PT <=> \(2\sqrt{x}-x-1=0\)

\(\Leftrightarrow2\sqrt{x}=x+1\)

Bình phương 2 vế ta được : \(4x=x^2+2x+1\)

\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy x = 1

\(9\left(3x+1\right)=\left(3x+1\right)\Leftrightarrow9\left(3x+1\right)-\left(3x+1\right)=0\)

\(\Leftrightarrow8\left(3x+1\right)=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)

\(\Rightarrow\frac{x+15}{x}=\frac{4}{3}\Rightarrow3.\left(x+15\right)=4.x\Rightarrow3x+45=4x\Rightarrow4x-3x=0-45\Rightarrow x=-45\)

\(\frac{1}{6}-\left(\frac{2}{3}\right)^2+\frac{5}{18}\)

\(=\frac{1}{6}-\frac{4}{9}+\frac{5}{18}\)

\(=\frac{3}{18}-\frac{8}{18}+\frac{5}{18}\)

\(=\frac{3-8+5}{18}\)

\(=0\)

e có nhầm đề không nhỉ, chỗ a+b+c không có mũ 2 hả e.

nếu không có, ta lấy 1 trường hợp cụ thể của a=b=c=1 thì đã thấy đề có vấn đề r e

Ta có \(A=\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{102}{5^{102}}+\frac{103}{5^{103}}\)

=> 5A = \(\frac{3}{5^2}+\frac{4}{5^3}+...+\frac{102}{5^{101}}+\frac{103}{5^{102}}\)

Khi đó 5A - A = \(\left(\frac{3}{5^2}+\frac{4}{5^3}+...+\frac{102}{5^{101}}+\frac{103}{5^{102}}\right)-\left(\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{102}{5^{102}}+\frac{103}{5^{103}}\right)\)

=> 4A = \(\frac{3}{5^2}+\left(\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{102}}\right)-\frac{103}{5^{103}}\)

=> 4A = \(\frac{3}{5^2}+\frac{\frac{1}{5^2}-\frac{1}{5^{102}}}{4}-\frac{103}{5^{103}}\)

=> A = \(\frac{3}{5^2.4}+\left(\frac{1}{5^2}-\frac{1}{5^{102}}\right).\frac{1}{16}-\frac{103}{5^{103}.4}\)

=> A = \(\frac{3}{100}+\frac{1}{5^2}.\frac{1}{16}\left(1-\frac{1}{5^{100}}\right)-\frac{103}{5^{103}.4}=\frac{3}{100}+\frac{1}{400}\left(1-\frac{1}{5^{100}}\right)-\frac{103}{5^{103}.4}\)

       \(=\frac{3}{100}+\frac{1}{400}-\frac{1}{400.5^{100}}-\frac{103}{5^{103}.4}=\frac{13}{400}-\frac{1}{400.5^{100}}-\frac{103}{5^{103}.4}< \frac{13}{400}\left(\text{ĐPCM}\right)\)

Vậy \(A< \frac{13}{400}\)(đpcm)

Trả lời :

Ta có:

25⁵⁰ = (5²)⁵⁰ = 5¹⁰⁰

2³⁰⁰ = 2^{3 . 100} = 8¹⁰⁰

Do 5 < 8 => 25⁵⁰ < 2³⁰⁰

Vậy 25⁵⁰ < 2³⁰⁰.

ta có:25^{50 và} 2³⁰⁰

^{\(\Rightarrow\)}2³⁰⁰=(2⁶)⁵⁰\(\Rightarrow\)64⁵⁰

Mà 25⁵⁰<64⁵⁰

\(\Rightarrow25^{50}\)< \(2^{300}\)