Bài 3: Rút gọn phân thức

sửa đề:

\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

ĐKXĐ: \(x\ge0\);\(x\ne\pm1\)

\(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^3}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)

\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{x^2+xy+y^2}{x\left(x^3-y^3\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{xy-y^2}{x\left(x^3-y^3\right)}\right)\)

\(=\dfrac{x\left(x^3-y^3\right)}{x^3-xy^2}.\dfrac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x^3-y^3\right)}\\ =\dfrac{x^2-y^2}{x\left(x^2-y^2\right)}=\dfrac{1}{x}\)

mình viết trên máy tinh hơi xấu bạn thông cảm nhé!!!Nếu ko chê có thể xem cách giải này!

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

=\(\dfrac{x^2-5+\left(x+1\right)\cdot\left(x+2\right)+x^2-x+1}{x^3+1}\)

=\(\dfrac{x^2-5+x^2+2\cdot x+x+2+x^2-x+1}{x^3+1}\)

=\(\dfrac{3\cdot x^2+2\cdot x-2}{x^3+1}\)

mình cx ko bt còn rút gọn nữa hay ko đâu ak

Khi x<2 , ta có \(x-2< 0\) nên | x - 2 | = -(x - 2) = -x + 2

Vậy M = 3x - x + 2 = 2x + 2

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

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

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

b) Khi \(x=\dfrac{2}{3}\Rightarrow\) \(A=\dfrac{-\dfrac{2}{3}-3}{3.\left(\dfrac{2}{3}\right)^2}=\dfrac{-11}{4}\)

c) Để A < 0 thì

\(\dfrac{-x-3}{3x^2}< 0\)

=> -x -3 <0

<=> -x < 3

\(\Rightarrow x>3\)

Đặt \(\left\{{}\begin{matrix}\left(x-1\right)^3=a\\x^3=b\\\left(x+1\right)^3=c\end{matrix}\right.\)\(\Rightarrow3x\left(x^2+2\right)=a+b+c\)

Thì bài toán trở thành

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

Giờ thế ngược lại làm tiếp xẽ ra nhé.

\(\dfrac{4X\left(X^2-4X+4\right)}{\left(X-2\right)\left(X+2\right)}\)= \(\dfrac{4X\left(X-2\right)^2}{\left(X-2\right)\left(X+2\right)}\)= \(\dfrac{4X\left(X-2\right)}{X+2}\)

Ta có: A = \(\dfrac{27-12x}{x^2+9}\) = \(\dfrac{\left(4x^2+36\right)-\left(4x^2+12x+9\right)}{x^2+9}\)

= \(\dfrac{4\left(x^2+9\right)-\left(2x+3\right)^2}{x^2+9}\)

= \(4-\dfrac{\left(2x+3\right)^2}{x^2+9}\)

Vì \(\left(2x+3\right)^2\) \(\ge\) 0

\(x^2+9\) > 0

=> \(\dfrac{\left(2x+3\right)^2}{x^2+9}\) \(\ge\) 0

=> \(4-\dfrac{\left(2x+3\right)^2}{x^2+9}\) \(\le\) 4

Dấu bằng xảy ra <=> \(\left(2x+3\right)^2\) = 0

<=> 2x +3 = 0

<=> x = \(\dfrac{-3}{2}\)

Vậy GTLN của A = 4 khi x = \(\dfrac{-3}{2}\)

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