\(9x^2-6x^2-3\)

\(=3x^2-3\)

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

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

\(9x^2-6x^2-3\)

\(=3x^2-3\)

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

\(=3.\left(x-1\right).\left(x+1\right)\)

\(9x^2-6x^2-3\)

\(=3x^2-3\)

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

\(=3.\left(x-1\right).\left(x+1\right)\)

Nguồn: kudo shinichi

\(A=-x^2+6x-6=-\left(x^2-6x+9\right)+3=-\left(x-3\right)^2+3\le3 \)

Vậy GTLN của A là 3 khi x = 3

\(B=-2x^22+5x-10=-\left(4x^2-5x+\frac{25}{16}\right)-\frac{135}{16}=-\left(2x-\frac{5}{4}\right)-\frac{135}{16}\le-\frac{135}{16}\)

Vậy GTLN của B là \(-\frac{135}{16}\)khi x = \(\frac{5}{8}\)

\(C=-5x^2+x+15=-5\left(x^2-\frac{1}{5}x+\frac{1}{100}\right)+\frac{301}{20}=-5\left(x-\frac{1}{10}\right)^2+\frac{301}{20}\le\frac{301}{20}\)

Vậy GTLN của C là \(\frac{301}{20}\)khi x = \(\frac{1}{10}\)

a) Đề là 4x²-9y² mới đúng chớ 

=(2x)² - (3y)²

=(2x-3y)(2x+3y)

b) = x⁴ + 2x³y + x²y² - y⁴- 2xy³ -x²y²

= ( x⁴-y⁴) + ( 2x³y - 2xy³)

=(x²+y²)(x²-y²)+ 2xy(x²-y²)

=(x²-y²)(x²+2xy+y²)

=(x-y)(x+y)³

Câu hỏi của Trà My - Toán lớp 8 - Học toán với OnlineMath

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+....+\frac{1}{47.48.49.50}\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}.\frac{6533}{39200}=\frac{6533}{117600}\)

\(12x-9x^2-3=0\)

\(9x^2-12x+3=0\)

\(3\left(3x^2-4x+1\right)=0\)

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

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)

\(12x-9x^2-3=0\)

\(-9x^2+12x-3=0\)

\(-3\left(3x^2-4x+1\right)=0\)

\(3x^2-4x+1=0\)

\(3x^2-3x-x+1=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

Vậy...

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

\(=\left[\left(x-1\right)\left(x+1\right)\right]\left[\left(x^2-x+1\right)\left(x^2+x+1\right)\right]\)

\(=\left(x^2-1\right)\left[\left(x^2-1\right)^2-x^2\right]\)

\(=\left(x^2-1\right)\left[x^4-2x^2+1-x^2\right]\)

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

you nhia la gi