1. a) \(CMR:A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{a}{b}-\dfrac{b}{a}\) ≥ 0
b) Tìm min \(B=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}+\dfrac{a}{b}+\dfrac{b}{a}\)
2. Cho x,y t/m: \(x^2+y^2=25\)
Tìm max M\(=3x+4y\)
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Khách
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Những câu hỏi liên quan
6 tháng 5 2022
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
9 tháng 12 2021
\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
9 tháng 12 2021
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
AH
Akai Haruma
Giáo viên
17 tháng 4 2021
Lời giải:
Xét:
$\frac{a}{a^2+1}-\left(\frac{16}{25}-\frac{3}{25}a\right)=\frac{(a-2)^2(3a-4)}{25(a^2+1)}\geq 0$ với mọi $a\geq \frac{4}{3}$
$\Rightarrow \frac{a}{a^2+1}\geq \frac{16}{25}-\frac{3}{25}a$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế, suy ra:
$A\geq \frac{48}{25}-\frac{3}{25}(a+b+c)=\frac{6}{5}$
Vậy $A_{\min}=\frac{6}{5}$.
Giá trị này đạt tại $a=b=c=2$
27 tháng 4 2021
có cách nào không gượng ép như thế này không ạ
kiểu như phân tích chọn điểm rơi để tìm cách thêm bớt ấy ạ
3 tháng 4 2022
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
23 tháng 12 2023
a: \(1-\left(5\dfrac{4}{9}+a-7\dfrac{7}{18}\right):15\dfrac{3}{4}=0\)
=>\(\left(5+\dfrac{4}{9}+a-7-\dfrac{7}{18}\right):\dfrac{63}{4}=1\)
=>\(\left(a-2+\dfrac{1}{18}\right)=\dfrac{63}{4}\)
=>\(a-\dfrac{35}{18}=\dfrac{63}{4}\)
=>\(a=\dfrac{63}{4}+\dfrac{35}{18}=\dfrac{637}{36}\)
b: \(B=\left(\dfrac{2}{15}+\dfrac{5}{3}-\dfrac{3}{5}\right):\left(4\dfrac{2}{3}-2\dfrac{1}{2}\right)\)
\(=\dfrac{2+5\cdot5-3^2}{15}:\left(4+\dfrac{2}{3}-2-\dfrac{1}{2}\right)\)
\(=\dfrac{2+4^2}{15}:\left(2+\dfrac{2}{3}-\dfrac{1}{2}\right)\)
\(=\dfrac{18}{15}:\dfrac{13}{6}=\dfrac{6}{5}\cdot\dfrac{6}{13}=\dfrac{36}{65}\)
AH
Akai Haruma
Giáo viên
5 tháng 10 2018
Bài 1:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{2ab}+\frac{1}{a^2+b^2}\geq \frac{4}{2ab+a^2+b^2}=\frac{4}{a+b)^2}=4(1)\)
Áp dụng BĐT AM-GM:
\(1=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}\Rightarrow \frac{3}{2ab}\geq 6(2)\)
\(a^4+b^4\geq \frac{(a^2+b^2)^2}{2}\geq \frac{(\frac{(a+b)^2}{2})^2}{2}=\frac{1}{8}\) \(\Rightarrow \frac{a^4+b^4}{2}\geq \frac{1}{16}(3)\)
Từ \((1);(2);(3)\Rightarrow P\geq 4+6+\frac{1}{16}=\frac{161}{16}\)
Vậy \(P_{\min}=\frac{161}{16}\). Dấu bằng xảy ra tại $a=b=0,5$
AH
Akai Haruma
Giáo viên
6 tháng 10 2018
Bài 2:
Áp dụng BĐT Cauchy-Schwarz:
\(2\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)\geq 2. \frac{4}{x^2+y^2+2xy}=\frac{8}{(x+y)^2}=\frac{9}{2}\)
Áp dụng BĐT AM-GM:
\(\frac{80}{81xy}+5xy\geq 2\sqrt{\frac{80}{81}.5}=\frac{40}{9}\)
\(\frac{4}{3}=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{4}{9}\Rightarrow \frac{1}{81ab}\geq \frac{1}{36}\)
Cộng những BĐT vừa cm được ở trên với nhau:
\(\Rightarrow A\geq \frac{9}{2}+\frac{40}{9}+\frac{1}{36}=\frac{323}{36}\)
Vậy \(A_{\min}=\frac{323}{36}\Leftrightarrow a=b=\frac{2}{3}\)
Câu 1:
a, Giả sử \(A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{a}{b}-\dfrac{b}{a}\ge0\)
\(\Leftrightarrow A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge0\)
Mà \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow A\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\cdot\dfrac{a}{b}-2\cdot\dfrac{b}{a}+2\ge0\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}-2\cdot\dfrac{a}{b}+1\right)+\left(\dfrac{b^2}{a^2}-2\cdot\dfrac{b}{a}+1\right)\ge0\\ \Leftrightarrow\left(\dfrac{a}{b}-1\right)^2+\left(\dfrac{b}{a}-1\right)^2\ge0\left(\text{luôn đúng}\right)\)
Dấu \("="\Leftrightarrow a=b\)
b, \(B=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\right)+2+\left(\dfrac{a^2}{b^2}+2+\dfrac{b^2}{a^2}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-4\)
\(B=\left(\dfrac{a^4}{b^4}-2\cdot\dfrac{a^2}{b^2}+1\right)+\left(\dfrac{b^4}{a^4}-2\cdot\dfrac{b^2}{a^2}+1\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-2\\ \Leftrightarrow B=\left(\dfrac{a^2}{b^2}-1\right)^2+\left(\dfrac{b^2}{a^2}-1\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\dfrac{a}{b}+\dfrac{b}{a}-4\\ \Leftrightarrow B\ge0+0+0+\dfrac{a^2+b^2}{ab}-4\ge\dfrac{2ab}{ab}-4=2-4=-2\)
Dấu \("="\Leftrightarrow\left(a;b\right)\in\left\{\left(1;-1\right);\left(-1;1\right)\right\}\)
Câu 2:
\(\left(x^2+y^2\right)\left(3^2+4^2\right)\ge\left(3x+4y\right)^2=M^2\\ \Leftrightarrow M^2\le25\cdot25\\ \Leftrightarrow M\le25\)
Dấu \("="\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{25}{25}=1\Leftrightarrow\left\{{}\begin{matrix}x^2=9\\y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy \(M_{max}=25\Leftrightarrow\left(x;y\right)=\left(3;4\right)\)