3.CB3 của (x^2+3) - CB2 của (8+2x-x^2) = x
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a: \(=\dfrac{x+2-x-1}{\left(x+2\right)\left(x+1\right)}=\dfrac{1}{\left(x+2\right)\left(x+1\right)}\)
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\(a,2\left(x^3-1\right)-2x^2\left(x+2x^4\right)+x\left(4x^5+4\right)=6\\ \Leftrightarrow2x^3-2-2x^3-4x^6+4x^6+4x-6=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow x=2\\ b,\left(2x\right)^2\left(4x-2\right)-\left(x^3-8x^3\right)=15\\ \Leftrightarrow4x^2\left(4x-2\right)+7x^3-15=0\\ \Leftrightarrow16x^3-8x^2+7x^3-15=0\\ \Leftrightarrow23x^3-8x^2-15=0\\ \Leftrightarrow23x^3-23x^2+15x^2-15x+15x-15=0\\ \Leftrightarrow\left(x-1\right)\left(23x^2+15x-15\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\left(23x^2+15x-15>0\right)\end{matrix}\right.\)
Bài 1:
a: Ta có: \(2\left(x^3-1\right)-2x^2\left(2x^4+x\right)+x\left(4x^5+4\right)=6\)
\(\Leftrightarrow2x^3-2-4x^6-2x^3+4x^6+4x=6\)
\(\Leftrightarrow4x=8\)
hay x=2
b: Ta có: \(\left(2x\right)^2\cdot\left(4x-2\right)-\left(x^3-8x^3\right)=15\)
\(\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^3=15\)
\(\Leftrightarrow16x^3-8x^2+7x^3=15\)
\(\Leftrightarrow23x^3-8x^2-15=0\)
\(\Leftrightarrow23x^3-23x^2+15x^2-15=0\)
\(\Leftrightarrow23x^2\left(x-1\right)+15\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(23X^2+15x+15\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
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1) \(\dfrac{3x}{4x-8}\)
\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)
2) \(\dfrac{2x}{x^2-9}\)
\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)
(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)
\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)
5) \(\dfrac{x-2}{x^2+3}\)
Do \(x^2+3>0\forall x\in R\)
Vậy biểu thức trên xác định với mọi x
6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)
\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
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\(\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(=\left(2x\right)^3+1\)
\(=8x^3+1\)
⇒ Chọn C
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a: ĐKXĐ: x<>2; x<>0
b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
c: M>=-3
=>(x+1+6x)/2x>=0
=>(7x+1)/x>=0
=>x>0 hoặc x<=-1/7
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Tổng 2 đa thức:
\(\begin{array}{l}A(x) + B(x) = - 8{x^5} + 6{x^4} + 2{x^2} - 5x + 1 + 8{x^5} + 8{x^3} + 2x - 3\\ = ( - 8 + 8){x^5} + 6{x^4} + 8{x^3} + 2{x^2} + ( - 5 + 2)x + (1 - 3)\\ = 6{x^4} + 8{x^3} + 2{x^2} - 3x - 2\end{array}\)
Vậy bậc của hai đa thức là tổng là: 4.
Hiệu 2 đa thức:
\(\begin{array}{l}A(x) - B(x) = - 8{x^5} + 6{x^4} + 2{x^2} - 5x + 1 - (8{x^5} + 8{x^3} + 2x - 3)\\ = - 8{x^5} + 6{x^4} + 2{x^2} - 5x + 1 - 8{x^5} - 8{x^3} - 2x + 3\\ = ( - 8 - 8){x^5} + 6{x^4} - 8{x^3} + 2{x^2} + ( - 5 - 2)x + (1 + 3)\\ = - 16{x^5} + 6{x^4} - 8{x^3} + 2{x^2} - 7x + 4\end{array}\)
Vậy bậc của hai đa thức là hiệu là: 5