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10 tháng 8 2019
\(\text{a) }x^2-9x+20\)
\(=x^2-4x-5x+20\)
\(=\left(x^2-4x\right)-\left(5x-20\right)\)
\(=x\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-4\right)\left(x-5\right)\)
\(\text{b) }x^2+9x+20\)
\(=x^2+4x+5x+20\)
\(=\left(x^2+4x\right)+\left(5x+20\right)\)
\(=x\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(x+5\right)\)
\(\text{c) }x^2+x-20\)
\(=x^2+5x-4x-20\)
\(=\left(x^2+5x\right)-\left(4x+20\right)\)
\(=x\left(x+5\right)-4\left(x+5\right)\)
\(=\left(x+5\right)\left(x-4\right)\)
\(\text{d) }x^2-x-20\)
\(=x^2+4x-5x-20\)
\(=\left(x^2+4x\right)-\left(5x+20\right)\)
\(=x\left(x+4\right)-5\left(x+4\right)\)
\(=\left(x+4\right)\left(x-5\right)\)
23 tháng 2 2017
\(\frac{6x-5}{10x+1}=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{49.51}\)
\(\Rightarrow\frac{6x-5}{10x+1}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right)\)
\(\Rightarrow\frac{6x-5}{10x+1}=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(\Rightarrow\frac{6x-5}{10x+1}=\frac{1}{2}\left(1-\frac{1}{51}\right)\)
\(\Rightarrow\frac{6x-5}{10x+1}=\frac{1}{2}.\frac{50}{51}\)
\(\Rightarrow\frac{6x-5}{10x+1}=\frac{25}{51}\)
\(\Rightarrow\left(6x-5\right).51=\left(10x+1\right).25\)
\(\Rightarrow306x-255=250x+25\)
\(\Rightarrow56x=280\)
\(\Rightarrow x=5\)
Vậy x = 5
NM
1
10 tháng 1 2018
Ta có:
\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)
Xét \(M=x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1\)
\(\Rightarrow x^4M=x^{28}+x^{24}+x^{20}+x^{16}+...+x^8+x^4\)
\(\Rightarrow x^4M-M=\left(x^{28}+x^{24}+x^{20}+...+x^8+x^4\right)-\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)\)
\(\Rightarrow\left(x^4-1\right)M=x^{28}-1\)
\(\Rightarrow M=\dfrac{x^{28}-1}{x^4-1}\)
Xét \(N=x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1\)
\(\Rightarrow x^2N=x^{28}+x^{26}+x^{24}+x^{20}+...+x^4+x^2\)
\(\Rightarrow x^2N-N=\left(x^{28}+x^{26}+x^{24}+...+x^4+x^2\right)-\left(x^{26}+x^{24}+x^{22}+...+x^2+1_{ }\right)\)
\(\Rightarrow\left(x^2-1\right)N=x^{28}-1\)
\(\Rightarrow N=\dfrac{x^{28}-1}{x^2-1}\)
Ta có:
\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)
\(=\dfrac{M}{N}=\dfrac{\dfrac{x^{28}-1}{x^4-1}}{\dfrac{x^{28}-1}{x^2-1}}\)
\(=\dfrac{x^{28}-1}{x^4-1}.\dfrac{x^2-1}{x^{28}-1}=\dfrac{x^2-1}{x^4-1}\)
\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{1}{x^2+1}\)
Chúc bạn học tốt!
7 tháng 2 2017
Ta có:
\(\left(x-1\right)\left(x+1\right)=8\\ < =>x^2-1=8\\ < =>x^2=9.\)
Ta được:
\(-12.x^2=-12.9=-108\)
Vậy: đáp án là -108
\(x^{20}+x+1=\left(x^{20}-x^2\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^{18}-1\right)+x^2+x+1\)
\(=x^2\left(x^6-1\right)\left(x^{12}+x^6+1\right)+x^2+x+1\)
\(=\left(x^{14}+x^8+x^2\right)\left(x^6-1\right)+x^2+x+1\)
\(=\left(x^{14}+x^8+x^2\right)\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^{18}-x^{17}+x^{15}-x^{14}+x^{12}-x^{11}+x^9-x^8+x^6-x^5+x^3-x^2+1\right)\left(x^2+x+1\right)\)