uki

chờ mik tí

a: Tỉ lệ phần trăm chất béo có trong 100gam gạo là:

1,3:100=1,3%

b:

1,5kg=1500(gam)

Trong 1,5kg thì khối lượng chất béo là:

1500:100*1,3=19,5(gam)

a) Tỉ lệ phần trăm khối lượng chất béo có trong 100 g gạo là:
(1,3 :100) x 100% = 1,3% 
b) Đổi: 1,5 kg = 1500 (g)
Khối lượng chất béo có trong 1,5 kg gạo là:
1500 x 1,3% = 19,5 (g)
Đáp số: a) 1,3%.
              b) 19,5 gam chất béo.

ĐKXĐ: n<>-1/5

Để \(\dfrac{4n-12}{5n+1}\) nguyên thì \(4n-12⋮5n+1\)

=>\(20n-60⋮5n+1\)

=>\(20n+4-64⋮5n+1\)

=>\(-64⋮5n+1\)

=>\(5n+1\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16;32;-32;64;-64\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{5};\dfrac{1}{5};-\dfrac{3}{5};\dfrac{3}{5};-1;\dfrac{7}{5};-\dfrac{9}{5};3;-\dfrac{17}{5};\dfrac{31}{5};-\dfrac{33}{5};\dfrac{63}{5};-13\right\}\)

mà n nguyên

nên \(n\in\left\{0;-1;3;-13\right\}\)

Lời giải:

Gọi số cần tìm là $a$ (điều kiện $a>20$). Theo bài ra:

$185-20\vdots a$

$\Rightarrow 165\vdots a$

$250-19\vdots a$

$\Rightarrow 231\vdots a$

$\Rightarrow a=ƯC(165,231)$

$\Rightarrow ƯCLN(165,231)\vdots a$

$\Rightarrow 33\vdots a$

Mà $a>20$ nên $\Rightarrow a=33$

1: ĐKXĐ: \(n\ne-\dfrac{1}{2}\)

Để \(\dfrac{3n+2}{2n+1}\) nguyên thì \(3n+2⋮2n+1\)

=>\(6n+4⋮2n+1\)

=>\(6n+3+1⋮2n+1\)

=>\(1⋮2n+1\)

=>\(2n+1\in\left\{1;-1\right\}\)

=>\(n\in\left\{0;-1\right\}\)(nhận)

2:

ĐKXĐ: n<>-1/5

Để \(\dfrac{8n+12}{5n+1}\) là số nguyên thì

\(8n+12⋮5n+1\)

=>\(40n+60⋮5n+1\)

=>\(40n+8+52⋮5n+1\)

=>\(52⋮5n+1\)

=>\(5n+1\in\left\{1;-1;2;-2;4;-4;13;-13;26;-26;52;-52\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{5};\dfrac{1}{5};-\dfrac{3}{5};\dfrac{3}{5};-1;\dfrac{12}{5};-\dfrac{14}{5};5;-\dfrac{27}{5};\dfrac{51}{5};-\dfrac{53}{5}\right\}\)

mà n nguyên

nên \(n\in\left\{0;-1;5\right\}\)

\(\left(2024-x\right)^2=1-y^2\)

=>\(\left(2024-x\right)^2+y^2=1\)

mà x,y nguyên

nên \(\left(2024-x\right)^2+y^2=0+1=1+0\)

TH1: \(\left\{{}\begin{matrix}\left(2024-x\right)^2=0\\y^2=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2024-x=0\\y\in\left\{1;-1\right\}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2024\\y\in\left\{1;-1\right\}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}\left(2024-x\right)^2=1\\y^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2024-x\in\left\{1;-1\right\}\\y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{2023;2025\right\}\\y=0\end{matrix}\right.\)

\(15\%+1,1:\left(\dfrac{2}{5}-1\dfrac{1}{2}\right)-\left(-\dfrac{1}{3}\right)^2\)

\(=\dfrac{3}{2}-\dfrac{1}{9}+1,1:\left(0,4-1,5\right)\)

\(=\dfrac{27-2}{18}+1,1:\left(-1,1\right)\)

\(=\dfrac{25}{18}-1=\dfrac{7}{18}\)

1/3 . 6/(-7) = -6/21 = -2/7

\(\dfrac{1}{3}\). \(\dfrac{6}{-7}\) = \(\dfrac{ }{7}\)

   \(\dfrac{2}{-7}\) =  \(\dfrac{ }{7}\)

        \(◻\) = \(\dfrac{2}{-7}\) x 7

       \(◻\)  = \(-2\) 

-2/3 . -5/8 = 10/24 = 5/12

\(\dfrac{-2}{3}\).\(\dfrac{-5}{8}\) = \(\dfrac{ }{12}\)

\(\dfrac{5}{12}\) = \(\dfrac{◻}{12}\)

\(◻\) = \(\dfrac{5}{12}\) \(\times\) 12

\(◻\)  = 5