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a) = (xyz+xy) +(z+1) +(yz+zx)+(x+y)
= xy(z+1) +(z+1)+z(x+y)+(x+y)
= (z+1)(xy+1)+(x+y)(Z+1)
=(z+1)(xy+1+x+y)
a)=x2-5x-2x+10=x(x-5)-2(x-5)=(x-5)(x-2)
b)=4x2-4x+x-1=4x(x-1)+(x-1)=(x-1)(4x+1)
c)=x2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)
\(b,9x^2+90x+225-\left(x-y\right)^2\)
\(=\left(3x+15\right)^2-\left(x-y\right)^2\)
\(=\left(3x+15-x+y\right)\left(3x+15+x-y\right)\)
\(=\left(2x+y+15\right)\left(4x-y+15\right)\)
áp dụng BĐT (a - b)² ≥ 0 → a² + b² ≥ 2ab ta có:
x² + y² ≥ 2xy
x² + 1 ≥ 2x
y² + z² ≥ 2yz
y² + 1 ≥ 2y
z² + x² ≥ 2xz
z² + 1 ≥ 2z
Cộng theo vế → 3(x² + y² + z²) + 3 ≥ 2(x + y + z + xy + yz + zx) = 2.6 = 12
→ x² + y² + z² ≥ 9/3 = 3
→ đpcm (dấu = xảy ra khi x = y = z = 1)
Bài 1:
Ta chứng minh bất đẳng thức phụ sau:
\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng).
Áp dụng vào bài toán:
\(x^2+y^2+z^2\ge xy+yz+zx\)\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)(1)
Sử dụng BĐT Cauchy, ta được:
\(x^2+1\ge2x;\)\(y^2+1\ge2y;\)\(z^2+1\ge2z\)
Cộng theo vế: \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)(2)
Cộng (1) với (2) theo vế: \(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)\)
Thay \(x+y+z+xy+yz+zx=6\)
Suy ra: \(3\left(x^2+y^2+z^2\right)+3\ge12\Leftrightarrow x^2+y^2+z^2\ge3\)(đpcm).
Bài 2:
Ta có: \(a^4+b^4-a^3b-ab^3=a^3\left(a-b\right)+b^3\left(b-a\right)\)
\(=a^3\left(a-b\right)-b^3\left(a-b\right)=\left(a-b\right).\left(a^3-b^3\right)\)
\(=\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(luôn đúng)
Suy ra \(a^4+b^4\ge a^3b+ab^3\)(1)
Áp dụng BĐT Cauchy, ta có: \(a^4+b^4\ge2\sqrt{a^4b^4}=2a^2b^2\)(2)
Cộng (1) với (2) theo vế, ta được:
\(2\left(a^4+b^4\right)\ge ab^3+a^3b+2a^2b^2\)(đpcm).
1)2( x2+y2+z2)>=2(xy+yz+xz )
x2+1>=2x
y2+1>=2y
z2+1>=2z
=>3(x2+y2+z2)>= 2(x+y+z+xy+yz+xz)=12
=> x2+y2+z2>=3
2) ta co a^4+b^4 >=2a^b^2 voi moi a,b
lai co a^4 +b^4 - ab^3-a^3b
=a^3(a-b)-b^3(a-b)
=(a-b)(a^3-b^3)
=(a-b)^2(a^2+b^2+ab)>=0 voi moi a,b
=> 2(a^4+b^4)>= ab^3+a^3b+2a^2b^2 voi moi a,b
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)