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\(...\Rightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)
\(\Rightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15\)
\(\Rightarrow45x+9=15\Rightarrow45x=6\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)
[(x+2)(x+5)][(x+3)(x+4)] -24 = ( x\(^2\) + 7x + 10)( x\(^2\) + 7x + 12) -24
Đặt : x\(^2\) + 7x + 10 = a ta được:
a * (a+2) - 24 = a\(^2\) + 2a -24 = a\(^2\) + 2a +1 - 5\(^2\) = (a+1)\(^2\) - 5\(^2\)
= (a + 1 -5)( a + 1 +5)
= (a-4)(a+6)
thay a ta được:
(a-4)(a-6) = ( x\(^2\) + 7x + 10 - 4)( x\(^2\) + 7x + 10 - 6)
= (x\(^2\) + 7x + 6)(x\(^2\) + 7x +4)
= (x+1)(x+6)(x\(^2\) + 7x + 4)
NHA!
Nhớ ghi dấu ngoặc tránh giải sai.
\(a.\) \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}\)
Ta có:
\(2x+6=2\left(x+3\right)\)
\(x^2-9=\left(x-3\right)\left(x+3\right)\)
nên \(MTC:\) \(2\left(x-3\right)\left(x+3\right)\)
Do đó: \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}=\frac{x+4}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+4\right)\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}+\frac{2.3}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2+x-12+6}{2\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2-2x+3x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x\left(x-2\right)+3\left(x-2\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-2\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{x-2}{2\left(x-3\right)}\)
\(9\left(x-3\right)^2-4\left(x+1\right)^2\)
\(=9\left(x^2+9-6x\right)-4\left(x^2+1+2x\right)\)
\(=9x^2+81-54x-4x^2-4-8x\)
\(=\left(9x^2-4x^2\right)-\left(54x+8x\right)+\left(81-4\right)\)
\(=5x^2-62x+77\)
\(=5x^2-5x-77x+77\)
\(=5x\left(x-1\right)-77\left(x-1\right)\)
\(=\left(5x-77\right)\left(x-1\right)\)
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