\(\left(a\right)\dfrac{5}{3}-\dfrac{1}{2}:x=\dfrac{2}{7}\\ \dfrac{1}{2}:x=\dfrac{5}{3}-\dfrac{2}{7}=\dfrac{29}{21}\\ x=\dfrac{1}{2}:\dfrac{29}{21}=\dfrac{21}{58}\\ \left(b\right)\dfrac{1}{4}\cdot x+\dfrac{7}{5}=\dfrac{5}{3}\\ \dfrac{1}{4}\cdot x=\dfrac{5}{3}-\dfrac{7}{5}=\dfrac{4}{15}\\ x=\dfrac{4}{15}:\dfrac{1}{4}=\dfrac{16}{15}\\ \left(c\right)5\cdot x-2024=3\cdot x\\ 5\cdot x-3\cdot x=2024\\ x\cdot\left(5-3\right)=2024\\ x\cdot2=2024\\ x=2024:2=1012\)

a) 

Có : 990 ml + 9 ml + 1 ml = 1000 ml = 1 l

Vậy chọn B.

B.1

$(x+6)\left(\frac{x^2+3}{2}-1\right)=0$

$\Leftrightarrow \left[\begin{array}{} x+6=0\\ \frac{x^2+3}{2}-1=0 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=-6\\ \frac{x^2+3}{2}=1 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=-6\\ x^2+3=2 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=-6\\ x^2=-1\text{ (vô lí)} \end{array} \right.$

$\Rightarrow x=-6$

#$\mathtt{Toru}$

\(\dfrac{x}{8}\) + \(\dfrac{2}{3}\) = \(\dfrac{7}{6}\)

 \(\dfrac{x}{8}\)        = \(\dfrac{7}{6}\) - \(\dfrac{2}{3}\)

  \(\dfrac{x}{8}\)       = \(\dfrac{1}{2}\) 

   \(x\)        = \(\dfrac{1}{2}\) x 8 

    \(x\)       = 4

\(\dfrac{x}{8}+\dfrac{2}{3}=\dfrac{7}{6}\)

        \(\dfrac{x}{8}=\dfrac{7}{6}-\dfrac{2}{3}\)

        \(\dfrac{x}{8}=\dfrac{1}{2}=\dfrac{4}{8}\)

⇒ x = 4

Vậy x = 4

\(y=\dfrac{x^3}{3}-x^2+\left(m-1\right)x+3\)

\(\Rightarrow y'=x^2-2x+\left(m-1\right)\)

Xét hàm số y' trên khoảng đóng \(\left[1;3\right]\). Do \(a>0\) và \(y'\le0,\forall x\in\left[1;3\right]\) nên phương trình \(y'=0\) luôn có 2 nghiệm thực phân biệt

Ta có: \(\Delta_{y'}=4-4\left(m-1\right)=-4m+8>0\) \(\Rightarrow m< 2\)

Phương trình y'=0 có 2 nghiệm thực phân biệt là:

\(x_1=\dfrac{2+\sqrt{-4m+8}}{2}=1+\sqrt{-m+2}\) ; \(x_2=\dfrac{2-\sqrt{-4m+8}}{2}=1-\sqrt{-m+2}\)

Để ý \(y'\le0\) trên đoạn \(\left[1-\sqrt{-m+2};1+\sqrt{-m+2}\right]\) (do \(a>0\)), nên \(\left[1-\sqrt{-m+2};1+\sqrt{-m+2}\right]\supseteq\left[1;3\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-\sqrt{-m+2}\le1\\1+\sqrt{-m+2}\ge3\end{matrix}\right.\Leftrightarrow m\le-2\)

Vậy với \(m\le-2\) thì \(y'\le0,\forall x\in\left[1;3\right]\)

\(\overline{ab8}+\overline{ab}=949\\ \overline{ab}\times10+8+\overline{ab}=949\\\overline{ab} \times\left(10+1\right)=949-8\\ \overline{ab}\times11=941\\ \overline{ab}=941:11\\ \overline{ab}=941:11\\ \overline{ab}=\dfrac{941}{11}\) 

Mà: ab là số tự nhiên nên ta loại 

\(y'=x^2-2x+(m-1) \geq0 \forall x \in (-1,+\infty) \)

\(\Rightarrow x^2-2x\ge1-m\forall x\in\left(-1,\infty\right)\)

Đặt \(f\left(x\right)=x^2-2x\Rightarrow f'\left(x\right)=2x-2\)

\(\Rightarrow f'\left(x\right)=0\Rightarrow x=1\)

bbt:

\(\dfrac{6}{8}=\dfrac{15}{x}\\ \Rightarrow6\times x=15\times8\\ \Rightarrow6\times x=120\\ \Rightarrow x=120:6\\ \Rightarrow x=20\)

Vậy \(x=20\)