Lời giải:

Đổi 30p = 0,5 giờ

Thời gian xe đi: $\frac{AB}{40}$ (h) 

Thời gian xe về: $\frac{AB}{45}$ (h) 

Theo bài ra ta có: $\frac{AB}{40}-\frac{AB}{45}=0,5$

$\Leftrightarrow AB.\frac{1}{360}=0,5$

$\Leftrightarrow AB=0,5: \frac{1}{360}=180$ (km)

Lời giải:

Diện tích mặt xung quanh hộp sữa:

$2.3,14.5.10 = 314$ (cm²)

Diện tích dán keo:

$2.3,14.5.1=31,4$ (cm²)

Diện tích nhãn: $314-31,4=282,6$ (cm²)

\(4746+765=5511\)

5511 nhé

Lời giải:

$49,358-32,16+39,452-9,358+2,16+0,548$

$=(49,358-9,358)-(32,16-2,16)+(39,452+0,548)$

$=40-30+40=10+40=50$

Lời giải:
$2S=\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{6.7}+...+\frac{1}{2022.2023}+\frac{1}{2022.2023}$

$< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{2021.2022}+\frac{1}{2022.2023}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2022}-\frac{1}{2023}$

$=1-\frac{1}{2023}=\frac{2022}{2023}$

$\Rightarrow S< \frac{1011}{2023}$

Lời giải:

$41,54-4,18+23,17+8,46-5,82-3,17$

$=(41,54+8,46)-(4,18+5,82)+(23,17-3,17)$

$=50-10+20=40+20=60$

\(41,54-4,18+23,17+8,46-5,82-3,17\)

\(=\left(41,54+8,46\right)-\left(4,18+5,82\right)-\left(23,17-3,17\right)\)

\(=50-10-10\)

\(=30\)

\(0,24\times450+0,8\times15\times3+3\times4\times8\)

\(=0,24\times100\times4,5+0,8\times3\times15+3\times8\times4\)

\(=24\times4,5+2,4\times10\times1,5+24\times4\)

\(=24\times4,5+24\times1,5+24\times4\)

\(=24\times\left(4,5+1,5+4\right)\)

\(=24\times10\)

\(=240\)

Lời giải:

$0,24\times 450+0,8\times 15\times 3+3\times 4\times 8$

$=108+36+96 = 240$

7.

$(3\times x-1)\times (\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16})=\frac{15}{13}$

$(3\times x-1)\times \frac{8+4+2+1}{16}=\frac{15}{13}$

$(3\times x-1)\times \frac{15}{16}=\frac{15}{13}$

$3\times x-1=\frac{15}{13}: \frac{15}{16}=\frac{16}{13}$

$3\times x=\frac{16}{13}+1=\frac{29}{13}$

$x=\frac{29}{13}:3=\frac{29}{39}$

9.

$x+\frac{1}{1\times 2}+\frac{1}{2\times 3}+....+\frac{1}{99\times 100}=\frac{101}{100}$

$x+\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+....+\frac{100-99}{99\times 100}=\frac{101}{100}$

$x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=\frac{101}{100}$

$x+1-\frac{1}{100}=\frac{101}{100}$

$x=\frac{101}{100}+\frac{1}{100}-1=\frac{1}{50}$

0,36x950+0,18x726x2+3x324x0,12

=0,36x950+0,18x2x726+3x0,12x324

=0,36x950+0,36x726+0,36x324

=0,36x(950+726+324)

=0,36x2000

=720

1.

Áp dụng định lý Viet:

$x_1+x_2=\frac{7}{2}$

$x_1x_2=\frac{-3}{2}$
Khi đó:

$B=x_1^2x_2+x_2^2x_1-3x_1x_2=x_1x_2(x_1+x_2)-3x_1x_2$

$=\frac{-3}{2}.\frac{7}{2}-3.\frac{-3}{2}=\frac{-3}{4}$

2.

Để pt có 2 nghiệm $x_1,x_2$ thì:

$\Delta'=(m+1)^2-3(2m-1)\geq 0$

$\Leftrightarrow m^2-4m+4\geq 0$

$\Leftrightarrow (m-2)^2\geq 0\Leftrightarrow m\in\mathbb{R}$
Áp dụng định lý Viet:

$x_1+x_2=\frac{2(m+1)}{3}$

$x_1x_2=\frac{2m-1}{3}$
Để PT có 2 nghiệm $x_1,x_2<2$ thì:
\(\left\{\begin{matrix} x_1+x_2< 4\\ (x_1-2)(x_2-2)>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x_1+x_2<4\\ x_1x_2-2(x_1+x_2)+4>0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \frac{2(m+1)}{3}<4\\ \frac{2m-1}{3}-2.\frac{2(m+1)}{3}+4>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} m<5\\ m< \frac{7}{2}\end{matrix}\right.\Leftrightarrow m< \frac{7}{2}\)

Vậy..........