Question

1) Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2010}=\dfrac{2010}{a}\)và a + b + c \(\ne\) \(-2010\). Tính a,b,c

2) Cho 3 tỉ số \(\dfrac{a}{b+c};\dfrac{b}{c+a};\dfrac{c}{a+b}\). Tính giá trị mỗi tỉ số.

3) Tìm GTNN của:

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

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

4) Tìm GTLN của:

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

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

5) Tìm n, biết:

\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1=2010}\)

Answer 1

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