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Bài 1:
\(A=a^4-2a^3+3a^2-4a+5\)
\(=(a^4-2a^3+a^2)+2a^2-4a+5\)
\(=(a^4-2a^3+a^2)+2(a^2-2a+1)+3\)
\(=(a^2-a)^2+2(a-1)^2+3\)
\(=a^2(a-1)^2+2(a-1)^2+3=(a-1)^2(a^2+2)+3\)
Vì \((a-1)^2\geq 0,\forall a\in\mathbb{R}; a^2+2>0, \forall a\)
\(\Rightarrow A=(a-1)^2(a^2+2)+3\geq 0+3=3\)
Vậy \(A_{\min}=3\Leftrightarrow (a-1)^2=0\Leftrightarrow a=1\)
Bài 2:
a)
\(M=3xyz+x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)\)
\(=3xyz+x^2y+x^2z+yx^2+yz^2+zx^2+zy^2\)
\(=(x^2y+xy^2+xyz)+(y^2z+yz^2+xyz)+(zx^2+z^2x+xyz)\)
\(=xy(x+y+z)+yz(y+z+x)+xz(z+x+y)\)
\(=(x+y+z)(xy+yz+xz)\)
b)
\(Q=(a+b+c)^3-a^3-b^3-c^3\)
\(=[(a+b)+c]^3-a^3-b^3-c^3\)
\(=(a+b)^3+c^3+3(a+b)^2c+3(a+b)c^2-a^3-b^3-c^3\)
\(=a^3+b^3+3ab^2+3a^2b+c^3+3(a+b)^2c+3(a+b)c^2-a^3-b^3-c^3\)
\(=3ab(a+b)+3(a+b)c(a+b+c)\)
\(=3(a+b)[ab+c(a+b+c)]=3(a+b)[a(b+c)+c(b+c)]\)
\(=3(a+b)(b+c)(a+c)\)
a) \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)
\(=\left[\left(x+a\right)\left(x+4a\right)\right]\cdot\left[\left(x+2a\right)\left(x+3a\right)\right]+a^4\)
\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)
\(=\left(x^2+5ax+5a^2-a^2\right)\left(x^2+5ax+5a^2+a^2\right)+a^4\)\
\(=\left(x^2+5ax+5a^2\right)^2-a^4+a^4\)
\(=\left(x^2+5ax+5a^2\right)^2\)
b) Đặt \(a=x^2+y^2+z^2\); \(b=xy+yz+xz\)
\(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)
\(=a\left(a+2b\right)+b^2\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)
\(=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)
a) \left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4(x+a)(x+2a)(x+3a)(x+4a)+a4
=\left[\left(x+a\right)\left(x+4a\right)\right]\cdot\left[\left(x+2a\right)\left(x+3a\right)\right]+a^4=[(x+a)(x+4a)]⋅[(x+2a)(x+3a)]+a4
=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4=(x2+5ax+4a2)(x2+5ax+6a2)+a4
=\left(x^2+5ax+5a^2-a^2\right)\left(x^2+5ax+5a^2+a^2\right)+a^4=(x2+5ax+5a2−a2)(x2+5ax+5a2+a2)+a4\
=\left(x^2+5ax+5a^2\right)^2-a^4+a^4=(x2+5ax+5a2)2−a4+a4
=\left(x^2+5ax+5a^2\right)^2=(x2+5ax+5a2)2
b) Đặt a=x^2+y^2+z^2a=x2+y2+z2; b=xy+yz+xzb=xy+yz+xz
\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2(x2+y2+z2)(x+y+z)2+(xy+yz+zx)2
=a\left(a+2b\right)+b^2=a(a+2b)+b2
=a^2+2ab+b^2=\left(a+b\right)^2=a2+2ab+b2=(a+b)2
=\left(x^2+y^2+z^2+xy+yz+zx\right)^2=(x2+y2+z2+xy+yz+zx)2
a: \(=\left(a+b+c\right)^2+2\left(a+b+c\right)\left(b-c\right)+\left(b-c\right)^2\)
\(=\left(a+b+c+b-c\right)^2=\left(a+2b\right)^2\)
b: \(=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^4-4\right)\)
\(=\left(x^4+4\right)\left(x^4-4\right)=x^8-16\)
d: \(=x^2+y^2+z^2+2\left(xy+yz+xz\right)+2x^2+2y^2+2z^2-2\left(xy+yz+xz\right)-3\left(x^2+y^2+z^2\right)\)
=0
\(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
\(\Leftrightarrow\)\(x^2+y^2+z^2+3-2x-2y-2z\ge0\)
\(\Leftrightarrow\)\(\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(z^2-2z+1\right)\ge0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)
Dáu "=" xảy ra \(\Leftrightarrow\) \(x=y=z=1\)
a,b,c,d > 0 ta có:
- a < b nên a.c < b.c
- c < d nên c.b < d.b
Áp dụng tính chất bắc cầu ta được: a.c < b.c < b.d hay a.c < b.d (đpcm)
a ) \(VT=\left(x+y+z\right)^2-\left(x-y-z\right)^2\)
\(=\left(x+y+z-x+y+z\right)\left(x+y+z+x-y-z\right)\)
\(=4x\left(y+z\right)=VP\)
b ) \(VT=\left(2a+b\right)^2-\left(a+b\right)^2-3a^2\)
\(=\left(2a+b-a-b\right)\left(2a+b+a+b\right)-3a^2\)
\(=a\left(3a+2b\right)-3a^2\)
\(=3a^2+2ab-3a^2=2ab=VP\)
a) \(\left(x+y+z\right)^2-\left(x-y-z\right)^2=4x\left(y+z\right)\)
\(\Rightarrow x^2+y^2+z^2+2xy+2xz+2yz-\left(x^2+y^2+z^2-2xy-2xz+2yz\right)=4x\left(y+z\right)\)\(\Rightarrow x^2+y^2+z^2+2xy+2xz+2yz-x^2-y^2-z^2+2xy+2xz-2yz=4x\left(y+z\right)\)\(\Leftrightarrow4xy+4xz=4x\left(y+z\right)\)
\(\Leftrightarrow4x\left(y+z\right)=4x\left(y+z\right)\).
b) \(\left(2a+b\right)^2-\left(a+b\right)^2-3a^2=2ab\)
\(\Rightarrow\left(2a\right)^2+2.2a.b+b^2-\left(a^2+2ab+b^2\right)-3a^2=2ab\)
\(\Rightarrow4a^2+4ab+b^2-a^2-2ab-b^2-3a^2=2ab\)
\(\Leftrightarrow2ab=2ab\)