Câu khó thế

um khó mà giúp tui đi mn

\(\dfrac{x^2-3x}{2x^2-3x-9}=\dfrac{x^2+3x}{A}\)

\(\Rightarrow A=\dfrac{\left(x^2+3x\right)\left(2x^2-3x-9\right)}{x^2-3x}\)

\(\Rightarrow A=\dfrac{x\left(x+3\right)\left(2x^2-3x-9\right)}{x\left(x-3\right)}\)

\(\Rightarrow A=\dfrac{\left(x+3\right)\left(2x^2-3x-9\right)}{\left(x-3\right)}\)

mà \(x=-\dfrac{3}{2}\)

\(\Rightarrow A=\dfrac{\left(-\dfrac{3}{2}+3\right)\left(2\left(-\dfrac{3}{2}\right)^2-3\left(-\dfrac{3}{2}\right)-9\right)}{\left(-\dfrac{3}{2}-3\right)}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(2.\dfrac{9}{4}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(\dfrac{9}{2}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(\dfrac{9}{2}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}=0\)

A = 0

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

\(=-3x\left(x^2+4x+4\right)+x^2+3x+x+3-\left(4x^2-12x+9\right)\)

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

\(=-3x^3-15x^2+4x-6\)

ai lam cau b cko tui dc ko

\(x^2-2xy+5y^2+4y+1\)

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

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

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

\(x^2-2xy+5y^2+4y+1=x^2-2xy+y^2+4y^2+4y+1=\left(x-y\right)^2+\left(2y+1\right)^2\)

Bài 1:

a, \(x^3\) + y³ + \(x\) + y

= (\(x^3\) + y³) + (\(x\)  + y)

= (\(x\) + y)(\(x^2\) - \(xy\) + y²)  + (\(x\) + y)

= (\(x\) + y)( \(x^2-xy+y^2\)+1)

b, \(x^3\) + 4\(x^2\)y + 4\(xy^2\) - 9\(x\)

= \(x\)(\(x^2\) + 4\(xy\) + 4y² - 9)

= \(x\)[ (\(x\) + 2y)² - 3²)

= \(x\)[ (\(x\) + 2y - 3).( \(x\) + 2y + 3)]

1) \(A=4x-x^2+3\)

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

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

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

Mà: \(-\left(x-2\right)^2\le0\forall x\) nên: \(A=-\left(x-2\right)^2+7\le7\)

Dấu "=" xảy ra:

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

\(\Rightarrow x=2\)

Vậy: \(A_{max}=7\) khi \(x=2\)

2) \(B=x-x^2\)

\(B=-x^2+x\)

\(B=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)

\(B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)

Mà: \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\) nên \(B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Dấu "=" xảy ra:
\(-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{2}\)

Vậy: \(B_{max}=\dfrac{1}{4}\) với \(x=\dfrac{1}{2}\)

Em viết đủ đề nha

thiếu đề bn ơi