Tìm x biết: \(\left(x+3\right)^{2n+1}=\left(2x-3\right)^{2n+1}\)
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a: \(=2x^{2n+1-2n}-2\cdot x^{2n}\cdot3\cdot x^{2-2n}+3\cdot x^{2n-1+1-2n}-9\cdot x^{2n-1+2-2n}\)
\(=2x-6x^2+3-9x\)
\(=-6x^2-7x+3\)
b: \(=\left(5x\right)^3-\left(2y\right)^3=125x^3-8y^3\)
\(a=x^{2n};b=x^{2n-1}\Rightarrow\frac{a}{b}=x\)
\(\left(2.a+3b\right)\left(\frac{1}{b}-\frac{3x^2}{a}\right)=\left(2x-6x^2+3-9x\right)=-\left(6x^2+7x-3\right)\)
Hai dòng giống nhau chẳng hiểu%
\(a,\left(2x-3\right)n-2n\left(n+2\right)\)
\(=n\left(2x-3-2n-4\right)\)
\(=-7n\)
Vì \(-7⋮7\Rightarrow-7n⋮7\) => ĐPCM
\(b,n\left(2n-3\right)-2n\left(n+1\right)\)
\(=n\left(2n-3-2n-2\right)\)
\(=-5n⋮5\) (ĐPCM)
Rút gọn
\(a,\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+33x-10x-55-6x^2-14x-9x-21\)
\(=-76\)
\(b,\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)
\(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1\)
\(=9\)
\(c,3x^2\left(x^2+2\right)+4x\left(x^2-1\right)-\left(x^2+2x+3\right)\left(3x^2-2x+1\right)\)
\(=3x^4+6x^2+4x^3-4x-3x^4+2x^3-x^2-6x^3+4x^2-2x-9x^2+6x-3\)
= -3
\(\left(x+3\right)^{2n+1}=\left(2x-3\right)^{2n+1}\)
\(\Rightarrow\left(x+3\right)=\left(2x-3\right)\)
\(\Rightarrow x+3=2x-3\)
\(\Rightarrow x+3-2x+3=0\)
\(\Rightarrow-x+6=0\)
\(\Rightarrow x=6\)
Vậy x=6
\(\left(x+3\right)^{2n+1}=\left(2x-3\right)^{2n+1}\)
\(\Rightarrow x+3=2x-3\)
\(x+3-2x+3=0\)
\(\left(x-2x\right)+\left(3+3\right)=0\)
\(-x+6=0\)
\(-x=0-6\)
\(-x=-6\)
\(\Rightarrow x=6\)
Vậy \(x=6\).