Lời giải:

$N=(a-2)(a+3)-(a-3)(a+2)=(a^2+a-6)-(a^2-a-6)=2a$ không có cơ sở để khẳng định đó là bội của $50$ bạn nhé.

= \(\left(\dfrac{2}{3}\right)^3-4.\left(-\dfrac{7}{4}\right)^2+\left(-\dfrac{2}{3}\right)^3\)

= \(\dfrac{8}{27}-4.\dfrac{49}{16}+\left(-\dfrac{8}{27}\right)\)

= \(\left[\dfrac{8}{27}+\left(-\dfrac{8}{27}\right)\right]-4.\dfrac{49}{16}\)

= \(-\dfrac{49}{4}\)

   \(\left(\dfrac{2}{3}\right)^3-4.\left(-1\dfrac{3}{4}\right)^2+\left(\dfrac{-2}{3}\right)^3\)

= \(\left[\left(\dfrac{2}{3}\right)^3+\left(\dfrac{-2}{3}\right)^3\right]-4.\left(-1\dfrac{3}{4}\right)^2\)

= \(\left(\dfrac{8}{9}+\dfrac{-8}{9}\right)-4.\left(-1\dfrac{3}{4}\right)^2\)

= \(0-4.\left(\dfrac{-7}{4}\right)^2\)

= \(0-4.\dfrac{49}{16}\)

= \(0-\dfrac{49}{4}\)

= \(\dfrac{-49}{4}\)

\(P=-\left(2x+1\right)^2-7\left(y-3,5\right)^2+\dfrac{2}{3}\)

 vì \(\left\{{}\begin{matrix}-\left(2x+1\right)^2\le0,\forall x\\-7\left(y-3,5\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow P=-\left(2x+1\right)^2-7\left(y-3,5\right)^2+\dfrac{2}{3}\le\dfrac{2}{3}\)

Dấu "=" xảy ra khi

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3,5=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(GTLN\left(P\right)=\dfrac{2}{3}\left(tạix=-\dfrac{1}{2};y=\dfrac{7}{2}\right)\)

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\(C=\dfrac{5}{3-\left(4x+1\right)^2}\)

Điều kiện xác định khi 

\(3-\left(4x+1\right)^2\ne0\Leftrightarrow\left[{}\begin{matrix}4x+1\ne\sqrt[]{3}\\4x+1\ne-\sqrt[]{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\sqrt[]{3}-1}{4}\\x\ne\dfrac{-\sqrt[]{3}-1}{4}\end{matrix}\right.\)

Ta có :

\(\left(4x+1\right)^2\ge0,\forall x\)

\(\Leftrightarrow3-\left(4x+1\right)^2\le3\)

\(\Leftrightarrow C=\dfrac{5}{3-\left(4x+1\right)^2}\ge\dfrac{5}{3}\)

Vậy \(GTNN\left(C\right)=\dfrac{5}{3}\left(tạix=-\dfrac{1}{4}\right)\)

\(B=\left(2x\right)^2+2\left(y-1\right)^2-5\)

vì \(\left\{{}\begin{matrix}\left(2x\right)^2\ge0,\forall x\\2\left(y-1\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\Rightarrow B=\left(2x\right)^2+2\left(y-1\right)^2-5\ge-5\)

Dấu "=" xảy tại khi

\(\left\{{}\begin{matrix}2x=0\\2\left(y-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Vậy \(GTNN\left(B\right)=-5\left(tạix=0;y=1\right)\)

\(\text{#040911}\)

Vì \(-\dfrac{5}{12}< 0\)

\(\Rightarrow-\dfrac{5}{12}< \dfrac{a}{5}\text{ }\forall\text{ }a\)

\(\dfrac{a}{5}< \dfrac{1}{4}\)

\(\Rightarrow a=1\)

Vậy, để thỏa mãn \(-\dfrac{5}{12}< \dfrac{a}{5}< \dfrac{1}{4}\) thì \(a=1.\)

1

Bài 4:

b. Ta có:

$(2-x)^2\geq 0$ với mọi $x$

$(y-1)^2\geq 0$ với mọi $y$

$\Rightarrow B=(2-x)^2+2(y-1)^2-5\geq 0+2.0-5=-5$

Vậy $B_{\min}=-5$. Giá trị này đạt tại $2-x=y-1=0$

$\Lefrightarrow x=2; y=1$

c.

Ta thấy: $(4x+1)^2\geq 0$ với mọi $x$

$\Rightarrow 3-(4x+1)^2\leq 3$

$\Rightarrow C=\frac{5}{3-(4x+1)^2}\geq \frac{5}{3}$

Vậy $C_{\min}=\frac{5}{3}$. Giá trị này đạt tại $4x+1=0\Leftrightarrow x=\frac{-1}{4}$

Bài 5:

c. 

Vì:

$(2x+1)^2\geq 0$ với mọi $x$

$(y-3,5)^2\geq 0$ với mọi $y$

$\Rightarrow -P= (2x+1)^2+7(y-3,5)^2-\frac{2}{3}\geq 0+7.0-\frac{2}{3}=\frac{-2}{3}$

$\Rightarrow P\leq \frac{2}{3}$

Vậy $P_{\max}=\frac{2}{3}$. Giá trị này đạt tại $2x+1=y-3,5=0$

$\Leftrightarrow x=\frac{-1}{2}; y=3,5$

\(\left|\dfrac{1}{-5}\right|-\dfrac{\left(-2\right)^2}{\left|-5\right|}-\dfrac{\left|-2\right|}{5}\\=\dfrac{1}{5}-\dfrac{4}{5}-\dfrac{2}{5}\\ =\dfrac{1-4-2}{5}\\ =\dfrac{-5}{5}=-1\)

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

\(\dfrac{6n+1}{2n+1}\left(n\in Z\right)\\ =\dfrac{3\left(2n+1\right)-2}{2n+1}=3-\dfrac{2}{2n+1}\)

Để biểu thức nhận gt nguyên thì : \(\dfrac{2}{2n+1}\in Z\)

\(=>2n+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\\ =>2n\in\left\{0;-2;1;-3\right\}\\ =>n\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2}\right\}\)

Do n nguyên -> Kết luận : n = 0 hoặc n = -1