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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, \(\left(x+3\right)^3-\left(x+2\right)\left(x-2\right)-6x^2-20\)
\(=x^3+9x^2+27x+27-\left(x^2-4\right)-6x^2-20\)
\(=x^3+9x^2+27x+27-x^2+4+6x^2+20\)
\(=x^3+14x^2+27x+51\)
b, \(\left(2x+3\right)\left(4x^2-6x+9\right)-\left(2x-3\right)\left(4x^2+6x+9\right)\)
\(=8x^3-12x^2+18x+12x^2-18x+18-\left(8x^3+12x^2+18x-12x^2-18x-18\right)\)
\(=8x^3+18-8x^3+18=36\)
c, \(\left(2x-1\right)\left(4x^2+2x+1\right)\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(=\left(8x^3+4x^2+2x-4x^2-2x-1\right)\left(8x^3-4x^2+2x+4x^2-2x+1\right)\)
\(=\left(8x^3-1\right)\left(8x^3+1\right)=\left(8x^3\right)^2-1\)
\(=64x^5-1\)
d, \(\left(x+4\right)\left(x^2-4x+16\right)-\left(50+x^2\right)\)
\(=x^3-4x^2+16x+4x^2-16x+64-50-x^2\)
\(=x^3-x^2+14\)
Chúc bạn học tốt!!!
Bài 1:
a: \(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2+12x+9\right)-5+20x\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)
\(=4x^2-8x-16-5+20x-4x^2-12x-9\)
\(=-30\)
b: \(B=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-\left(x^3+4x^2-246x\right)-175\)
\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)
\(=-175\)
d: \(D=25x^2-20x+4-36x^2-12x-1+11\left(x^2-4\right)-48+32x\)
\(=-11x^2-32x+3-48+32x+11x^2-44\)
=-89
toàn hằng đẳng thức (1) và (2) thôi mà bạn, đọc SGK 8 tập 1 là hiểu ngay. Có gì khó hiểu hỏi nhé!
a, x2-6x +9 = (x-3)2
b, 4x2+4x +1 = (2x)2+2.2x.1 +12=(2x+1)2
c, 9x2 -12x +4 = (3x-2)2
d, 25x2 -10x +1= (5x -1)2
e, x4-4x2+4 = (x2 -2)2
f, x2 +8x +16 = (x+4)2
a)x3-7x+6
=x3+0x2-7x+6
=x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2-2x+3x-6)
=(x-1)[x(x-2)+3(x-2)]
=(x-1)(x+3)(x-2)
Lời giải:
a) ĐKXĐ: $x\neq \pm 1$
\(\frac{x^4-4x^2+3}{x^4+6x^2-7}=\frac{x^2(x^2-1)-3(x^2-1)}{x^2(x^2-1)+7(x^2-1)}=\frac{(x^2-3)(x^2-1)}{(x^2-1)(x^2+7)}=\frac{x^2-3}{x^2+7}\)
b) ĐKXĐ: Với mọi $x\in\mathbb{R}$
\(\frac{x^4+x^3-x-1}{x^4+x^4+2x^2+x+1}=\frac{(x^4-x)+(x^3-1)}{(x^4+x^3+x^2)+(x^2+x+1)}=\frac{x(x^3-1)+(x^3-1)}{x^2(x^2+x+1)+(x^2+x+1)}\)
\(=\frac{(x^3-1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{(x-1)(x^2+x+1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{x^2-1}{x^2+1}\)
c) ĐK: $x\neq 1;-2$
\(\frac{x^3+3x^2-4}{x^3-3x+2}=\frac{x^2(x-1)+4(x^2-1)}{x^2(x-1)+x(x-1)-2(x-1)}=\frac{(x-1)(x^2+4x+4)}{(x-1)(x^2+x-2)}\)
\(=\frac{(x-1)(x+2)^2}{(x-1)(x-1)(x+2)}=\frac{x+2}{x-1}\)
d) ĐK: $x^2+3x-1\neq 0$
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{(x^2+3x)^2-1}{(x^2+3x)^2-2x^2-6x+1}\)
\(=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x)^2-2(x^2+3x)+1}=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x-1)^2}=\frac{x^2+3x+1}{x^2+3x-1}\)
\(1,x^2-2x+1=x^2-2\cdot x\cdot1+1^2=\left(x-1\right)^2\)
\(2,x^2+2x+1=x^2+2\cdot x\cdot1+1^2=\left(x+1\right)^2\)
\(3,x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)
\(4,x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)
\(5,4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)
\(6,4x^2+4x+1=\left(2x\right)^2+2\cdot2x\cdot1+1^2=\left(2x+1\right)^2\)
\(7,x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
\(8,x^2-x+\dfrac{1}{4}=x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)