(-3).8/8.6 rút gọn

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

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)

chịu

\(\dfrac{8-2x}{x^2+x-20}=-\dfrac{2\left(4-x\right)}{\left(4-x\right)\left(x+5\right)}=\dfrac{-2}{x+5}\)

Để biểu thức trên nhận giá trị dương khi 

\(x+5< 0\)do -2 < 0 

\(\Leftrightarrow x< -5\)

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)

\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)

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

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)

a: Sửa đề: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

ĐKXĐ: x∉{0;2;-2;3}

Ta có: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

\(=\left\lbrack\frac{-\left(x+2\right)}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right\rbrack:\frac{x\left(x-3\right)}{x^2\cdot\left(2-x\right)}\)

\(=\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\frac{x-3}{x\left(2-x\right)}\)

\(=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{-x\left(x-2\right)}{x-3}\)

\(=\frac{-4x^2-8x}{x+2}\cdot\frac{-x}{x-3}=\frac{-4x\left(x+2\right)}{x+2}\cdot\frac{-x}{x-3}=\frac{4x^2}{x-3}\)

b: Để A>0 thì \(\frac{4x^2}{x-3}>0\)

=>x-3>0

=>x>3

c: |x-7|=4

=>\(\left[\begin{array}{l}x-7=4\\ x-7=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=11\left(nhận\right)\\ x=3\left(loại\right)\end{array}\right.\)

Thay x=11 vào A, ta được:

\(A=\frac{4\cdot11^2}{11-3}=\frac{4\cdot121}{8}=\frac{121}{2}\)

a: Đặt \(A=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)

\(=\left(\frac{x^2-2x}{2\left(x^2+4\right)}+\frac{2x^2}{x^2\left(x-2\right)+4\left(x-2\right)}\right)\cdot\frac{x^2-x-2}{x^2}\)

\(=\left(\frac{x^2-2x}{2\left(x^2+4\right)}+\frac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\frac{x^2-x-2}{x^2}\)

\(=\frac{x\left(x-2\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\frac{\left(x-2\right)\left(x+1\right)}{x}=\frac{x\left(x^2-4\right)+4x^2}{2\left(x^2+4\right)}\cdot\frac{x+1}{x}\)

\(=\frac{x^3-4x+4x^2}{2\left(x^2+4\right)}\cdot\frac{x+1}{x}=\frac{x\left(x^2+4x-4\right)}{2\left(x^2+4\right)}\cdot\frac{x+1}{x}=\frac{\left(x^2+4x-4\right)\left(x+1\right)}{2\left(x^2+4\right)}\)

b: Thay x=1/2 vào A, ta được:

\(A=\frac{\left\lbrack\left(\frac12\right)^2+4\cdot\frac12-4\right\rbrack\left(\frac12+1\right)}{2\left\lbrack\left(\frac12\right)^2+4\right\rbrack}=\frac{\left(\frac14+2-4\right)\cdot\frac32}{2\left(\frac14+4\right)}=\frac{\frac32\cdot\frac{-7}{4}}{2\cdot\frac{17}{4}}=\frac{-21}{8}:\frac{17}{2}=-\frac{21}{8}\cdot\frac{2}{17}=\frac{-21}{68}\)

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

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

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

\(1-\left(\frac{a+x}{ax-x^2}+\frac{2a+3x}{x^2-a^2}\right):\frac{a^4-4x^4}{a^4x-a^2x^3}\)

\(=1-\left(\frac{a+x}{x\left(a-x\right)}+\frac{2a+3x}{\left(x-a\right)\left(x+a\right)}\right):\frac{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}{a^2x\left(a^2-x^2\right)}\)

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

\(=1-\frac{a^2+2\cdot ax+x^2-2ax-3x^2}{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}\cdot a^2=1-a^2\cdot\frac{a^2-2x^2}{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}\)

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