Cho a3 + b3 - 3ab = -18. Chứng minh rằng -9 < a + b < -1
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Helo Dung :)
\(P=\left(\frac{x-4}{x^3-1}+\frac{1}{x-1}\right)\div\left(1-\frac{x-8}{x^2+x+1}\right)\)
ĐKXĐ : \(x\ne1\)
a) \(=\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\div\left(\frac{x^2+x+1}{x^2+x+1}-\frac{x-8}{x^2+x+1}\right)\)
\(=\left(\frac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\div\left(\frac{x^2+x+1-x+8}{x^2+x+1}\right)\)
\(=\frac{x^2+2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\div\frac{x^2+9}{x^2+x+1}\)
\(=\frac{x^2-x+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}\times\frac{x^2+x+1}{x^2+9}\)
\(=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x^2+9\right)}=\frac{x+3}{x^2+9}\)
b) \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Với x = 2 =>\(P=\frac{2+3}{2^2+9}=\frac{5}{13}\)
c) Đến đây rồi sao ? Bậc của tử thấp hơn bậc của mẫu thì chia hết kiểu gì ? :D
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( 3x - 1 )2 - ( x + 5 )2 = 0
⇔ [ ( 3x - 1 ) - ( x + 5 ) ][ ( 3x - 1 ) + ( x + 5 ) ] = 0
⇔ ( 3x - 1 - x - 5 )( 3x - 1 + x + 5 ) = 0
⇔ ( 2x - 6 )( 4x + 4 ) = 0
⇔ 2x - 6 = 0 hoặc 4x + 4 = 0
⇔ x = 3 hoặc x = -1
<=> ( 3x - 1 - x - 5 )( 3x - 1 + x + 5 ) = 0
<=> ( 2x - 6 )( 4x + 4) = 0
=>
+) 2x - 6 = 0 => x = 3
+) 4x +4 = 0 => x = -1
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\(\left(x-2\right)\left(x+2\right)\left(x+2\right)^2=\left(x+2\right)\left[\left(x-2\right)\left(x+2\right)\right]\)
\(=\left(x+2\right)\left(x^2-4\right)=x^3-4x+2x^2-8\)
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Ta có : x2 + 4x - y2 + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y )( x + 2 + y )