Đúng đề k?

\(n=1\Rightarrow n\cdot n+14\cdot2n+11=1^2+14\cdot2\cdot1+11=1+28+11=40⋮̸3\)

\(S^2=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\\ \text{Ta có:}\\ \dfrac{1}{2}< \dfrac{2}{3}\\ \dfrac{3}{4}< \dfrac{4}{5}\\ \dfrac{5}{6}< \dfrac{6}{7}\\ ...\\ \dfrac{199}{200}< \dfrac{200}{201}\\ \Rightarrow S^2< \left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{200}{201}\right)\\ \Leftrightarrow S^2< \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{199}{200}\cdot\dfrac{200}{201}\\ \Leftrightarrow S^2< \dfrac{1\cdot2\cdot3\cdot...\cdot200}{2\cdot3\cdot4\cdot...\cdot201}\\ \Leftrightarrow S^2< \dfrac{1}{201}< \dfrac{1}{200}\)

Vậy ...

1/4 nhé 

tick nhé bạn MIN YINO dễ thương

\(\left|2x-1\right|\le5\\ \Leftrightarrow-5\le2x-1\le5\\ \Leftrightarrow-4\le2x\le6\\ \Leftrightarrow-2\le x\le3\)

\(A=-x^2-y^2+xy+2x+2y\\ =-2x^2-2y^2+2xy+4x+4y\\ =\left(-x^2+2xy-y^2\right)+\left(-x^2+4x-4\right)+\left(-y^2+4y-4\right)+8\\ =-\left(x^2-2xy+y^2\right)-\left(x^2-4x+4\right)-\left(y^2-4y+4\right)+8\\ =-\left(x-y\right)^2-\left(x-2\right)^2-\left(y-2\right)^2+8\\ =-\left[\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\right]+8\\ \left(x-y\right)^2\ge0\forall x,y;\left(x-2\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y\\ \Rightarrow\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\ge0\\ \Leftrightarrow-\left[\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\right]\le0\\ \Leftrightarrow-\left[\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\right]+8\le8\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-2\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-2=0\\y-2=0\end{matrix}\right.\\ \Leftrightarrow x=y=2\)

Vậy \(MAX_A=8\text{ khi }x=y=2\)

a) \(m-5\ne0\Leftrightarrow m\ne5\)

b)

\(m=-2\\ \Rightarrow\left(-2-5\right)x+3=0\\ \Leftrightarrow-7x+3=0\\ \Leftrightarrow-7x=-3\\ \Leftrightarrow x=\dfrac{3}{7}\)

\(VT=a\cdot\left(b+c\right)-a\cdot\left(b+d\right)\\ =a\cdot\left[\left(b+c\right)-\left(b+d\right)\right]\\ =a\cdot\left(b+c-b-d\right)\\ =a\cdot\left(c-d\right)=VP\)

\(f\left(1\right)=1^2+b\cdot1+c=1+b+c\\ \Leftrightarrow1+b+c=2\\ \Leftrightarrow b+c=1\\ f\left(-3\right)=\left(-3\right)^2+b\cdot\left(-3\right)+c=9-3b+c\\ \Leftrightarrow9-3b+c=0\\ \Leftrightarrow-3b+c=-9\\ \left(b+c\right)-\left(-3b+c\right)=1-\left(-9\right)\\ \Leftrightarrow b+c+3b-c=1+9\\ \Leftrightarrow4b=10\\ \Leftrightarrow b=2,5\\ \Rightarrow2,5+c=1\\ \Leftrightarrow c=-1,5\)