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a)\(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+2xy+y^2=a^2\Rightarrow x^2+y^2=a^2-2xy\Rightarrow x^2+y^2=a^2-2b\)
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
2. Đặt c + d = x
Ta có: \(a+b+c+d=0\Rightarrow a+b+x=0\Rightarrow a^3+b^3+c^3+d^3=3abx\)
\(\Rightarrow a^3+b^3+c^3+d^3+3cd\left(c+d\right)=3ab\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(ab-cd\right)\left(c+d\right)\)
Câu 4:
\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}+a^{1008}\)
\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}-2a^{1008}b^{1008}-2b^{1008}c^{1008}-2c^{1008}a^{1008}=0\)
\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2=0\)
\(\Rightarrow a^{1008}=b^{1008},b^{1008}=c^{1008},c^{1008}=a^{1008}\)
\(\Rightarrow a=b,b=c,c=a\) (vì a,b,c > 0 nên \(a\ne-b,b\ne-c,c\ne-a\) )
\(\Rightarrow a-b=0,b-c=0,a-c=0\)
Thay vào A ta tính được A = 0
\(A=\left(x+y\right)^2-2xy=25-12=13\)
\(B=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=5\left(25-18\right)=35\)
\(C=x^2-y^2\Rightarrow C^2=x^4+y^4-2x^2y^2=\left(x^2+y^2\right)^2-4x^2y^2\)
\(C^2=\left[\left(x+y\right)^2-2xy\right]^2-4\left(xy\right)^2=\left(25-12\right)^2-4.36=25\Rightarrow C=\pm5\)
\(D=\frac{x^2+y^2}{xy}=\frac{\left(x+y\right)^2-2xy}{xy}=\frac{25-12}{6}=\frac{13}{6}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
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