x^4 / 1-x + x^3 + x^2 + x +1 = a / a-x
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1: =(8+a^3)(8-a^3)=64-a^6
2: =x^3-6x^2+12x-8-x(x^2-1)+6x^2-18x
=x^3-6x-8-x^3+x
=-5x-8
3: =x^3+3x^2+3x+1-x^3+1-3x^2-3x
=2
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\(A=\sqrt{x}+1\) (đã thu gọn)
\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)
\(A=x-\sqrt{x}+1=\sqrt{x}\cdot\sqrt{x}-\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)+1\)
\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)
\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)
\(A=1-\sqrt{x}\) (đã thu gọn)
\(A=x-2\sqrt{x}-1=\sqrt{x}\left(\sqrt{x}-2\right)-1\)
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\(A=\dfrac{6x}{5x-20}-\dfrac{x}{x^2-8x+16}\)
\(ĐKXĐ:x\ne\pm4\)
\(\Leftrightarrow A=\dfrac{6x}{5\left(x-4\right)}-\dfrac{x}{\left(x-4\right)^2}\)
\(\Leftrightarrow A=\dfrac{6x^2-24x-5x}{5\left(x-4\right)^2}\)
\(\Leftrightarrow\dfrac{6x^2-29x}{5\left(x-4\right)^2}\)
\(\Leftrightarrow\dfrac{x\left(6x-29\right)}{5\left(x-4\right)^2}\)
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\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)
\(A=\left(x^3+3x^2+3x+1\right)-\left(x^2+6x+9\right)\left(x+1\right)-4x^2+8\)
\(A=\left(x^3+3x^2+3x+1\right)-\left(x^3+x^2+6x^2+6x+9x+9\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-x^3-x^2-6x^2-6x-9x-9+4x^2+8\)
\(A=-12x\)
Thay \(x=-\dfrac{1}{6}\) vào \(A\) ta có:
\(A=-12\times\left(-\dfrac{1}{6}\right)=2\)
Vậy \(A=2\) khi \(x=-\dfrac{1}{6}\)
\(B=\left(x-1\right)^3-+\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)
\(B=\left(x^3-3x^2+3x-1\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)+\left(3x^2-48\right)\)
\(B=x^3-3x^2+3x-1-x^3+2x^2-4x-2x^2+4x-8+3x^2-48\)
\(B=3x-57\)
Thay \(x=-2\) vào \(B\) ta có:
\(B=3\times\left(-2\right)-57=-6-57=-63\)
Vậy \(B=-63\) khi \(x=-2\)