\(1\dfrac{1}{7}\times1\dfrac{1}{8}\times1\dfrac{1}{9}\times1\dfrac{1}{10}\times1\dfrac{1}{11}\times1\dfrac{1}{12}\times1\dfrac{1}{13}\)

\(=\dfrac{8}{7}\times\dfrac{9}{8}\times\dfrac{10}{9}\times\dfrac{11}{10}\times\dfrac{12}{11}\times\dfrac{13}{12}\times\dfrac{14}{13}\)

\(=\dfrac{8\times9\times10\times11\times12\times13\times14}{7\times8\times9\times10\times11\times12\times13}\)

\(=\dfrac{1\times1\times1\times1\times1\times1\times2}{1\times1\times1\times1\times1\times1\times1}\)

\(=2\)

\(P=\left|2018-x\right|+\left|2019-x\right|\)

\(=\left|x-2018\right|+\left|2019-x\right|>=\left|x-2018+2019-x\right|=1\)

Dấu '=' xảy ra khi 2018<=x<=2019

\(100\times4364\times959\times736:\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(100\times4364\times959\times736:\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)

\(100\times4364\times959\times736:0\)

\(436400\times959\times736:0\)

\(41812400\times736:0\)

\(=>0\)

1: \(\left(2x+1\right)^2+\left(2x-1\right)^2\)

\(=4x^2+4x+1+4x^2-4x+1\)

\(=8x^2+2\)

2: \(-\left(x+1\right)^2-\left(x-1\right)^2\)

\(=-\left(x^2+2x+1\right)-\left(x^2-2x+1\right)\)

\(=-x^2-2x-1-x^2+2x-1\)

\(=-2x^2-2\)

3: \(\left(x+2y\right)^2-\left(x-2y\right)^2\)

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

\(=2x\cdot4y=8xy\)

giúp mình với ạ

\(4x^2-5x+a⋮2x-3\)

=>\(4x^2-6x+x-1,5+a+1,5⋮2x-3\)

=>a+1,5=0

=>a=-1,5

Nguyễn Lê Phước Thịnh 6x đâu ra vậy bạn

a: Tỉ số giữa số học sinh giỏi kì 2 so với tổng số học sinh là:

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

Số học sinh lớp 6A là: \(5:\left(\dfrac{1}{3}-\dfrac{2}{9}\right)=5:\dfrac{1}{9}=45\left(bạn\right)\)

b: Số học sinh giỏi kì 2 là \(45\cdot\dfrac{1}{3}=15\left(bạn\right)\)

Đặt D(x)=0

=>\(x^2-7x+10=0\)

=>\(x^2-2x-5x+10=0\)

=>x(x-2)-5(x-2)=0

=>(x-2)(x-5)=0

=>\(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Số con gà là :

\(150\times60\%=150\times\dfrac{3}{5}=90\) (con)

Số con vịt là :

\(150-90=60\) (con)

Đáp số : con gà : 90 con ; con vịt : 60 con

yahdhhd

\(S=\dfrac{1}{299\cdot297}-\dfrac{1}{297\cdot295}-...-\dfrac{1}{3\cdot1}\)

\(=\dfrac{1}{299\cdot297}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{295\cdot297}\right)\)

\(=\dfrac{1}{299\cdot297}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{295}-\dfrac{1}{297}\right)\)

\(=\dfrac{1}{299\cdot297}-\dfrac{1}{2}\left(1-\dfrac{1}{297}\right)\)

\(=\dfrac{1}{299\cdot297}-\dfrac{1}{2}\cdot\dfrac{296}{297}\)

\(=\dfrac{1}{299\cdot297}-\dfrac{146}{297}=\dfrac{1-146\cdot299}{297\cdot299}=\dfrac{-43653}{88803}\)