a)\(x+\dfrac{2}{3}x\dfrac{5}{6}=7\)

\(x+\dfrac{10}{18}=7\)

\(x=7-\dfrac{10}{18}\)

\(x=\dfrac{58}{9}\)

b)\(9-xx\dfrac{1}{2}=\dfrac{3}{4}\)

\(xx\dfrac{1}{2}=9-\dfrac{3}{4}\)

\(xx\dfrac{1}{2}=\dfrac{33}{4}\)

\(x=\dfrac{33}{4}:\dfrac{1}{2}\)

\(x=\dfrac{33}{2}\)

c)\(18-x:\dfrac{2}{3}=\dfrac{4}{5}-\dfrac{1}{10}\)

\(18-x:\dfrac{2}{3}=\dfrac{7}{10}\)

\(x:\dfrac{2}{3}=18-\dfrac{7}{10}\)

\(x:\dfrac{2}{3}=\dfrac{173}{10}\)

\(x=\dfrac{173}{10}x\dfrac{2}{3}\)

\(x=\dfrac{173}{15}\)

a)\(27.65+27.35+300=27.\left(65+35\right)+300\)

\(=27.100+300=2700+300=3000\)

b)\(3838:\left[\left(190-6.5^2\right):4+3\right]\)

\(=3838:\left[\left(190-6.25\right):4+3\right]\)

\(=3838:\left[\left(190-150\right):4+3\right]\)

\(=3838:\left[40:4+3\right]=3838:\left[10+3\right]\)

\(=3838:13=\dfrac{3838}{13}\)

c)\(2022-x=2021\)

\(x=2022-2021=1\)

d)\(26+14:\left(x-5\right)=33\)

\(14:\left(x-5\right)=33-26=7\)

\(x-5=14+7=2\)

\(x=2+5=7\)

e)đề hỏi làm j thế bạn

\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)

\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)

\(x-2023=1\)

\(x=2024\)

Vậy..............

\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)

\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)

0,856m=?m

\(...=\dfrac{3}{2}x\left(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\right)\)

\(=\dfrac{3}{2}x\left(\dfrac{256}{256}+\dfrac{64}{256}+\dfrac{16}{256}+\dfrac{4}{256}+\dfrac{1}{256}\right)\)

\(=\dfrac{3}{2}x\dfrac{341}{256}=\dfrac{1023}{512}\)

a) Số chữ số từ 1 đến 9 : \(\left(9-1\right):1+1=9\left(chữ.số\right)\)

Số chữ số từ 10 đến 99 \(\left(\left(99-10\right):1+1\right).2=180\left(chữ.số\right)\)

Chữ số 100 có \(3\left(chữ.số\right)\)

Vậy có \(9+180+3=192\left(chữ.số\right)\) thỏa đề bài

Câu b bạn xóa các số theo thứ tự 1;2;3.. sao cho đủ 100 chữ số là được.

Lời giải:
20 phút đầu người đó đi được: $7\times 20:60=\frac{7}{3}$ (km) 

Quãng đường còn lại dài: $4-\frac{7}{3}=\frac{5}{3}$ (km) 

Người đó đi quãng đường còn lại trong: 

$\frac{5}{3}: 6=\frac{5}{18}$ (giờ)

Người đó đi quãng đường AB sau: $\frac{20+10}{60}+\frac{5}{18}=\frac{7}{9}$ (giờ)

\(\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)

\(=\left(2^3\cdot3^8+3^6\cdot5\cdot3^2\right):\left[9^2\cdot\left(10-1\right)\right]\)

\(=\left(2^3\cdot3^8+3^8\cdot5\right):\left(9^2\cdot9\right)\)

\(=\left[3^8\cdot\left(2^3+5\right)\right]:9^3\)

\(=3^8\cdot13:3^6\)

\(=3^2\cdot13\)

\(=117\)

\(\left(2^3.9^4+9^3.45\right):\left(9^2.10-9^2\right)\)

\(=\left(2^3.3^8+3^6.3^2.5\right):\left(3^4.10-3^4\right)\)

\(=\left(2^3.3^8+3^8.5\right):\left(3^4.\left(10-1\right)\right)\)

\(=3^8\left(2^3+5\right):3^4.9=3^8\left(2^3+5\right):3^6\)

\(=3^2.13=9.13=117\)

2n + 19 chia hết cho 2n + 5 

⇒ 2n + 5 + 14 chia hết cho 2n + 5 

⇒ 14 chia hết cho 2n + 5 

⇒ 2n + 5 ϵ Ư(14) 

Mà n nguyên nên 2n + 5 ϵ { 1 ; -1 ; 7; -7)

Ta có bảng sau:

Vậy: n ϵ {-2 ; -3; 1; -6}

Lần sau bạn lưu ý ghi đầy đủ đề. 

Tìm $n\in\mathbb{Z}$ sao cho $2n+19\vdots 2n+5$

Lời giải:

$2n+19\vdots 2n+5$

$\Rightarrow (2n+5)+14\vdots 2n+5$

$\Rightarrow 14\vdots 2n+5$

$\Rightarrow 2n+5\in\left\{1;7; -1; -7\right\}$ (do $2n+5$ lẻ) 

$\Rightarrow n\in\left\{-2; 1; -3; -6\right\}$

1) \(f\left(x\right)=6x^2-15x+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{25}{36}\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+4-\dfrac{25}{6}\)

\(\Rightarrow f\left(x\right)=6\left(x-\dfrac{5}{6}\right)^2-\dfrac{1}{6}\ge-\dfrac{1}{6}\left(6\left(x-\dfrac{5}{6}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{1}{6}\left(tạix=\dfrac{5}{6}\right)\)

2) \(f\left(x\right)=4x^2-13x+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}-\dfrac{169}{64}\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)+5-\dfrac{169}{16}\)

\(\Rightarrow f\left(x\right)=4\left(x-\dfrac{13}{8}\right)^2-\dfrac{89}{16}\ge-\dfrac{89}{16}\left(4\left(x-\dfrac{13}{8}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{89}{16}\left(tạix=\dfrac{13}{8}\right)\)