.1999}{1995.1995}< 1\)

Vậy \(\frac{1991.1999}{1995.1995}< 1.\)

\(A=2019\cdot2019\)

\(B=2017\cdot2021=\left(2019-2\right)\left(2019+2\right)=2019\cdot2019+2019\cdot2-2019\cdot2-2\cdot2=2019\cdot2019-2\cdot2\)

Vậy A > B

thank you bn
 

\(A=2018\times2020+2021\) và \(B=2019\times2019+2021\)

\(A=2018\times2019+2018+2021\)

\(B=2018\times2019+2019+2021\)

Vì \(2019>2018\Rightarrow A< B\)

Ta có :

2018 x 2020 = 2018 x ( 2019 + 1 ) = 2018 + 2018 x 2019 < 2019 + 2018 x 2019 = 2019 x ( 2018 + 1 )

= 2019 x 2019

=> 2018 x 2020 < 2019 x 2019

=> 2018 x 2020 + 2021 < 2019 x 2019 + 2021

=> A < B

Giải:

a) 2019 + 2021 - 1 

= 4040 - 1

= 4039 

b) 2020 x 2019 + 2018

= 4078380 + 2018

= 4080398

Học tốt!!!

 \(Ta\)có :\(a\)=\(\frac{2017\cdot2018-1}{2017.2018}\)=\(\frac{2017.2018}{2017.2018}\)-\(\frac{1}{2017.2018}\)=1-\(\frac{1}{2017.2018}\)

          \(b\)=\(\frac{2019.2020-1}{2019.2020}\)=\(\frac{2019.2020}{2019.2020}\)-\(\frac{1}{2019.2020}\)=1-\(\frac{1}{2019.2020}\)

Vì \(\frac{1}{2018.2019}\)> \(\frac{1}{2019.2020}\)nên \(a\)< \(b\)(sử dụng phần bù)

so sánh a và b biết a=2017×2018−12017×20182017×20182017×2018−1​và b =2019×2020−12019×20202019×20202019×2020−1​
 

Bằng nhau

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