Vì \(\left\{{}\begin{matrix}\left(2x-y\right)^2\ge0\forall x,y\in R\\\left(x-1\right)^2\ge0\forall x\in R\end{matrix}\right.\) nên \(\left(2x-y\right)^2+\left(x+1\right)^2=0\)

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

\(A=\left(x-2\right)^{2021}+\left(y-2\right)^{2020}\\ =\left(1-2\right)^{2021}+\left(2-2\right)^{2020}=-1^{2021}+0^{2020}=-1+0=-1\)

Gọi số cần tìm là \(\overline{abc}\)

Nếu thêm chữ số 1 vào trước số đó thì số mới gấp 9 lần số ban đầu nên ta có: \(\overline{1abc}=9.\overline{abc}\)

\(\Leftrightarrow1000+\overline{abc}=9.\overline{abc}\)

\(\Leftrightarrow1000=9.\overline{abc}-\overline{abc}=8.\overline{abc}\)

\(\Leftrightarrow\overline{abc}=1000:8=125\)

Vậy số cần tìm là 125.

a, \(D=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(D=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)

\(D=\dfrac{2\left(2\sqrt{x}-3\right)-\left(\sqrt{x}-1\right)}{2\sqrt{x}-3}:\left(\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)

\(D=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\dfrac{2x+3\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(D=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}:\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(D=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}:\dfrac{2\sqrt{x}+1}{2\sqrt{x}-3}\)

\(D=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}.\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)

a, \(3x^2-2x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-\dfrac{2}{3}x+\dfrac{1}{9}=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b, \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(2x-1-x-2\right)\left(2x-1+x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}-\dfrac{3y}{x+1}=5\\\dfrac{4x-6}{y+2}+\dfrac{5y}{x+1}=3\end{matrix}\right.\)

DKXD: \(x\ne1,y\ne-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}-\dfrac{3y}{x+1}=5\\\dfrac{2\left(2x-3\right)}{y+2}+\dfrac{5y}{x+1}=3\end{matrix}\right.\)

Đặt \(\dfrac{2x-3}{y+2}=a,\dfrac{y}{x+1}=b\)

Ta có hệ pt: \(\left\{{}\begin{matrix}a-3b=5\\2a+5b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{34}{11}\\b=-\dfrac{7}{11}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}=\dfrac{34}{11}\\\dfrac{y}{x+1}=-\dfrac{7}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11\left(2x-3\right)=34\left(y+2\right)\\-7\left(x+1\right)=11y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}22x-33=34y+68\\-7x-7=11y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}22x-34y=101\\-7x-11y=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{97}{42}\\y=-\dfrac{31}{21}\end{matrix}\right.\) (tmdk)

Ta có hệ thức độc lập là: \(A^2=\dfrac{v^2}{\omega^2}+\dfrac{a^2}{\omega^4}\)

\(\Rightarrow a^2=\left(A^2-\dfrac{v^2}{\omega^2}\right)\omega^4=A^2\omega^4-v^2\omega^2=\omega^2\left(A^2\omega^2-v^2\right)\)

\(\Rightarrow a=\sqrt{\omega^2\left(A^2\omega^2-v^2\right)}=\omega\left(A^2\omega^2-v^2\right)\)

1. You don't have to mop the floor twice every day.

2. She likes sweeping the floors.

3. It takes Jennie 2 hours to tidy her bedroom.

4. He is never late for school.

5. In Vietnam, an extended family usually consists of three or four generations.

6. Parents should share experience with their children so that they don't make wrong decisions.

M là trung điểm AB \(\Rightarrow AM=BM=\dfrac{AB}{2}=\dfrac{10}{2}=5cm\)

N là trung điểm AC \(\Rightarrow AN=CN=8cm\)

Ta có:

M là trung điểm AB, N là trung điểm AC

\(\Rightarrow\) MN là đường trung bình \(\Delta ABC\)

\(\Rightarrow MN=\dfrac{BC}{2}=7cm\Rightarrow BC=14cm\)

16. beautiful

17. pollution

18. warmer - hottest

19. teeth

20. once

\(16x^2\left(x-5\right)+5-x=0\)

\(\Leftrightarrow16x^2\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(16x^2-1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[\left(4x\right)^2-1^2\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(4x-1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\4x-1=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)