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a: \(=x^2-4x+4+y^2+2y+1\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
b: \(=x^2+10x+25+x^2-2xy+y^2\)
\(=\left(x+5\right)^2+\left(x-y\right)^2\)
c: \(=a^2+2ab+b^2+4b^2+4b+1\)
\(=\left(a+b\right)^2+\left(2b+1\right)^2\)
d: \(=2\left(x^2+b^2\right)\)
Giải:
1) \(\left(x-6\right)\left(x^2+6x+36\right)-\left(x+4\right)^3=\left(x-2\right)^3+\left(x+5\right)\left(x^2-10x+25\right)-\left(2x^3+6x^2\right)\)
\(\Leftrightarrow x^3-216-\left(x^3+12x^2+48x+64\right)=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow x^3-216-x^3-12x^2-48x-64=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow-280-12x^2-48x=-12x^2+12x+117\)
\(\Leftrightarrow-280-48x-12x-117=0\)
\(\Leftrightarrow-397-60x=0\)
\(\Leftrightarrow-60x=397\)
\(\Leftrightarrow x=-\dfrac{397}{60}\)
Vậy ...
2) \(\left(2x+3\right)^3-\left(2x+5\right)\left(4x^2-10x+25\right)=\left(6x-1\right)^2-\left(x-2\right)\left(x^2+2x+4\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-\left(8x^3+125\right)=36x^2-12x+1-\left(x^3-8\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3-125=36x^2-12x+1-x^3+8+x^3\)
\(\Leftrightarrow54x-98=-12x+9\)
\(\Leftrightarrow54x+12x=9+98\)
\(\Leftrightarrow66x=107\)
\(\Leftrightarrow x=\dfrac{107}{66}\)
Vậy ...
b1:
câu a,f áp dụng a2-b2=(a-b)(a+b)
câu b,c áp dụng a3-b3=(a-b)(a2+ab+b2)
câu d: \(x^2+2xy+x+2y=x\left(x+2y\right)+\left(x+2y\right)=\left(x+1\right)\left(x+2y\right)\)
câu e: \(7x^2-7xy-5x+5y=7x\left(x-y\right)-5\left(x-y\right)=\left(7x-5\right)\left(x-y\right)\)
câu g xem lại đề
a)\(A=\left(x-5\right)^2\ge0\)
\(\Rightarrow Min=0\)dấu \(=\)xảy ra khi \(x=5\)
a) \(A=x^2-10x+25\)
\(A=\left(x^2-10x+25\right)+0\)
\(A=\left(x-5\right)^2+0\)
Mà \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow A\ge0\)
Dấu "=" xảy ra khi : \(x-5=0\Leftrightarrow x=5\)
Vậy ...
a) 2xy + 3zy + 6y + xz
= (2xy + 6y) + (xz + 3zy)
= 2y(x + 3y) + z(x + 3y)
= (x + 3y)(2y + z)
b) x2 - 10x + 25
= x2 - 2.x.5 + 52
= (x - 5)2
c) x2 + 6x + 9 - y2
= x2 + 2.x.3 + 32 - y2
= (x - 3)2 - y2
= (x - 3 - y)(x - 3 + y)
d) x3 - 4x2 - xy2 + 4x
= x(x2 - 4x + 4 - y2)
= x[(x - 2)2 - y2]
= x(x - 2 - y)(x - 2 + y)
\(\text{a) }A=x^2-10x+25\\ A=x^2-2\cdot x\cdot5+5^2\\ A=\left(x-5\right)^2\\ Do\text{ }\left(x-5\right)^2\ge0\forall x\\ \Leftrightarrow A\ge0\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-5\right)^2=0\\ \Leftrightarrow x-5=0\\ \Leftrightarrow x=5\\ \text{Vậy }A_{\left(Min\right)}=0\text{ }khi\text{ }x=5\)
\(\text{b) }B=x^2+y^2-x+6y+10\\ B=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\\ B=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \left(y+3\right)^2\ge0\forall y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;y\\ \text{Dấu "=" xảy ra khi: }\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\\\left(y+3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\\ \text{ Vậy }B_{\left(Min\right)}=\dfrac{3}{4}\text{ }khi\text{ }x=\dfrac{1}{2};y=-3\)
\(\text{c) }C=2x^2-6x+10\\ C=\left(2x^2-6x+\dfrac{9}{2}\right)+\dfrac{11}{2}\\ C=2\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{2}\\ C=2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{11}{2}\\ C=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\\ Do\text{ }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\ge\dfrac{11}{2}\\ \text{Dấu "=" xảy ra khi: }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }C_{\left(Min\right)}=\dfrac{11}{2}khi\text{ }x=\dfrac{3}{2}\)
\(\)
b)
\(B=x^2+y^2-x+6y+10=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\left(10-9-\dfrac{1}{4}\right)\)\(B=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Câu hỏi của Hồ Thu Giang - Toán lớp 8 - Học toán với OnlineMath
Em tham khảo tại đây nhé.
c) \(x^2+x-ax-a\)
\(=x\left(x+1\right)-a\left(x+1\right)\)
\(=\left(x+1\right)\left(x-a\right)\)
d) \(2xy-ax+x^2-2ay\)
\(=2y\left(x-a\right)+x\left(x-a\right)\)
\(=\left(x-a\right)\left(2y+x\right)\)
e) \(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
f) \(25-10x-4y^2+x^2\)
\(=\left(x^2-10x+25\right)-\left(2y\right)^2\)
\(=\left(x-5\right)^2-\left(2y\right)^2\)
\(=\left(x-5-2y\right)\left(x-5+2y\right)\)
g) \(x^3-6xy+9y^2-36\)
h) \(4x^2-9y^2+4x-6y\)
\(=\left(2x\right)^2-\left(3y\right)^2+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
k) \(-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(-x+y+5\right)\)
i) \(4x^2-25y^2-6x+15y\)
\(=\left(2x\right)^2-\left(5y\right)^2-3\left(2x-5y\right)\)
\(=\left(2x-5y\right)\left(2x+5y\right)-3\left(2x-5y\right)\)
\(=\left(2x-5y\right)\left(2x+5y-3\right)\)
a, \(x\left(y+z\right)^2+y\left(x+z\right)^2+z\left(x+y\right)^2+4xyz\)
\(=x\left(y+z\right)^2+x^2\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(xy+xz+z^2+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b, \(yz\left(y+z\right)+xz\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+xz^2-x^2z-x^2y-xy^2\)
\(=yz\left(y+z\right)-x\left(y+z\right)\left(y-z\right)-x^2\left(y+z\right)\)
\(=\left(y+z\right)\left(yz-xy+xz-x^2\right)\)
\(=\left(y+z\right)\left[y\left(z-x\right)+x\left(z-x\right)\right]\)
\(=\left(y+z\right)\left(y+x\right)\left(z-x\right)\)
\(x^2-25=\left(x-5\right)\left(x+5\right)\\ x^2+10x+25=\left(x+5\right)^2\\ x^2-6x+xy-6y=x\left(x-6\right)+y\left(x-6\right)=\left(x+y\right)\left(x-6\right)\\ x^2-2x-y^2+1=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)