\(x^5=x^4+x^3+x^2+x+2\)
Tìm x
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1) \(x^3+6x^2+11x+6\)
\(=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
2) \(A=n^3\left(n^2-7\right)^2-36n\)
\(A=n\left[n^2\left(n^2-7\right)^2-36\right]\)
\(A=n\left\{\left[n\left(n^2-7\right)\right]^2-6^2\right\}\)
\(A=n\left(n^3-7n-6\right)\left(n^3-7n+6\right)\)
\(A=n\left(n^3-7n-6\right)\left(n^3-n-6n+6\right)\)
\(A=n\left(n^3-7n-6\right)\left[n\left(n-1\right)\left(n+1\right)-6\left(n-1\right)\right]\)
\(A=n\left(n^3-7n-6\right)\left(n-1\right)\left(n^2+n-6\right)\)
\(A=n\left(n-1\right)\left(n^3-7n-6\right)\left(n^2+3n-2n-6\right)\)
\(A=n\left(n-1\right)\left(n^3-7n-6\right)\left[n\left(n+3\right)-2\left(n+3\right)\right]\)
\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n^3-7n-6\right)\)
\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n^3-n-6n-6\right)\)
\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left[n\left(n-1\right)\left(n+1\right)-6\left(n+1\right)\right]\)
\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left(n^2+n-6\right)\)
\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left(n^2+3n-2n-6\right)\)
\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left[n\left(n+3\right)-2\left(n+3\right)\right]\)
\(A=n\left(n-1\right)\left(n-2\right)\left(n+3\right)\left(n+1\right)\left(n+3\right)\left(n-2\right)\)
\(A=\left(n-1\right)n\left(n+1\right)\left(n-2\right)^2\left(n+3\right)^2\)
Rồi sao nữa còn nghĩ :))
a) \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)ĐKXĐ : \(x\ne1;-3\)
\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}=\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{2x^2+6x+4}{\left(x-1\right)\left(x+3\right)}=\frac{2x^2-7x+5}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow2x^2+6x+4=2x^2-7x+5\)
\(\Leftrightarrow2x^2+5x+4-2x^2+7x-5=0\)
\(\Leftrightarrow12x-1=0\)
\(\Leftrightarrow x=\frac{1}{12}\)( thỏa mãn ĐKXĐ )
b) c) tương tự
b= (15+13):2= 14
a=(15-13):2=1
theo bài tìm 2 số khi biết tổng và hiệu nhé bn
\(\left(2x-6\right)\left(x^2+2\right)=\left(2x-6\right)\left(8x-10\right)\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2\right)-\left(2x-6\right)\left(8x-10\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2-8x+10\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x^2-6x-2x-12\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x-6\right)\left(x-2\right)=0\)
\(\Rightarrow x\in\left\{3;6;2\right\}\)
\(\left(5x-1\right)^2=\left(3x+5\right)^2\)
\(\Leftrightarrow\left(5x-1\right)^2-\left(3x+5\right)^2=0\)
\(\Leftrightarrow\left(5x-1-3x-5\right)\left(5x-1+3x+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(8x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\8x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)
a= (7+3):2=5
b=(7-3):2=2
theo tính chất tổng và hiệu của 2 số nhé bn
\(x^5=x^4+x^3+x^2+x+1\)
\(\Leftrightarrow x^5-x^4-x^3-x^2-x-2=0\)
\(\Leftrightarrow x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2\)
\(\Leftrightarrow x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x^4+x^3+x^2+x+1\right)\left(x-2\right)\)
Xét: \(x^4+x^3+x^2+x+1\)
\(=x\left(x^3+1\right)+\left(x^3+1\right)+x^2=\left(x+1\right)\left(x^3+1\right)+x^2=\left(x+1\right)^2\left(x^2+x+1\right)+x^2>0\)
\(\Rightarrow x-2=0\Leftrightarrow x=2\)
\(x^5=x^4+x^3+x^2+x+1\)
\(\Leftrightarrow x^5-x^4-x^3-x^2-x-2=0\)
\(\Leftrightarrow x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2\)
\(\Leftrightarrow x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x^4+x^3+x^2+x+1\right)\left(x-2\right)\)
Xét : \(x^4+x^3+x^2+x+1\)
\(=x\left(x^3+1\right)+\left(x^3+1\right)+x^2=\left(x+1\right)\left(x^3+1\right)+x^2\)
\(=\left(x+1\right)^2\left(x^2+x+1\right)+x^2>0\)
\(\Rightarrow x-2=0\Leftrightarrow x=2\)
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