Bài 3: Rút gọn phân thức

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 = \(\dfrac{-3}{2}\)

câu này trên google

bạn nên tra google trước khi đăng

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

= \(\dfrac{2+x}{2-x}\)- \(\dfrac{4x^2}{^{^{ }}4-x^2}\)- \(\dfrac{2-x}{2+x}\): \(\dfrac{x^2-3x}{2x^2-x^3}\)

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

= \(\dfrac{4+4x+x^{2^{ }}-4x^{2^{ }}-4+4x-x^2}{^{ }4-x^2}\). \(\dfrac{2x^{2^{ }}-x^3}{x^2-3x}\)

= \(\dfrac{8x-4x^2}{4-x^2}\). \(\dfrac{2x^2-x^3}{x^2-3x}\)

= \(\dfrac{8-4x^2}{2+x}\). \(\dfrac{x^2}{x-3}\)

mình viết nhầm hàng thứ 2 phải là -(2-x)(2-x)

= - - :

= .

= .

= .

= .

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.

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

\(\Rightarrow x>3\)

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

Vậy M = 3x - x + 2 = 2x + 2

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

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

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

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

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

a , ĐKXĐ : bạn tự làm nhé

Đặt \(\sqrt{a}=x\) , khi đó biểu thức trỏ thành:

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

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

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

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

= \(\dfrac{4x^2+4}{x}\)= \(\dfrac{4a+4}{\sqrt{a}}\)

Mấy câu sau dễ rồi

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

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

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

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