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ko biết mk làm có đúng ko nhma có gì sai thì đừng trách mk nhé
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{63}{a^2+b^2+c^2}\)
\(6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{a}{ac}\right)+2021\ge\dfrac{54}{ab+bc+ac}+2021\ge\dfrac{54}{a^2+b^2+c^2}+2021\)
<=>\(\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{2021}{9}\)
\(p^2=\left(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\right)^2\)
áp dụng bđt \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
\(p^2\le3.\left(\dfrac{1}{3\left(2a^2+b^2\right)}+\dfrac{1}{3\left(2b^2+c^2\right)}+\dfrac{1}{3\left(2c^2+a^2\right)}\right)=\dfrac{1}{2a^2+b^2}+\dfrac{1}{2b^2+c^2}+\dfrac{1}{2c^2+a^2}\)
\(< =>p^2\le\dfrac{9}{2a^2+b^2+2b^2+c^2+2c^2+a^2}\)
<=> \(p^2\le3.\dfrac{1}{a^2+b^2+c^2}=\dfrac{2021}{3}< =>p\le\sqrt{\dfrac{2021}{3}}\)
dấu bằng xảy ra khi \(a=b=c=\sqrt{\dfrac{3}{2021}}\)
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2021\)
\(\Rightarrow2021\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\sqrt{2021.3}=\sqrt{6063}\)
Từ đó:
\(\sqrt{3\left(2a^2+b\right)}=\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}\ge\sqrt{\left(2a+b\right)^2}=2a+b\)
\(\Rightarrow\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\dfrac{1}{2a+b}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)
Tương tự: \(\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(\Rightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{\sqrt{6063}}{3}\)
\(P_{max}=\dfrac{\sqrt{6063}}{3}\) khi \(a=b=c=\dfrac{3}{\sqrt{6063}}\)
Đề bài mình sửa lại : A = a2021 - b2021 + c2021 - (a - b + c)2021
Ta có \(\sqrt{a}-\sqrt{b}+\sqrt{c}=\sqrt{a-b+c}\)
\(\Leftrightarrow a+b+c-2\sqrt{ab}-2\sqrt{bc}+2\sqrt{ca}=a-b+c\)
\(\Leftrightarrow b-\sqrt{ab}-\sqrt{bc}+\sqrt{ca}=0\)
\(\Leftrightarrow\sqrt{b}\left(\sqrt{b}-\sqrt{a}\right)-\sqrt{c}\left(\sqrt{b}-\sqrt{a}\right)=0\)
\(\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right).\left(\sqrt{b}-\sqrt{a}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=c\\b=a\end{matrix}\right.\)
Với b = c
A = a2021 - b2021 + c2021 - (a - b + c)2021
= a2021 - a2021
= 0
Tương tự với b = a ta được A = 0
Vậy A = 0
ơ đang chờ mấy bạn top bxh vô trả lời mà hỏng thấy đou
hộ mình với:(
Mình có nghĩ ra cách này mọi người xem giúp mình với
f(x) = \(ax^2+bx+c\)
Ta có f(0) = 2 => c = 2
Ta đặt Q(x) = \(ax^2+bx+c-2020\)
và G(x) = \(ax^2+bx+c+2021\)
f(x) - 2020 chia cho x - 1 hay Q(x) chia cho x - 1 được số dư
\(R_1\) = Q(1) = \(a.1^2+b.1+c-2020=a+b+c-2020\)
Mà Q(x) chia hết cho x-1 nên \(R_1\) = 0
hay \(a+b+c-2020=0\). Mà c = 2 => a + b = 2018 (1)
G(x) chia cho x + 1 số dư
\(R_2\) = G(-1) = \(a.\left(-1\right)^2+b.\left(-1\right)+c+2021=a-b+2+2021\)
Mà G(x) chia hết cho x + 1 nên \(R_2\)=0
hay \(a-b+2+2021=0\) => \(a-b=-2023\) (2)
Từ (1) và (2) suy ra: \(\left\{{}\begin{matrix}a+b=2018\\a-b=-2023\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}a=-\dfrac{5}{2}\\b=\dfrac{4041}{2}\end{matrix}\right.\)
\(a+b+c=\sqrt{6063}\Leftrightarrow\dfrac{a}{\sqrt{2021}}+\dfrac{b}{\sqrt{2021}}+\dfrac{c}{\sqrt{2021}}=\sqrt{3}\)
Đặt \(\left(\dfrac{a}{\sqrt{2021}};\dfrac{b}{\sqrt{2021}};\dfrac{c}{\sqrt{2021}}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{3}\)
\(P=\dfrac{2x}{\sqrt{2x^2+1}}+\dfrac{2y}{\sqrt{2y^2+1}}+\dfrac{2z}{\sqrt{2z^2+1}}\)
Ta có đánh giá:
\(\dfrac{x}{\sqrt{2x^2+1}}\le\dfrac{3\sqrt{15}x+2\sqrt{5}}{25}\)
Thật vậy, BĐT tương đương:
\(\left(\sqrt{3}x-1\right)^2\left(9x^2+10\sqrt{3}x+2\right)\ge0\) (luôn đúng)
Tương tự và cộng lại:
\(P\le\dfrac{6\sqrt{15}\left(x+y+z\right)+12\sqrt{5}}{25}=\dfrac{6\sqrt{5}}{5}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Rightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow ab^2+a^2b+ac^2+a^2c+bc^2+b^2c+2abc=0\)
\(\Leftrightarrow ab^2+a^2b+ac^2+bc^2+a^2c+abc+b^2c+abc=0\)
\(\Leftrightarrow\left(a+b\right)ab+c^2\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(c^2+ab+bc+ac\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
Vậy ta có các trường hợp: \(a=-b,c=0\)hoặc \(b=-c,a=0\)hoăc \(a=-c,b=0\).
Với từng trường hợp ta đều có đpcm.
a)
\(P=a\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}+\frac{a}{b}=a\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}\)
=\(a\sqrt{\frac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}=a\sqrt{\frac{\left[a\left(a+1\right)+1\right]^2}{\left[a\left(a+1\right)\right]^2}}+\frac{a}{a+1}\)
\(=a.\frac{a\left(a+1\right)+1}{a\left(a+1\right)}+\frac{a}{a+1}=a+\frac{1}{a+1}+\frac{a}{a+1}=a+1\)
Vay P=a+1
phan b,c ap dung phan a la ra
CM bài toán phụ: \(x+y+z=0\)
CM: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) với x,y,z dương
Ta có: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)
\(=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\cdot\frac{x+y+z}{xyz}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)
\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng vào ta được: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)
\(Q=2021-\frac{1}{2021}=...\)
Áp dụng BĐT Bunhiacopxki ta có:
\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)
\(=3\left(2a+2b+2c\right)=3.2\left(a+b+c\right)=6.2021=12126\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{12126}\)
Dấu ''='' xảy ra khi \(a=b=c=\dfrac{2021}{3}\)
\(P=\sqrt{a+b}+\sqrt{b+c}\sqrt{c+a}\)
Aps dụng Bunhia-cốpxki : \(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)
\(=6\left(a+b+c\right)\)
\(=6.2021=12126\Leftrightarrow P=\sqrt{12126}\)
Vậy \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\dfrac{2021}{3}\)
(Refer ;-;)
Ta có : \(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
\(T=\frac{a^{2021}+b^{2021}+c^{2021}}{\left(a+b+c\right)^{2021}}=\frac{b^{2021}+b^{2021}+b^{2021}}{\left(b+b+b\right)^{2021}}=\frac{3b^{2021}}{\left(3b\right)^{2021}}=\frac{3}{3^{2021}}=\frac{1}{3^{2020}}\)