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a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )
\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )
\(\dfrac{2x^2-x}{x-1}=\dfrac{x-2x^2}{1-x}\)
b)Ta có : \(\dfrac{1}{x-1}=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x+1}{x^3-1}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2-2x}{x^3-1}\)
\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )
c) MTC = ( x+ 2)2(x - 2)2
Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}\)
\(\dfrac{x}{x^2-4}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x^2-2^2\right)}{\left(x+2\right)^2\left(x-2\right)^2}=\dfrac{x^3-4x}{\left(x+2\right)^2\left(x-2\right)^2}\)
d) MTC = xyz( x - y)( y - z)( x - z)
Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Cộng các phân thức lại ta có :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Theo bài ra , ta có :
\(\dfrac{x+5}{25}+1+\dfrac{x+6}{24}+1+\dfrac{x+7}{23}=0\)
\(\Leftrightarrow\dfrac{x+5+25}{25}+\dfrac{x+6+24}{24}+\dfrac{x+7+23}{23}=0\)
\(\Leftrightarrow\dfrac{x+30}{25}+\dfrac{x+30}{24}+\dfrac{x+30}{23}=0\)
\(\Leftrightarrow\left(x+30\right)\left(\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}\right)=0\)
Vì \(\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}\ne0\)
\(\Leftrightarrow x+30=0\)
\(\Leftrightarrow x=-30\)
Vậy S={-30}
Chúc bạn học tốt =))
Ta có :
\(\dfrac{x+5}{25}+\dfrac{x+6}{24}+\dfrac{x+7}{23}=-3\)
=> \(\left(x+5\right).\dfrac{1}{25}+\left(x+5+1\right).\dfrac{1}{24}+\left(x+5+2\right).\dfrac{1}{23}=-3\)
=>\(\left(x+5\right).\dfrac{1}{25}+\left(x+5\right).\dfrac{1}{24}+\dfrac{1}{24}+\left(x+5\right).\dfrac{1}{23}+2.\dfrac{1}{23}\)= -3
=> (x + 5).\(\left(\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}\right)\) + \(\dfrac{1}{24}+\dfrac{2}{23}\) = -3
=> (x + 5). \(\dfrac{1727}{13800}\) + \(\dfrac{71}{552}\) = -3
=> (x + 5). \(\dfrac{1727}{13800}\) = -3 - \(\dfrac{71}{552}\)
=> (x + 5). \(\dfrac{1727}{13800}\) = \(\dfrac{-1727}{552}\)
=> x + 5 = -25
=> x = -25-5
=> x = -30
Vậy x = -30
quá tồi \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
Ta có :
\(VT=\left(\dfrac{1}{2}xy-\dfrac{1}{3}y\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy^2+\dfrac{1}{9}y^2\right)\)
\(=\dfrac{1}{8}x^3y^3+\dfrac{1}{12}x^2y^3+\dfrac{1}{18}xy^3-\dfrac{1}{12}x^2y^3-\dfrac{1}{18}xy^3-\dfrac{1}{27}y^3\)
\(=\dfrac{1}{8}x^3y^3-\dfrac{1}{27}y^3=VT\)
\(\Rightarrow dpcm\)
Vậy : ..............