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\(...\Rightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)
\(\Rightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15\)
\(\Rightarrow45x+9=15\Rightarrow45x=6\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)
\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)
`(x-2)(2x+3)-3x=11`
`<=> 2x^2 - 4x + 3x - 6 - 3x -11 = 0`
`<=> 2x^2 - 4x - 17 = 0`
`<=> 2x^2 - 4x + 2 - 19 = 0`
`<=> 2 (x-1)^2 = 19`
`<=> (x-1)^2 = 19/2`
`<=> x - 1 = sqrt{19/2}` hoặc `x - 1 = -sqrt{19/2}`
`<=> x = sqrt{19/2} +1` hoặc `x = -sqrt{19/2} + 1`
`<=> x = (2 + sqrt{38})/2` hoặc `x = (2 - sqrt{38})/2`
[(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: \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}=\frac{x+4}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+4\right)\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}+\frac{2.3}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2+x-12+6}{2\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2-2x+3x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x\left(x-2\right)+3\left(x-2\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-2\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{x-2}{2\left(x-3\right)}\)
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