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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) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)
a. \(x^2-4x+4=x^2-2.x.2+2^2=\left(x-2\right)^2\)
b. \(x^2-4y^2=x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)
c. \(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\)
d. \(x^3-3x^2+3x-1\)
\(=x^3-1^3-3x^2+3x\)
\(=\left(x-1\right)\left(x^2-x+1\right)-3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x-1\right)\left(x^2-4x+1\right)\)
e. \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)
g. \(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
a) 4x2+4x+1
= (2x+1)2
b) x2-16x+64
= (x-8)2
c) 4x2-9y2
= (2x+3y)(2x-3y)
d) ( x-3).(x^2+3x+9)
1) \(A=x\left(x-6\right)+10=x^2-6x+10=x^2-6x+9+1=\left(x-3\right)^2+1\ge1>0\)
Dấu "=" xảy ra khi: \(x=3\)
\(B=x^2-2x+9y^2-6y+3\)
\(B=\left(x^2-2x+1\right)+\left(9y^2-6y+1\right)+1\)
\(B=\left(x-1\right)^2+\left(3y-1\right)^2+1\ge1>0\)
Dấu "=" xảy ra khi: \(x=y=1\)
2) \(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi: \(x=2\)
\(B=4x^2+4x+11=4x^2+4x+1+10=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi: \(x=-\dfrac{1}{2}\)
\(C\) mk nghĩ đề sai
\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(C=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)\)
\(C=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)\)
\(C=\left(x^2+5x+5\right)^2-1\)
\(C=\left(x^2+5x+\dfrac{25}{4}-\dfrac{5}{4}\right)^2-1\)
\(C=\left[\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}\right]^2-1\ge\dfrac{9}{16}\)
Dấu "=" xảy ra khi: \(x=-\dfrac{5}{2}\)
\(D=4x-x^2+1=-\left(x^2-4x-1\right)=-\left(x^2-4x+4-5\right)=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\le5\)
Dấu "=" xảy ra khi: \(x=2\)
\(E=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)=-\left(x+4\right)^2+21\le21\)
Dấu "=" xảy ra khi: \(x=-4\)
a. (a2 - b2)2 - (a2 + b2)2
= (a2 - b2 - a2 - b2)(a2 - b2 + a2 + b2)
= -2b2 . 2a2
b. a6 - b6
<=> (a3)2 - (b3)2
<=> (a3 - b3)(a3 + b3)
\(a,\left(a^2-b^2\right)^2-\left(a^2+b^2\right)^2\\ =a^4-2a^2b^2+b^4-a^4-2a^2b^2-b^4\\ =-4a^2b^2\)
\(b,a^6-b^6=a^2\left(a^3-b^3\right)=a^2\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(c,-4x^2+9y^2=\left(3y-2x\right)\left(3y+2x\right)\\ d,\left(x+1\right)^3-\left(2-x\right)^3\\ =\left(x+1-2+x\right)\left[\left(x+1\right)^2+\left(x+1\right)\left(2-x\right)+\left(2-x\right)^2\right]\\ =\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)\\ =\left(2x-1\right)\left(x^2-x+7\right)\)
\(e,8+\left(4x-3\right)^3\\ =\left(8+4x-3\right)\left[64-8\left(4x-3\right)+\left(4x-3\right)^2\right]\\ =\left(4x+5\right)\left(64-32x+24+16x^2-24x+9\right)\\ =\left(4x+5\right)\left(16x^2-56x+97\right)\)
\(g,81-\left(9-x^2\right)^2\\ =\left(9-9+x^2\right)\left(9+9-x^2\right)\\ =x^2\left(18-x^2\right)\left[=x^2\left(\sqrt{18}-x\right)\left(\sqrt{18}+x\right)\right]\)
Chỗ trong ngoặc nếu bạn chưa học căn thì ko cần ghi nha