\(40\%=\dfrac{40}{100}=\dfrac{2}{5}\)

Số tấn thóc thửa 1 :

\(1.\dfrac{2}{5}=0,4\left(tấn\right)=400\left(kg\right)\)

Số tấn thóc thửa 2 :

\(1.\dfrac{2}{5}=0,4\left(tấn\right)=400\left(kg\right)\)

Số tấn thóc thửa 3 :

\(1-\left(0,4+0,4\right)=0,2\left(tấn\right)=200\left(kg\right)\)

a/

\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)

\(\Rightarrow x=12\)

\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)

\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)

Vậy x = 2023

\(A=\dfrac{3}{5.6}+\dfrac{3}{6.7}+...+\dfrac{3}{91.92}\)

\(\Rightarrow A=3\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{91.92}\right)\)

\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{92}\right)\)

\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{92}\right)\)

\(\Rightarrow A=3.\dfrac{87}{460}=\dfrac{261}{460}\)

\(2018\equiv-1\left(mod2019\right)\)

\(\Rightarrow2018^{2019}\equiv-1^{2019}=-1\) (mod 2019)

\(\Rightarrow2018^{2019}\equiv-1\) (mod 2019)

\(\Rightarrow2018^{2018}+1⋮2019\)

mọi người giúp em nhanh với

ko giải đâu

đùa thôi =)

5⁷. \(\dfrac{1}{5^5}\) = 5² = 25

5⁷ . 1/5⁵

= 5⁷/5⁵

= 5²

= 25

Bài 4 :

Ta có :

\(\widehat{ABC}=100^o\)

\(\widehat{BCD}=40+60=100^o\)

\(\Rightarrow\widehat{ABC}=\widehat{BCA}=100^o\) ở vị trí sole trong

\(\Rightarrow AB//CD\left(1\right)\)

Ta lại có :

\(\widehat{MCD}=60^o\)

Kẻ thêm từ MN qua trái 1 đường thẳng tạo thành 1 góc \(\widehat{CMx}\)

\(\Rightarrow\widehat{CMx}=180-120=60^o\)

\(\widehat{MCD}=\widehat{CMx}=60^o\) ở vị trí sole trong

\(\Rightarrow MN//CD\left(2\right)\)

\(\left(1\right).\left(2\right)\Rightarrow\Rightarrow AB//MN\)

1. Bài 4.

Ta có: AB//CD ( góc so le trong)

Mặt khác: góc MCD + góc CMN = 180^o nên 2 góc trên là 2 góc trong cùng phía bù nhau

==>CD//MN do AB//CD ==> AB//MN (đpcm)

2. Bài 5

Từ C kẻ đoạn thẳng CF // với AB và DE

Ta có góc BCF = góc ABC = 40^o (so le trong) (1)

góc FCE = góc CED = 30^o (so le trong) (2)

Từ 1 và 2 suy ra góc BCE = góc BCF + góc FCE = 40^o +30^o =70^o

\(\sqrt[]{x-2}=12\)

\(\Rightarrow x-2=12^2=144\)

\(\Rightarrow x=144+2=146\)

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