\(\left(x^2-2x+3\right)\left(\frac{1}{2x}-5\right)\)

\(=\frac{x^2}{2x}-5x^2-\frac{2x}{2x}+10x+\frac{3}{2x}-15\)

\(=\frac{x^2}{2x}-5x^2-16+10x+\frac{3}{2x}\)

\(=-5x^2+\frac{x^2}{2x}+\frac{20x^2}{2x}+\frac{3}{2x}-16\)

\(=-5x^2+\frac{x^2+20x+3}{2x}-16\)

học tốt

(x^2-2x+3)(1/2x-5)=1/2x^3-5x^2-x^2+10x+3/2x-15=1/2x^3-6x^2+11,5x-15

a,   \(7x^2-14xy^2+7y^4=7\left(x^2-2xy^2+y^4\right)=7\left(x-y^2\right)^2\)

\(7.\left(x^2-2.x.y^2+y^4\right)\) \(=7.\left(x-y^2\right)^2\)

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

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

\(=\left(1-2x\right)^2-\left(3x-2\right)^2=0\)

\(\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)

\(\left(3-5x\right)\left(x-1\right)=0\)

\(\Rightarrow3-5x=0\) \(x-1=0\)

\(\Rightarrow x=\frac{3}{5}\)  or \(x=1\)

b)\(\left(x-2\right)^3+\left(5-2x\right)^3\)

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

\(\left(3-x\right)\left(x^2-4x+4-5x+2x^2+10-4x+25-20x+4x^2\right)\)

(\(\left(3-x\right)\left(7x^2-33x+39\right)\)

..............

B=x^2+2xy+y^2-4x-4y+1

=(x+y)^2-4(x+y)+1

=3^2-4.3+1=-2

B= (x²+2xy+y²) - (4x+4y) +1

B= (x+y)² - 4(x+y) +1

Thay x+y=3 vào B

ta được: B= 3² - 4.3 +1

               B= 9-12+1

               B= -2

\(\left(a^2+b^2+c^2\right)^2-\left(a^2-b^2-c^2\right)^2\)

\(=\left[\left(a^2+b^2+c^2\right)-\left(a^2-b^2-c^2\right)\right]\cdot\left[\left(a^2+b^2+c^2\right)+\left(a^2-b^2-c^2\right)\right]\)

\(=\left[a^2+b^2+c^2-a^2+b^2+c^2\right]\cdot\left[a^2+b^2+c^2+a^2-b^2-c^2\right]\)

\(=\left(2b^2+2c^2\right)\cdot2a^2\)

\(=2\left(a^2+b^2\right)\cdot2a^2\)

\(=4a^2\left(a^2+b^2\right)\)

\(\left(a^2+b^2+c^2\right)^2-\left(a^2-b^2+c^2\right)^2\)

\(=\left(a^2+b^2+c^2-a^2+b^2-c^2\right)\left(a^2+b^2+c^2+a^2-b^2+c^2\right)\)

\(=2b^2\left(2a^2+2c^2\right)\)

\(=4a^2b^2+4b^2c^2\)

=.= hok tốt!!

Đặt \(2x^2+3x+1=y\).Ta có:

\(\left(y-2\right)^2-5\left(y+2\right)+24=0\)

\(\Leftrightarrow y^2-4y+4-5y-10+24=0\)

\(\Leftrightarrow y^2-9y+18=0\)

\(\Leftrightarrow\left(y-3\right)\left(y-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=3\\y=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+3x+1=3\\2x^2+3x+1=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)\left(x+2\right)=0\\\left(2x+5\right)\left(x-1\right)=0\end{cases}}\)

Vậy PT có 4 nghiệm là:\(\frac{1}{2}\)\(,\)\(-2,-\frac{5}{2},1\)