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a) \(x^2-6x+9=\left(x-3\right)^2\)
b) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)
c) \(4x^2-\frac{1}{16}=\left(2x-\frac{1}{4}\right)\left(2x+\frac{1}{4}\right)\)
d) \(\left(a+b\right)^2-4=\left(a+b-2\right)\left(a+b+2\right)\)
e) \(\left(a^2+9\right)^2-36a^2=\left(a^2-6a+9\right)\left(a^2+6a+9\right)\)
\(=\left(a-3\right)^2\cdot\left(a+3\right)^2\)
a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)
b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)
c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)
a) x2 -6x +9 = (x-3)2
b) x2+4x +4= (x+2)2
c) 4x2+4x+1= (2x+1)2
d) 4x2+12xy+9y2 = (2x+3y)2
e) x2-8x+16 = (x-4)2
Đây chính là hằng đẳng thức nhé bn....
\(a,\left(x+3\right)^2\)
\(b,\left(x+\frac{1}{2}\right)^2\)
\(c,\left(xy^2+1\right)^2\)
a, \(x^2-6x+9=\left(x-3\right)^2\)
b, \(x^2-12x+36=\left(x-4\right)^2\)
c, \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)
d, \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
e, \(x^4-8x^2+16=\left(x^2-4\right)^2=\left[\left(x-2\right)\left(x+2\right)\right]^2\)
f, \(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)
g, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)
h, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2\)
\(=\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)
a) \(x^2+6x+9=\left(x+3\right)^2\)
b) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)
c) \(2xy^2+x^2y^4+1=\left(xy^2+1\right)^2\)
a, \(\left(x+3\right)^2\)
b,\(\left(x+\frac{1}{2}\right)^2\)Mik giải thích tí nè, cái này =\(x^2+2.x.\frac{1}{2}+\frac{1}{4}\)=\(x^2+x+\frac{1}{4}\)
c,thì mik chịu.
\(A=x^2-8x+13=\left(x^2-8x+16\right)-3\ge-3\)Vậy \(Min_A=-3\) khi \(x+4=0\Leftrightarrow x=-4\)
\(B=2x^2+10x+5=2\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{5}{4}=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}\ge\dfrac{-5}{4}\)Vậy \(Min_B=-\dfrac{5}{4}\) khi \(x+\dfrac{5}{2}=0\Rightarrow=\dfrac{-5}{2}\)
\(C=4x-x^2=4-\left(4-4x+x^2\right)=4-\left(2-x\right)^2\le4\)Vậy \(Max_C=4\) khi \(2-x=0\Rightarrow x=2\)
Bài 1:
a, \(A=x^2-8x+13\)
\(A=x^2-4x-4x+16-3\)
\(A=\left(x-4\right)^2-3\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-4\right)^2\ge0\Rightarrow\left(x-4\right)^2-3\ge-3\)
Hay \(A\ge-3\) với mọi giá trị của \(x\in R\).
Để \(A=-3\) thì \(\left(x-4\right)^2-3=-3\Rightarrow x=4\)
Vậy......
Câu b tương tự
c, \(4x-x^2\)
\(C=-\left(x^2-4x\right)=-\left(x^2-2x-2x+4-4\right)\)
\(=-\left[\left(x-2\right)^2-4\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2-4\ge-4\)
\(\Rightarrow-\left[\left(x-2\right)^2-4\right]\le4\)
Hay \(A\le4\) với mọi giá trị của \(x\in R\).
Để \(A=4\) thì \(-\left[\left(x-2\right)^2-4\right]=4\Rightarrow x=2\)
Vậy......
Chúc bạn học tốt!!!
a) \(x^2-81=\left(x-9\right)\left(x+9\right)\)
b) \(4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
c) \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
d) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
e) \(6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x-3\right)^2\)
f) \(x^2-4x^2+4y^2+4xy=\left(x^2+4xy+4y^2\right)-4x^2=\left(x+2y\right)^2-4x^2\\ =\left(x+2y+2x\right)\left(x+2y-2x\right)=\left(3x+2y\right)\left(2y-x\right)\)
g) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)=2a\left(a^2+3b^2\right)\)
h) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)
Đã trả lời: Câu hỏi của Naryu Wikashi - Toán lớp 8 | Học trực tuyến
a. x2 - 6x + 9
= x2 - 2x3 + 32
= (x - 3)2
b. x2 + x + \(\frac{1}{4}\)
= x2 + 2x\(\frac{1}{2}\)+ \(\left(\frac{1}{2}\right)^2\)
= (x + \(\frac{1}{2}\))2
c. 4x2 - \(\frac{1}{16}\)
= (2x)2 - \(\left(\frac{1}{4}\right)^2\)
= (2x +\(\frac{1}{4}\))(2x - \(\frac{1}{4}\))
d. (a + b)2 - 4
= (a + b)2 - 22
= (a + b + 2)(a + b - 2)
e. (a2 + 9)2 - 36a2
= (a2 + 9)2 - (6a)2
= (a2 + 9 + 6a)(a2 + 9 - 6a)