a) x = \(\dfrac{1}{5}+\dfrac{2}{11}\)

    x = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)

    x = \(\dfrac{21}{55}\)

b) \(\dfrac{x}{15}\) = \(\dfrac{3}{5}+\dfrac{-2}{3}\)

    \(\dfrac{x}{15}=\dfrac{9}{15}+\dfrac{-10}{15}\)

    \(\dfrac{x}{15}=\dfrac{-1}{15}\)

      \(x=\dfrac{-1}{15}\cdot15\)

       x = -1

c) \(\dfrac{11}{8}+\dfrac{13}{6}=\dfrac{85}{x}\)

    \(\dfrac{33}{24}+\dfrac{52}{24}=\dfrac{85}{x}\)

              \(\dfrac{85}{24}=\dfrac{85}{x}\)

              \(\dfrac{85}{x}=\dfrac{85}{24}\)

               \(x=85:\dfrac{85}{24}\)

               \(x=85\cdot\dfrac{24}{85}\)

               \(x=24\)

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

a, ( \(\dfrac{1}{4}\) + \(\dfrac{-5}{13}\)) +( \(\dfrac{2}{11}\) + \(\dfrac{-8}{13}\) + \(\dfrac{3}{4}\))

= \(\dfrac{1}{4}\)  - \(\dfrac{5}{13}\) + \(\dfrac{2}{11}\) - \(\dfrac{8}{13}\) + \(\dfrac{3}{4}\)

= ( \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) + \(\dfrac{2}{11}\)

= 1 - 1 + \(\dfrac{2}{11}\)

= \(\dfrac{2}{11}\)

b, ( \(\dfrac{21}{31}\) + \(\dfrac{-16}{7}\)) +( \(\dfrac{44}{53}\) + \(\dfrac{10}{31}\)) + \(\dfrac{9}{53}\)

  =   \(\dfrac{21}{31}-\dfrac{16}{7}+\dfrac{44}{53}+\dfrac{10}{31}+\dfrac{9}{53}\)

    = ( \(\dfrac{21}{31}\) + \(\dfrac{10}{31}\))  + ( \(\dfrac{44}{53}\) + \(\dfrac{9}{53}\))  - \(\dfrac{16}{7}\)

    = 1 + 1 - \(\dfrac{16}{7}\)

     = \(\dfrac{14}{7}-\dfrac{16}{7}\)

     = - \(\dfrac{2}{7}\)

\(\text{#040911}\)

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

B = [1200 - ( 16 - 6 )³ ] : 40

B = [1200 - 10³ ] : 40

B = [1200 - 1000] : 40 

B = 200 : 40 

B = 5

B = [ 1200 - (4² - 2.3)³ ] : 40

B = [ 1200 - (16 - 6)³ ] : 40

B = [ 1200 - 1000]: 40

B = 200: 40 

B = 5

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)

đáp án B bạn nha 

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 $2x+1=y-3,5=0$

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