a: \(\dfrac{5}{4}-\dfrac{1}{3}+\dfrac{7}{6}\)

\(=\dfrac{15}{12}-\dfrac{4}{12}+\dfrac{14}{12}\)

\(=\dfrac{25}{12}\)

b: \(-\dfrac{7}{19}\cdot\dfrac{6}{11}+\dfrac{-7}{19}\cdot\dfrac{5}{11}+\dfrac{-12}{19}\)

\(=-\dfrac{7}{19}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{-12}{19}\)

\(=-\dfrac{7}{19}-\dfrac{12}{19}=-\dfrac{19}{19}=-1\)

có 32 từ mới đc các bạn HS nói trên

\(\dfrac{1}{5}x-\dfrac{3}{4}x=-\dfrac{9}{20}\)

\(\Rightarrow x\cdot\left(\dfrac{1}{5}-\dfrac{3}{4}\right)=-\dfrac{9}{20}\)

\(\Rightarrow x\cdot\dfrac{-11}{20}=\dfrac{-9}{20}\)

\(\Rightarrow x=\dfrac{-9}{20}:\dfrac{-11}{20}\)

\(\Rightarrow x=\dfrac{9}{11}\)

x.(1/5-3/4)=-9/20

x.(-11/20)=-9/20

x=-9/20.(20/-11)

x=9/11

6,25 -(-6,58) - 25%

= 6,25 + 6,58 - 25%

= 6,25 + 6,58 - 0,25

= (6,25 - 0,25) + 6,58

= 6 + 6,58

= 12,58

$6,25-(-6,58)-25%$

$=6,25+6,58-0,25$

$=(6,25-0,25)+6,58$

$=6+6,58=12,58$

Câu 1:

1:

a: \(A=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)

b: \(1\cdot2+2\cdot3+...+2022\cdot2023\)

\(=1\left(1+1\right)+2\left(1+2\right)+...+2022\left(1+2022\right)\)

\(=\left(1+2+...+2022\right)+\left(1^2+2^2+...+2022^2\right)\)

\(=\dfrac{2022\cdot2023}{2}+\dfrac{2022\left(2022+1\right)\left(2\cdot2022+1\right)}{6}\)

\(=1011\cdot2023+2022\cdot2023\cdot\dfrac{4045}{6}\)

\(=1011\cdot2023\left(1+2\cdot\dfrac{4045}{6}\right)\)

\(Q=\dfrac{1\cdot2+2\cdot3+...+2022\cdot2023}{2022\cdot2023\cdot2024}\)

\(=\dfrac{1011\cdot2023\left(1+\dfrac{4045}{3}\right)}{2022\cdot2023\cdot2024}=\dfrac{1}{2}\cdot\dfrac{1+\dfrac{4045}{3}}{2024}\)

\(=\dfrac{1}{2}\cdot\dfrac{4048}{3}:2024=\dfrac{1}{2}\cdot\dfrac{2}{3}=\dfrac{1}{3}\)

Câu 3:

1:

b: Số lần lấy được viên màu đỏ là 16 lần

=>Xác suất thực nghiệm là \(\dfrac{16}{40}=0,4\)

2: 2xy-4x-y=3

=>2x(y-2)-y+2=5

=>(2x-1)(y-2)=5

=>\(\left(2x-1;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(1;7\right);\left(3;3\right);\left(0;-3\right);\left(-2;1\right)\right\}\)

Bài 1:

a, \(\dfrac{7}{13}\)+ \(\dfrac{6}{13}\)

= \(\dfrac{13}{13}\)= 1

b,\(\dfrac{5}{6}\)- \(\dfrac{7}{3}\)x \(\dfrac{1}{14}\)

= \(\dfrac{5}{6}\)- \(\dfrac{1}{6}\)

= \(\dfrac{2}{3}\)

c, \(\dfrac{5}{17}\) x \(\dfrac{-13}{21}\) + \(\dfrac{5}{17}\) x \(\dfrac{-4}{21}\)

= \(\dfrac{5}{17}\) x \(\dfrac{-17}{21}\)

= \(\dfrac{-5}{21}\)

Bài 2:

a, x - \(\dfrac{3}{5}\)= \(\dfrac{2}{3}\)

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

    x = \(\dfrac{19}{15}\)

b, \(\dfrac{1}{4}\) - ( \(\dfrac{3}{4}\) + x ) = 2

     \(\dfrac{3}{4}\) + x = \(\dfrac{1}{4}\) - 2

      \(\dfrac{3}{4}\) + x = \(\dfrac{-7}{4}\)

             x = \(\dfrac{-7}{4}\) - \(\dfrac{3}{4}\)

             x = \(\dfrac{-5}{2}\)

Câu 1: D

Câu 2: C

Câu 4: C, D

Câu 5: C

Câu 7: C

Câu 8: D

Câu 9: D

Câu 10: D

Lời giải:

$3n+24\vdots n-4$

$\Rightarrow 3(n-4)+36\vdots n-4$

$\Rightarrow 36\vdots n-4$

$\Rightarrow n-4\in\left\{\pm 1; \pm 2; \pm 3; \pm 4; \pm 6; \pm 9; \pm 12; \pm 18; \pm 36\right\}$
$\Rightarrow n\in\left\{5; 3; 2; 6; 7; 1; 8; 0; 10; -2; 13; -5; 16; -8; 22; -14; 40; -32\right\}$

Lời giải:

\(A=2024.\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-2022}{2023}.\frac{-2023}{2024}\\ =2024.\frac{(-1)(-2)(-3)...(-2022)}{2.3.4...2023.2024}\\ =2024.\frac{1.2.3...2022}{2.3.4....2023.2024}\\ =2024.\frac{1}{2023.2024}=\frac{1}{2023}\)