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ta muốn làm nắm nhưng mi bợi thêm năm nữa đi nha đợi ta lên lớp 8 ta giải cho
a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2
1)
a) \(-7x\left(3x-2\right)\)
\(=-21x^2+14x\)
b) \(87^2+26.87+13^2\)
\(=87^2+2.87.13+13^2\)
\(=\left(87+13\right)^2\)
\(=100^2\)
\(=10000\)
2)
a) \(x^2-25\)
\(=x^2-5^2\)
\(=\left(x-5\right)\left(x+5\right)\)
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..........
3)
a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)
Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)
b)
Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)
Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)
\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a) \(x^2+2x-8\\ =\left(x^2+2x+1\right)-9\\ =\left(x+1\right)^2-3^2\\ =\left(x+1-3\right).\left(x+1+3\right)\\ =\left(x-2\right).\left(x+4\right)\)
b) \(12x^2-13x+3\\ =12x^2-4x-9x+3\\ =4x\left(3x-1\right)-3\left(3x-1\right)\\ =\left(3x-1\right).\left(4x-3\right)\)
a) \(x^2+2x-8\)
\(=x^2+2x+1-9\)
\(=\left(x+1\right)^2-9\)
\(=\left(x+1+3\right)\left(x+1-3\right)\)
\(=\left(x+4\right)\left(x-2\right)\)
b) \(12x^2-13x+3\)
\(=12x^2-4x-9x+3\)
\(=4x\left(3x-1\right)-3\left(3x-1\right)\)
\(=\left(3x-1\right)\left(4x-3\right)\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)\)
Đặt x2 + 7x + 11 = a, ta được
\(=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
d) \(x^5+x^4+1\)
\(=\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)-\left(x^3+x^2+x\right)\)
\(=x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)