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\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
Ta có: \(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)=\(\dfrac{a}{x-1}\)+\(\dfrac{bx+c}{x^2+1}\)
<=>\(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)=\(\dfrac{a\left(x^2+1\right)+\left(bx+c\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)
=>a(x2+1)+(bx+c)(x-1)=x2+2x-1
<=>ax2+a+bx2-bx+cx-c=x2+2x-1
<=>(a+b)x2+(c-b)x-(c-a)=x2+2x-1
=>\(\left\{{}\begin{matrix}a+b=1\\c-b=2\\c-a=1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}a+b=1\\b=c-2\\a=c-1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}c-1+c-2=1\\b=c-2\\a=c-1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}c=2\\b=0\\a=1\end{matrix}\right.\)
Vậy a=1,b=0,c=2
\(8x^3-1=\left(2x\right)^3-1^3=\left(2x-1\right)\left[\left(2x\right)^2+1.2x+1^2\right]\)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
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