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a) \(4\left(a+b\right)ab=3\left(a-b\right)^2+\left(a+b\right)^2\Leftrightarrow4\left(a+b\right)ab=4a^2+4b^2-4ab\Leftrightarrow\left(a+b\right)ab=a^2+b^2-ab\) (đúng)
=> đẳng thức được cm
b) nếu nghĩ ra thì tớ giải cho
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT=\left(a^2+4\right)\left(b^2+9\right)\)
\(\ge\left(\sqrt{a^2b^2}+\sqrt{4\cdot9}\right)^2=\left(ab+36\right)^2=VP\)
Xảy ra khi \(\dfrac{a^2}{4}=\dfrac{b^2}{9}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow b=\dfrac{3a}{2}\)
Khi đó \(A=\dfrac{a^2-ab+b^2}{a^2+ab+b^2}=\dfrac{a^2-a\cdot\dfrac{3a}{2}+\left(\dfrac{3a}{2}\right)^2}{a^2+a\cdot\dfrac{3a}{2}+\left(\dfrac{3a}{2}\right)^2}=\dfrac{7}{19}\)
xin lỗi bn nhưng bn có thể giải bằng cách khác ko , mk chưa học BĐT Cauchy-Schwart
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
3
Ta có: \(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+2a\left(b+c\right)+\left(b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow\text{Đ}PCM\)
2b)
Ta có: \(x^2+y^2-4x-2y+5=0\Leftrightarrow x^2+y^2-4x-2y+4+1=0\Leftrightarrow\left(x-2\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}}\)
c) \(x^4-11x^2+4x-21=0\Leftrightarrow x^4-10x^2+25-x^2+4x-4=0\)
\(\Leftrightarrow\left(x^2-5\right)^2-\left(x-2\right)^2=0\Leftrightarrow\left(x^2-x-5+2\right)\left(x^2+x-5-2\right)=0\)
đến đây tự làm
<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0< =>\)
a=b=c => 32020 = 3.a2019 <=> 32019 = a2019 => a=b=c=3
A= 12017 + 02018 + (-1)2019 = 0
\(Q=\left(a-b\right)^2=\left(a+b\right)^2-4ab=15^2-4.26=121\)
\(R=\left(a^2+b^2\right)^2-2a^2b^2=\left[\left(a+b\right)^2-2ab\right]^2-2\left(ab\right)^2=\left[15^2-2.26\right]^2-2.\left(26\right)^2=28577\)