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

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

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

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

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

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

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

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

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

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

Bạn xem lại đề

a) \(2x^2-3xy-2y^2=2\)

\(\Rightarrow2x^2+xy-4xy-2y^2=2\)

\(\Rightarrow x\left(2x+y\right)-2y\left(2x+y\right)=2\)

\(\Rightarrow\left(2x+y\right)\left(x-2y\right)=2\)

\(\Rightarrow\left(2x+y\right);\left(x-2y\right)\in\left\{-1;1;-2;2\right\}\)

Ta giải các hệ phương trình sau với x;y nguyên 

1) \(\left\{{}\begin{matrix}2x+y=-1\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-2\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}2x+y=1\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}4x+2y=2\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}5x=4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

3) \(\left\{{}\begin{matrix}2x+y=-2\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-4\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)

4)  \(\left\{{}\begin{matrix}2x+y=2\\x-2y=1\end{matrix}\right.\) \(\left\{{}\begin{matrix}4x+2y=4\\x-2y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(-1;0\right);\left(1;1\right)\right\}\)

b) \(xy-y+x=9\)

\(\Rightarrow y\left(x-1\right)+x-1+1=9\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=8\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;-9\right);\left(2;7\right);\left(-1;-5\right);\left(3;3\right);\left(-3;-3\right);\left(5;1\right);\left(-7;-2\right);\left(9;0\right)\right\}\)

a) Ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(dpcm\right)\)

b) Ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\left(1\right)\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\left(dpcm\right)\)