=2023phần2053345

\(\frac{2022\times2023-2020\times2023}{2022\times2023+2024\times7+2016}\)

\(=\frac{2023\times\left(2022-2020\right)}{2022\times2023+7\times\left(2023+1\right)+2016}\)

\(=\frac{2023\times2}{2023\times2022+7\times2023+7+2016}=\frac{2023\times2}{2023\times\left(2022+7+1\right)}=\frac{2}{2022+8}\)

\(=\frac{2}{2030}=\frac{1}{1015}\)

khó quá 

a) 11/12 x 9/19 - 22/24 x 6/19 + 11/12 x 16/19.

= 11/12 x 9/19 - 11/12 x 6/19 + 11/12 x 16/19.

=11/12 x (  9/19 -6/19 +  16/19)

=11/12x 1

=11/12

mình biết câu a/ thôi

\(\Leftrightarrow4sin^{2020}x\left(1-2sin^2x\right)=4cos^{2020}x\left(2cos^2x-1\right)+5cos2x=0\)

\(\Leftrightarrow4sin^{2020}x.cos2x=4cos^{2020}x.cos2x+5cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=...\\4sin^{2020}x=4cos^{2020}x+5\left(1\right)\end{matrix}\right.\)

Xét (1), ta có \(\left\{{}\begin{matrix}4sin^{2020}x\le4\\4cos^{2020}x+5\ge5\end{matrix}\right.\)

\(\Rightarrow4sin^{2020}x< 4cos^{2020}x+5\) với mọi x

\(\Rightarrow\left(1\right)\) vô nghiệm

Cứu với ;-;

help me !!!!!!!

\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)

`a, A = 3020 xx 3110 - 5 = 3020 xx 3109 + 3020 - 5`

`= 3020 xx 3109 + 3015 = B`.

`b, B = (2022-2)(2022+2) = 2022^2-4 < 2022^2 = A.`

B nha bn

Lý thuyết : Những số nào có chữ số tạn cùng là 0 hoặc 5 thì chia hết cho 5.

⇒ Đáp án B. 2020 + 2000 + 2030