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Bài này làm như sau
Ta có \(x+y+z=6\Rightarrow\left(x+y+z\right)^2=36\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=36\)
\(\Rightarrow2xy+2yz+2zx=36-12=24\left(x^2+y^2+z^2=12\right)\)
\(\Rightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
hay \(x=y=z\Rightarrow x=y=z=\frac{6}{3}=2\)
Vậy \(A=3\)
\(\)
1a) A = \(x^2-4x+2023=\left(x-2\right)^2+2019\)
Ta luôn có: (x - 2)2 \(\ge\)0 \(\forall\)x
=> (x - 2)2 + 2019 \(\ge\)2019 \(\forall\)x
Hay A \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra khi : (x - 2)2 = 0 => x - 2 = 0 => x = 2
Nên Amin = 2019 khi x = 2
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
\(1,\Rightarrow2^b\left(2^{a-b}-1\right)=256=2^8\left(a>b\right)\)
Do \(2^b\) chẵn, \(2^{a-b}-1\) lẻ, \(2^8\) chẵn nên \(2^{a-b}-1=1\Leftrightarrow2^{a-b}=2\Leftrightarrow a-b=1\)
\(\Leftrightarrow2^b\cdot1=2^8\Leftrightarrow b=8\Leftrightarrow a=9\)
Vậy \(\left(a;b\right)=\left(8;9\right)\)
Sao vậy