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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 \(A⋮B\)
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Bai 1
\(x^2+x-30=x^2+6x-5x-30=\left(x-5\right)\left(x+6\right)\)
Bai 2
a, \(\left(x-2\right)^2-x\left(x-5\right)=13\)
\(\Leftrightarrow x^2-4x+4-x^2+5x=13\)
\(\Leftrightarrow x+4=13\Leftrightarrow x=9\)
b, \(4x^3-100x=0\Leftrightarrow x\left(4x^2-100\right)=0\)
\(\Leftrightarrow x\left(2x-10\right)\left(2x+10\right)=0\Leftrightarrow x=0;\pm5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2,25x^2-12x-13\)
\(=25x^2-25x+13x-13\)
\(=25x\left(x-1\right)+13\left(x-1\right)\)
\(=\left(x-1\right)\left(25x+13\right)\)
\(3,2y^2-3y-5\)
\(=2y^2+2y-5y-5\)
\(=2y\left(y+1\right)-5\left(y+1\right)\)
\(=\left(y+1\right)\left(2y-5\right)\)
Còn bài 1 mik đang nghĩ, khi nào biết mik trả lời nha!!!
Chúc bn học giỏi!!!
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\(f\left(x\right)=x^4+8x^3+28x^2+48x-13\)
\(=\left(x^4+4x^3+7x^2\right)+\left(4x^3+16x^2+28x\right)+\left(5x^2+20x+35\right)-48\)
\(=x^2\left(x^2+4x+7\right)+4x\left(x^2+4x+7\right)+5\left(x^2+4x+7\right)-48\)
\(=\left(x^2+4x+7\right)\left(x^2+4x+5\right)-48\)
đặt t=\(x^2+4x+6\)khi đó g(t)=(t-1)(t+1)-48=t2-49=(t-7)(y+7)
vậy f(x)=(x2+4x-1)(x2+4x+13)
Trả lời:
Thay \(f\left(x\right)=0\), ta có:
\(0=x^4+8x^3+28x^2+48x-13\)
\(\Leftrightarrow-x^4-8x^3-28x^2-48x+13=0\)
\(\Leftrightarrow-x^4-4x^3-4x^3+x^2-16x^2-13x^2+4x-56x+13=0\)
\(\Leftrightarrow\left(-x^4-4x^3+x^2\right)+\left(-4x^3-16x^2+4x\right)+\left(-13x^2-56x+13\right)=0\)
\(\Leftrightarrow-x^2.\left(x^2+4x-1\right)-4x.\left(x^2+4x-1\right)-13.\left(x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(-x^2-4x-13\right).\left(x^2+4x-1\right)=0\)
Vì \(-x^2-4x-13=-x^2-4x-4-9\)
\(=-\left(x^2+4x+4\right)-9\)
\(=-\left(x+2\right)^2-9< 0\forall x\)
\(\Rightarrow x^2+4x-1=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)-5=0\)
\(\Leftrightarrow\left(x+2\right)^2=5=\left(\pm\sqrt{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=\sqrt{5}\\x+2=-\sqrt{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2+\sqrt{5}\\x=-2-\sqrt{5}\end{cases}}\)
Vậy đa thức có 2 nghiêm \(x\in\left\{-2+\sqrt{5},-2-\sqrt{5}\right\}\)
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\(\left(x+1\right)^2-3\left(x+13\right)=x^2+2x+1-3x-39\)
\(=x ^2-x-38\)
\(=x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{153}{4}\)
\(=\left(x-\frac{1}{2}\right)^2-\sqrt{\frac{153}{4}}^2\)
\(=\left(x-\frac{1}{2}-\sqrt{\frac{153}{4}}\right)\left(x-\frac{1}{2}+\sqrt{\frac{153}{4}}\right)\)
\(=\left(x-\frac{1+\sqrt{153}}{2}\right)\left(x-\frac{1-\sqrt{153}}{2}\right)\)
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a)\(\left(4x^3-xy^2+y^3\right)\left(x^2y+2xy^2-2y^3\right)\)
\(=x^2y\left(4x^3-xy^2+y^3\right)+2xy^2\left(4x^3-xy^2+y^3\right)\)
\(-2y^3\left(4x^3-xy^2+y^3\right)\)
\(=4x^5y-x^3y^3+x^2y^4+8x^4y^2-2x^2y^4+2xy^5\)
\(-8x^3y^3+2xy^5-2y^6\)
\(=-2y^6+4x^5y+\left(2xy^5+2xy^5\right)+8x^4y^2+\left(x^2y^4-2x^2y^4\right)\)
\(-\left(x^3y^3+8x^3y^3\right)\)
\(=-2y^6+4x^5y+4xy^5+8x^4y^2-x^2y^4-9x^3y^3\)
b)
(!) \(2\left(x+y\right)^2-7\left(x+y\right)+5\)
\(=2\left(x+y\right)^2-2\left(x+y\right)-5\left(x+y\right)+5\)
\(=2\left(x+y\right)\left(x+y-1\right)-5\left(x+y-1\right)\)
\(=\left(2x+2y-5\right)\left(x+y-1\right)\)
(!!) \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-x^2-y^2-z^2\)
\(=2\left(xy+yz+zx\right)\)