bằng 5 nhé 

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

\(\Leftrightarrow2.\left(x^2+2x-5x-10\right)=x^2-5x\)

\(\Leftrightarrow2x^2+4x-10x-20=x^2-5x\)

\(\Leftrightarrow2x^2-x^2+4x-10x+5x-20=0\)

\(\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow x^2-5x+4x-20=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=5\end{cases}}}\)

Vậy....

a, Xét tam giác ABC và tam giác OMN có 

^BAC = ^MON = 90⁰

\(\dfrac{AC}{ON}=\dfrac{BC}{MN}=\dfrac{8}{4}=\dfrac{10}{5}=2\)

Vậy tam giác ABC ~ tam giác OMN 

b, \(\dfrac{AB}{OM}=\dfrac{BC}{MN}=\dfrac{AC}{ON}\)( tỉ số đồng dạng ) 

\(Q=\dfrac{1}{2}\left(x^2-4xy+4y^2\right)+\dfrac{1}{2}\left(x^2-2x+1\right)+\dfrac{2021}{2}\)

\(Q=\dfrac{1}{2}\left(x-2y\right)^2+\dfrac{1}{2}\left(x-1\right)^2+\dfrac{2021}{2}\ge\dfrac{2021}{2}\)

\(Q_{min}=\dfrac{2021}{2}\) khi \(\left(x;y\right)=\left(1;\dfrac{1}{2}\right)\)

TA CÓ BỔ ĐỀ SAU:

\(a^3+b^3=a^3+3a^2b+3ab^2+b^3-3ab.\left(a+b\right)=\left(a+b\right)^3-3ab.\left(a+b\right)\)

ta có:

\(\left(x-2023\right)^3+\left(x-2021\right)^3=\left(2x-4044\right)^3\)

\(\Leftrightarrow\left(x-2023+x-2021\right)^3-3.\left(x-2023\right).\left(x-2021\right).\left(x-2023+x-2021\right)=\left(2x-4044\right)^3\)

\(\Leftrightarrow\left(2x-4044\right)^3-3.\left(x-2023\right).\left(x-2021\right).\left(2x-4044\right)=\left(2x-4044\right)^3\)

\(\Leftrightarrow\left(2x-4044\right)^3-\left(2x-4044\right)^3=3.\left(x-2023\right).\left(x-2021\right).\left(2x-4044\right)\)

\(\Leftrightarrow3.\left(x-2023\right).\left(x-2021\right).\left(2x-4044\right)=0\)

\(\Leftrightarrow x-2023=0;x-2021=0;2x-4044=0\)

\(\Rightarrow\) x=2023 hoạc x=2021 hoặc x=2022 là nghiệm của phương trình.

vậy................

KHÓ QUÉ