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(x2-8)2+36=x4-16x2+100=x4+20x2+100-36x2=(x2+10)2-36x2=(x2+10-6x)(x2+10+6x)
a: \(x^2+12x+36=0\)
=>\(x^2+2\cdot x\cdot6+6^2=0\)
=>\(\left(x+6\right)^2=0\)
=>x+6=0
=>x=-6
b: \(4x^2-4x+1=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>x=1/2
c: \(x^3+6x^2+12x+8=0\)
=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)
=>\(\left(x+2\right)^3=0\)
=>x+2=0
=>x=-2
\(\left(x^2-8\right)^2+36\)
\(=x^4-16x^2+64+36\)
\(=x^4-16x^2+100\)
\(=x^4+20x^2+100-36x^2\)
\(=\left(x^2+10\right)^2-36x^2\)
\(=\left(x^2+10-6x\right)\left(x^2+10+6x\right)\)
( x2 - 8)2 + 36 = x4 -16x2 +64 + 36
= (x4 +20x2 +100) -36x2
=( x2+10)2 -(6x)2
=(x2 + 10 -6x)(x2 +10 +6x)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
a. \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)
b. \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)
a: \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)
b: \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)
\(x^2-36+2xy+y^2=\left(x^2+2xy+y^2\right)-36=\left(x+y\right)^2-6^2=\left(x+y+6\right)\left(x+y-6\right)\)
\(x^4-2x^3-12x^2+12x+36=x^4+x^2+36-2x^3+12x-12x^2-x^2\)
\(=\left(x^2-x-6\right)^2-x^2=\left(x^2-6\right)\left(x^2-2x-6\right)\)
\(\left(x-8\right)^2+36=x^2-16x+64+36=x^2-16x+100>0\)
xem lại đề
(x-8)2+ 36
(x-8)2 +62
(x-8) +6
x = 8+6=14