\(a,x^2+20x+100=\left(x+10\right)^2\)

\(b,16x^2+24xy+9y^2=\left(4x+3y\right)^2\)

\(c,\left(2a+3b\right)\left(4a^2-6ab+9b^2\right)=8a^3+27a^3\)

\(d,\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)=125x^3-64y^3\)

z ở đâu hả bạn ?

\(3\left(5x-1\right)-x\left(x+1\right)+x^2=14\)

\(\Leftrightarrow15x-3-x^2-x+x^2=14\)

\(\Leftrightarrow\left(x^2-x^2\right)+\left(15x-x\right)-3=14\)

\(\Leftrightarrow14x=17\)

\(\Leftrightarrow x=\frac{17}{14}\)

Vậy  \(x=\frac{17}{14}\)

3(5x - 1) - x(x+1)+x ² = 14

➡️15x - 3 - x ² - x + x ² = 14

➡️(15x - x ) + ( -x ² + x ² ) - 3= 14

➡️14x -3 = 14

➡️14x = 14+3

➡️14x = 17

➡️x = 17/14

Hok tốt~

\(x^{14}-10x^{13}+10x^{12}-10x^{11}+...-10x+10=x^{14}-9x^{13}-x^{13}+9x^{12}+x^{12}-...-9x-x+9+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-...-x^2-x+x+1=1\)

\(\frac{1}{x+2}+\frac{5}{x-2}=\frac{3x-12}{x^2-4}\)

ĐKXĐ \(x\ne\pm2\)

\(\frac{1}{x+2}+\frac{5}{x-2}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow\frac{x-2+5\left(x+2\right)}{x^2-4}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow x-2+5x+10=3x-12\)

\(\Leftrightarrow6x+8=3x-12\)

\(\Leftrightarrow3x=20\Leftrightarrow x=\frac{20}{3}\left(tm\right)\)

Đặt  \(A=x^2+y^2+2x+4y+16\)

\(A=\left(x^2+2x+1\right)+\left(y^2+4y+4\right)+11\)

\(A=\left(x+1\right)^2+\left(y+2\right)^2+11\)

Mà  \(\left(x+1\right)^2\ge0\forall x\)

       \(\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow A\ge11\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x+1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-2\end{cases}}\)

Vậy  \(A_{Min}=11\Leftrightarrow\left(x;y\right)=\left(-1;-2\right)\)