\(21-5\left(x-4\right)=10\)

\(5\left(x-4\right)=21-10\)

\(5\left(x-4\right)=11\)

\(x-4=\dfrac{11}{5}\)

\(x=\dfrac{11}{5}+4=\dfrac{31}{5}\)

\(\left|2x-1\right|+\left(\dfrac{2}{3}-x\right)^{2024}=0\)

\(\left|2x-1\right|=-\left(\dfrac{2}{3}-x\right)^{2024}\)

Vì \(VT\ge0;VP\le0\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}2x-1=0\\\dfrac{2}{3}-x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)(Loại)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+11\right)^2-25\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

a) \(x\left(x^2+4x+5\right)-x^2\left(x+4\right)\)

\(=x^3+4x^2+5x-x^3-4x^2\)

\(=5x\)

b) \(\left(x-2\right)^2+\left(3-x\right)\left(x-1\right)\)

\(=x^2-4x+4+3x-3-x^2+x\)

\(=1\)

c) \(\left(x+2\right)^3-x\left(x^2+6x+12\right)\)

\(=x^3+6x^2+12x+8-x^3-6x^2-12x\)

\(=8\)

\(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

\(=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)

\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)\)

\(=\left(x+y-4\right)\left(x+y+3\right)\)

a) \(\left(x-3\right)^{2022}=\left(x-3\right)^{2023}\)

\(\left(x-3\right)^{2022}\left(x-3-1\right)=0\)

\(\left(x-3\right)^{2022}\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) \(\left(x+1\right)^{2024}+\left(x+1\right)^{2023}=2\)

\(\left[\left(x+1\right)^{2022}-1\right]\left[\left(x+1\right)^{2022}+2\right]=0\)

\(\Leftrightarrow x+1=1\)

\(\Leftrightarrow x=0\)

c) \(\left(x+3\right)^2+\left(x+3\right)=12\)

\(\left[\left(x+3\right)-3\right]\left[\left(x+3\right)+4\right]=0\)

\(x\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

Áp dụng định lý Pytago

\(AC^2=BC^2-AB^2\)

\(=10^2-6^2=64\)

\(\Rightarrow AC=8cm\)

\(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(9x^2-6x+4\right)\left(3x-1\right)\)

Ta có:

\(\widehat{A}=180^o-60^o-45^o=75^o\)(tổng 3 góc trong tam giác)

Kẻ đường cao AH

\(sin60=\dfrac{AH}{AB}\)

\(\Rightarrow AH=sin60.4=2\sqrt{3}\)

\(sin45=\dfrac{AH}{AC}\)

\(\Rightarrow AC=\dfrac{2\sqrt{3}}{sin45}=2\sqrt{6}\)

\(BC=\sqrt{4^2-\left(2\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{6}\right)^2-\left(2\sqrt{3}\right)^2}=2+2\sqrt{3}\)