a: Xét ΔBAD vuông tại A và ΔBMD vuông tại M có

BD chung

\(\widehat{ABD}=\widehat{MBD}\)

Do đó: ΔBAD=ΔBMD

b: ΔBAD=ΔBMD

=>DA=DM

mà DM<DC(ΔDMC vuông tại M)

nên DA<DC

c: ΔBAD=ΔBMD

=>BA=BM 

=>ΔBAM cân tại B

Ta có: ΔBAM cân tại B

mà BI là đường phân giác

nên BI\(\perp\)AM và I là trung điểm của AM

Ta có: BI\(\perp\)AM

ME\(\perp\)AM

Do đó: ID//ME

Xét ΔAME có

I là trung điểm của AM

ID//ME

Do đó: D là trung điểm của AE

Xét ΔAME có

AK,EI,MD là các đường trung tuyến

Do đó: AK,EI,MD đồng quy

gấp ạa

Nửa chu vi mảnh đất hình chữ nhật là:

90 : 2 = 45 (m)

Chiều rộng mảnh đất hình chữ nhật là:

45 : (2 + 3) x 2 = 18 (m)

Chiều dài mảnh đất hình chữ nhật là:

45 - 18 = 27 (m)

Diện tích mảnh đất hình chữ nhật là:

18 x 27 = 486 (m²)

Diện tích trồng rau là:

486 x 20% = 97,2 (m²)

Diện tích trồng hoa là:

486 : 9 x 2 = 108 (m²)

Diện tích trồng cây ăn quả là:

486 - 97,2 - 108 = 280,8 (m²)

Đáp số: 97,2m² trồng rau

              108m² trồng hoa

               208,8m² trồng cây ăn quả

2B. 

a) \(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}\)

\(=\dfrac{3}{8}+\dfrac{5}{8}\)

\(=\dfrac{8}{8}\)

\(=1\)

b) \(B=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}-\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{2\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)

\(=\dfrac{1}{3}\cdot\dfrac{2\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{3\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}\)

\(=\dfrac{1}{3}\cdot\dfrac{2}{3}\)

\(=\dfrac{2}{9}\)

3A:

\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}=\dfrac{-1}{10}>-\dfrac{1}{9}\)

3B:

\(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\cdot\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)

\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)

Vì 20<21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\)

=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)

=>\(B< -\dfrac{11}{21}\)

\(2,5-3x=5,5.2022^0\)

\(=>2,5-3x=5,5.1\)

\(=>2,5-3x=5,5\)

\(=>3x=2,5-5,5\)

\(=>3x=-3\)

\(=>x=\left(-3\right):3\)

\(=>x=\dfrac{-3}{3}=-1\)

Vậy...

\(#NqHahh\)

\(2,5-3x=5,5\cdot2022^0\)

\(2,5-3x=5,5\cdot1\)

\(2,5-3x=5,5\)

\(3x=2,5-5,5\)

\(3x=-3\)

\(x=-3:3\)

\(x=-1\)

Vậy \(x=-1\)

a: Vì O thuộc tia đối của tia AB

nên A nằm giữa O và B

=>OB=OA+AB=4+6=10(cm)

M là trung điểm của OA

=>\(OM=MA=\dfrac{OA}{2}=\dfrac{4}{2}=2\left(cm\right)\)

N là trung điểm của OB

=>\(ON=NB=\dfrac{OB}{2}=5\left(cm\right)\)

Vì OM<ON

nên M nằm giữa O và N

=>OM+MN=ON

=>MN+2=5

=>MN=3(cm)

b: \(MN=ON-OM=\dfrac{OB-OA}{2}=\dfrac{BA}{2}\)

=>MN không phụ thuộc vào điểm O

c: Gọi số điểm phải lấy thêm là n(điểm)

Tổng số điểm trên đoạn thẳng AB lúc này là n+2(điểm)

Số tam giác tạo thành là \(C^2_{n+2}\left(tamgiác\right)\)

Theo đề, ta có: \(C^2_{n+2}=465\)

=>\(\dfrac{\left(n+2\right)!}{\left(n+2-2\right)!\cdot2!}=465\)

=>(n+1)(n+2)=930

=>\(n^2+3n-928=0\)

=>\(\left[{}\begin{matrix}n=29\left(nhận\right)\\n=-32\left(loại\right)\end{matrix}\right.\)

Vậy: Số điểm phải lấy thêm là 29 điểm

a: \(-1,2+\dfrac{2}{3}+x=5\)

=>\(x=5+1,2-\dfrac{2}{3}=6,2-\dfrac{2}{3}\)

=>\(x=\dfrac{31}{5}-\dfrac{2}{3}=\dfrac{93}{15}-\dfrac{10}{15}=\dfrac{83}{15}\)

b: \(2\dfrac{4}{7}-3x=\dfrac{-4}{5}+\dfrac{2}{3}\)

=>\(\dfrac{18}{7}-3x=\dfrac{-12}{15}+\dfrac{10}{15}=\dfrac{-2}{15}\)

=>\(3x=\dfrac{18}{7}+\dfrac{2}{15}=\dfrac{270}{105}+\dfrac{14}{105}=\dfrac{284}{105}\)

=>\(x=\dfrac{284}{315}\)

c: \(\dfrac{1}{6}-\dfrac{3}{8}+1,75=3\dfrac{4}{3}-x\)

=>\(\dfrac{13}{3}-x=\dfrac{4}{24}-\dfrac{9}{24}+\dfrac{42}{24}=\dfrac{37}{24}\)

=>\(x=\dfrac{13}{3}-\dfrac{37}{24}=\dfrac{108}{24}-\dfrac{37}{24}=\dfrac{71}{24}\)

d: \(\dfrac{1}{6}-\dfrac{4}{9}+0,125=2\dfrac{4}{3}-2x\)

=>\(\dfrac{10}{3}-2x=\dfrac{-11}{72}\)

=>\(2x=\dfrac{10}{3}+\dfrac{11}{72}=\dfrac{240}{72}+\dfrac{11}{72}=\dfrac{251}{72}\)

=>\(x=\dfrac{251}{144}\)

e: \(2\dfrac{2}{3}-4x=\dfrac{-7}{5}+\dfrac{2}{3}\)

=>\(2+\dfrac{2}{3}-4x=\dfrac{-7}{5}+\dfrac{2}{3}\)

=>\(2-4x=-\dfrac{7}{5}\)

=>\(4x=2+\dfrac{7}{5}=\dfrac{17}{5}\)

=>\(x=\dfrac{17}{20}\)

f: \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

=>\(x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}=\dfrac{3}{6}-\dfrac{5}{6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

=>\(x=-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)

g: \(\left(\dfrac{3}{5}-\dfrac{4}{3}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{9}{7}\)

=>\(\dfrac{-11}{15}+\dfrac{5}{8}-x=\dfrac{9}{7}\)

=>\(\dfrac{-13}{120}-x=\dfrac{9}{7}\)

=>\(x=-\dfrac{13}{120}-\dfrac{9}{7}=\dfrac{-1171}{840}\)

a, \(-1,2+\dfrac{2}{3}+x=5\Leftrightarrow x=5+1,2-\dfrac{2}{3}=\dfrac{83}{15}\)

b, \(2\dfrac{4}{7}-3x=-\dfrac{4}{5}+\dfrac{2}{3}\Leftrightarrow\dfrac{18}{7}-3x=-\dfrac{2}{15}\Leftrightarrow3x=\dfrac{284}{105}\Leftrightarrow x=\dfrac{284}{315}\)

c, \(\dfrac{1}{6}-\dfrac{3}{8}+1,75=3\dfrac{4}{3}-x\Leftrightarrow-x+\dfrac{13}{3}=\dfrac{37}{24}\Leftrightarrow x=\dfrac{13}{3}-\dfrac{37}{24}=\dfrac{67}{24}\)

d, \(\dfrac{1}{6}-\dfrac{4}{9}+0,125=2\dfrac{4}{3}-2x\Leftrightarrow-2x+\dfrac{10}{3}=-\dfrac{-11}{72}\Leftrightarrow2x=\dfrac{251}{72}\Leftrightarrow x=\dfrac{251}{144}\)

e, \(2\dfrac{2}{3}-4x=-\dfrac{7}{5}+\dfrac{2}{7}\Leftrightarrow\dfrac{8}{3}-4x=-\dfrac{39}{35}\Leftrightarrow4x=\dfrac{397}{105}\Leftrightarrow x=\dfrac{397}{420}\)

f, \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{2}{3}\)

g, \(\left(\dfrac{3}{5}-\dfrac{4}{3}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{9}{7}\Leftrightarrow\dfrac{-11}{15}+\dfrac{5}{8}-x=\dfrac{9}{7}\Leftrightarrow\left(-\dfrac{13}{120}\right)-x=\dfrac{9}{7}\Leftrightarrow x=-\dfrac{1171}{840}\)

a) \(\dfrac{-3}{100}>\dfrac{-50}{100}=-\dfrac{1}{2}\)

\(\dfrac{-2}{3}< \dfrac{-1,5}{3}=-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{-3}{100}>\dfrac{-2}{3}\)

b) \(\dfrac{-3}{5}=\dfrac{-9}{15}\)

\(\dfrac{-2}{3}=\dfrac{-10}{15}\)

Mà:  - 9 > -10 

\(\Rightarrow-\dfrac{9}{15}>\dfrac{-10}{15}\)

hay `-3/5>-2/3` 

c) \(\dfrac{-5}{4}< \dfrac{-2}{4}=-\dfrac{1}{2}\)

\(-\dfrac{3}{8}>\dfrac{-4}{8}=-\dfrac{1}{2}\)

\(\Rightarrow-\dfrac{5}{4}< \dfrac{-3}{8}\) 

d) \(-\dfrac{2}{3}=\dfrac{1}{3}-1\)

\(-\dfrac{3}{4}=\dfrac{1}{4}-1\)

Vì: `1/3>1/4` 

`=>1/3-1>1/4-1` 

Hay `-2/3>-3/4` 

a: \(\dfrac{-3}{100}=\dfrac{-3\cdot3}{100\cdot3}=\dfrac{-9}{300};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot100}{3\cdot100}=\dfrac{-200}{300}\)

mà -9>-200

nên \(\dfrac{-3}{100}>\dfrac{-2}{3}\)

b: \(\dfrac{-3}{5}=\dfrac{-3\cdot3}{5\cdot3}=\dfrac{-9}{15};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot5}{3\cdot5}=\dfrac{-10}{15}\)

mà -9>-10

nên \(\dfrac{-3}{5}>\dfrac{2}{-3}\)

c: \(\dfrac{-5}{4}=\dfrac{-5\cdot2}{4\cdot2}=\dfrac{-10}{8};\dfrac{-3}{8}=\dfrac{-3}{8}\)

mà -10<-3

nên \(-\dfrac{5}{4}< -\dfrac{3}{8}\)

d: \(\dfrac{-2}{3}=\dfrac{-2\cdot4}{3\cdot4}=\dfrac{-8}{12};\dfrac{3}{-4}=\dfrac{-3}{4}=\dfrac{-3\cdot3}{4\cdot3}=\dfrac{-9}{12}\)

mà -8>-9

nên \(-\dfrac{2}{3}>\dfrac{3}{-4}\)

e: \(\dfrac{267}{-268}=\dfrac{-267}{268}>-1;-1=\dfrac{-1343}{1343}>\dfrac{-1347}{1343}\)

Do đó: \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

f: \(\dfrac{2022\cdot2023-1}{2022\cdot2023}=1-\dfrac{1}{2022\cdot2023}\)

\(\dfrac{2023\cdot2024-1}{2023\cdot2024}=1-\dfrac{1}{2023\cdot2024}\)

Ta có: 2022<2024

=>\(2022\cdot2023< 2023\cdot2024\)

=>\(\dfrac{1}{2022\cdot2023}>\dfrac{1}{2023\cdot2024}\)

=>\(-\dfrac{1}{2022\cdot2023}< -\dfrac{1}{2023\cdot2024}\)

=>\(\dfrac{-1}{2022\cdot2023}+1< \dfrac{-1}{2023\cdot2024}+1\)

=>\(\dfrac{2022\cdot2023-1}{2022\cdot2023}< \dfrac{2023\cdot2024-1}{2023\cdot2024}\)

g: \(\dfrac{2022\cdot2023}{2022\cdot2023+1}=1-\dfrac{1}{2022\cdot2023+1}\)

\(\dfrac{2023\cdot2024}{2023\cdot2024+1}=1-\dfrac{1}{2023\cdot2024+1}\)

Vì \(2022\cdot2023+1< 2023\cdot2024+1\)

nên \(\dfrac{1}{2022\cdot2023+1}>\dfrac{1}{2023\cdot2024+1}\)

=>\(\dfrac{-1}{2022\cdot2023+1}< \dfrac{-1}{2023\cdot2024+1}\)

=>\(\dfrac{-1}{2022\cdot2023+1}+1< \dfrac{-1}{2023\cdot2024}+1\)

=>\(\dfrac{2022\cdot2023}{2022\cdot2023+1}< \dfrac{2023\cdot2024}{2023\cdot2024+1}\)