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\(x^3-x^2-21x+45\)
\(=\left(x^3-3x^2\right)+\left(2x^2-6x\right)+\left(-15x+45\right)\)
\(=x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)\)
\(=\left(x^2+2x-15\right)\left(x-3\right)\)
\(=\left[\left(x^2-3x\right)+\left(5x-15\right)\right]\left(x-3\right)\)
\(=\left[x\left(x-3\right)+5\left(x-3\right)\right]\left(x-3\right)\)
\(=\left(x+5\right)\left(x-3\right)^2\)
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 \(A⋮B\)
câu a:
\(=x^2+6x-x+6\)
\(=\left(x^2-x\right)-\left(6x-6\right)\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x-6\right)\left(x-1\right)\)
câu b:
\(=x^2+5x-x-5\)
\(=x^2-x+5x-5\)
\(=x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x+5\right)\left(x-1\right)\)
a, x2 + 5x +6
= x2 - 6x-x +6
= x(x-6)-(x-6)
=( x-1)(x-6)
b, x2+4x-5
= x2+ 5x -x -5
= x(x+5)-(x+5)
=(x-1)(x+5)
Bài 1 :
a) \(A=x^2-6x+11\)
\(A=x^2-2\cdot x\cdot3+3^2+2\)
\(A=\left(x-3\right)^2+2\ge2\forall x\)
Dấu "=' xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
b) \(B=2x^2+10x-1\)
\(B=2\left(x^2+5x-\frac{1}{2}\right)\)
\(B=2\left[x^2+2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)
c) \(C=5x-x^2\)
\(C=-\left(x^2-5x\right)\)
\(C=-\left[x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right]\)
\(C=-\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\right]\)
\(C=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Bài 2 :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[x+\left(y+z\right)\right]^3-x^3-y^3-z^3\)
\(=x^3+3x^2\left(y+z\right)+3x\left(y+z\right)^2+\left(y+z\right)^3-x^3-y^3-z^3\)
\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+y^3+3y^2z+3yz^2+z^3-y^3-z^3\)
\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+3yz\left(y+z\right)\)
\(=3\left(y+z\right)\left[x^2+x\left(y+z\right)+yz\right]\)
\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
\(=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
\(=x^2+8x-x-8=x\left(x+8\right)-\left(x+8\right)=\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+3x+2-x^2+6x-9=11\)
=>x=2
chi tiết hơn đc khum ạ ?