\(=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{12}}{3^{12}\cdot5^6+7\cdot3^{12}\cdot5^6}=\dfrac{3^{12}\cdot5^4\left(3^2-1\right)}{3^{12}\cdot5^6\left(1+7\right)}=\dfrac{1}{25}\)

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 \("="\Leftrightarrow a=b=c=2\)

a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)

\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)

\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)

\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)

Câu hỏi của Hạ Anh Thư - Toán lớp 7 | Học trực tuyến

\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)

a) \(\dfrac{3^6\cdot3^8\cdot5^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+3^{12}\cdot5^6}=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{3^{12}\cdot5^6\cdot2}=\dfrac{3^{12}\cdot5^4\left(3^2-3\right)}{3^{12}\cdot5^6\cdot2}=\dfrac{3^2-3}{5^2\cdot2}=\dfrac{6}{50}=\dfrac{3}{25}\)

3/ Áp dụng bất đẳng thức AM-GM, ta có :

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)

Cộng 3 vế của BĐT trên ta có :

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)

\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

1) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2010}=\dfrac{2010}{a}=\dfrac{a+b+c+2010}{b+c+2010+a}=1\)

\(\dfrac{2010}{a}=1\Rightarrow a=2010\);

\(\dfrac{c}{2010}=1\Rightarrow c=2010\);

\(\dfrac{b}{c}=1\Rightarrow\dfrac{b}{2010}=1\Rightarrow b=2010\).

Vậy (a, b, c) = (2010; 2010; 2010)

3)

a) \(A=\sqrt{x+24}+\dfrac{4}{7}\)

Có: \(\sqrt{x+24}\ge0\forall x\in R\)

\(\Rightarrow\sqrt{x+24}+\dfrac{4}{7}\ge\dfrac{4}{7}\forall x\in R\)

\(\Rightarrow A\ge\dfrac{4}{7}\forall x\in R\)

Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x+24}=0\Rightarrow x+24=0\Rightarrow x=-24\)

Vậy GTNN của \(A=\dfrac{4}{7}\Leftrightarrow x=-24\)

b) \(B=\sqrt{2x+\dfrac{4}{13}}-\dfrac{13}{191}\)

Có: \(\sqrt{2x+\dfrac{4}{13}}\ge0\forall x\in R\)

\(\Rightarrow\sqrt{2x+\dfrac{4}{13}}-\dfrac{13}{191}\ge-\dfrac{13}{191}\forall x\in R\)

\(\Rightarrow B\ge-\dfrac{13}{191}\forall x\in R\)

Đẳng thức xảy ra \(\Leftrightarrow\sqrt{2x+\dfrac{4}{13}}=0\)

\(\Rightarrow2x+\dfrac{4}{13}=0\)

\(\Rightarrow2x=-\dfrac{4}{13}\)

\(\Rightarrow x=-\dfrac{2}{13}\)

Vậy GTNN của \(B=-\dfrac{13}{191}\Leftrightarrow x=-\dfrac{2}{13}\)

4)

a) \(A=-\sqrt{x+\dfrac{5}{41}}+\dfrac{7}{12}\)

Có: \(\sqrt{x+\dfrac{5}{41}}\ge0\forall x\in R\)

\(\Rightarrow-\sqrt{x+\dfrac{5}{41}}\le0\forall x\in R\)

\(\Rightarrow-\sqrt{x+\dfrac{5}{41}}+\dfrac{7}{12}\le\dfrac{7}{12}\forall x\in R\)

\(\Rightarrow A\le\dfrac{7}{12}\forall x\in R\)

Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x+\dfrac{5}{41}}=0\)

\(\Rightarrow x+\dfrac{5}{41}=0\)

\(\Rightarrow x=-\dfrac{5}{41}\)

Vậy GTLN của \(A=\dfrac{7}{12}\Leftrightarrow x=-\dfrac{5}{41}\)

b) \(B=\dfrac{-5}{13}-\sqrt{x-\dfrac{2}{3}}\)

Có: \(\sqrt{x-\dfrac{2}{3}}\ge0\forall x\in R\)

\(\Rightarrow-\sqrt{x-\dfrac{2}{3}}\le0\forall x\in R\)

\(\Rightarrow\dfrac{-5}{13}-\sqrt{x-\dfrac{2}{3}}\le\dfrac{-5}{13}\forall x\in R\)

\(\Rightarrow B\le\dfrac{-5}{13}\forall x\in R\)

Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x-\dfrac{2}{3}}=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\)

\(\Rightarrow x=\dfrac{2}{3}\)

Vậy GTLN của \(B=\dfrac{-5}{13}\Leftrightarrow x=\dfrac{2}{3}\)