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Em tham khảo: Câu hỏi của Edogawa G - Toán lớp 8 - Học toán với OnlineMath
Chứng minh rằng:
\(\left[\left(x+1\right)^{2n}+\left(x+2\right)^n-1\right]⋮\left(x^2+3x+2\right)\)
Ta có\(\left(x+1\right)^{2n}⋮\left(n+1\right)\)(1)
\(\left(x+2\right)^n-1=\left(x+1\right)\left[\left(x+2\right)^{n-1}+\left(n+2\right)^{n-2}+...+1\right]\)
\(\Rightarrow\left(x+2\right)^n-1⋮\left(x+1\right)\)(2)
Từ (1) và (2)\(\Rightarrow\left[\left(x+1\right)^{2n}+\left(x+2\right)^n-1\right]⋮\left(x+1\right)\) (*)
Lại có\(\left(x+1\right)^{2n}-1\)
\(=\left[\left(x+1\right)^n+1\right]\left[\left(x+1\right)^n-1\right]\)
\(=\left[\left(x+1\right)^n-1\right]\left(x+2\right)\left[\left(x+1\right)^{n-1}-\left(x+1\right)^{n-2}+........+1\right]\)
\(\Rightarrow\left(x+1\right)^{2n}-1⋮\left(x+2\right)\)
Mà \(\left(x+2\right)^n⋮\left(x+2\right)\)
\(\Rightarrow\left[\left(x+1\right)^{2n}+\left(x+2\right)^n-1\right]⋮\left(x+2\right)\)(**)
Ta lại có (x+1) và (x+2) nguyên tố cùng nhau (***)
Từ (*);(**) và(***) \(\Rightarrow\left[\left(x+1\right)^{2n}+\left(x+2\right)^n-1\right]⋮\left(x^2+3x+2\right)\)
\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)
\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)
\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)
\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)
\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)
\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)
a, = x^2+a+x^2a+a^2+a^2x^2+1/x^2-a-x^2a+a^2+a^2x^2+1
= (x^2+1).(a^2+a+1)/(x^2+1)(a^2-a+1) = a^2+a+1/a^2-a+1
=> phân thức trên ko phụ thuộc vào biến x
=> ĐPCM
Nếu đúng thì k mk nha
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a)
\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\left(đpcm\right)\)
b)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{x+5}\\ =\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}\\ =\dfrac{1}{x}\)
Ta có:
\(x\left(x+1\right)^4+x\left(x+1\right)^3+x\left(x+1\right)^2+\left(x+1\right)^2\)
\(=x\left(x+1\right)^4+x\left(x+1\right)^3+\left(x+1\right)^2.\left(x+1\right)\)
\(=x\left(x+1\right)^4+x\left(x+1\right)^3+\left(x+1\right)^3\)
\(=x\left(x+1\right)^4+\left(x+1\right)^3\left(x+1\right)\)
\(=x\left(x+1\right)^4+\left(x+1\right)^4=\left(x+1\right)^4\left(x+1\right)=\left(x+1\right)^5\)