a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\) 

   8 .x + 1 . x = 990

x . [ 8 +1 ] = 990

x . 9 = 990

x = 990 : 9

x = 110

các bạn giúp mình với mình đang vội.

a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)

=>\(8\cdot x+1\cdot x=3305+1\)

=>\(9x=3306\)

=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)

b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)

=>\(2^x\left(1+2+4+8\right)=480\)

=>\(2^x\cdot15=480\)

=>\(2^x=32\)

=>\(2^x=2^5\)

=>x+5

troi mk dang bao phamdangkhoi ma

\(x-\sqrt{x^2-1}=\frac{x^2-\left(x^2-1\right)}{x+\sqrt{x^2-1}}=\frac{1}{x+\sqrt{x^2-1}}=t\)\(\Rightarrow x+\sqrt{x^2-1}=\frac{1}{t}\)

Ta có: \(\left(1+t\right)^{2015}+\left(1+\frac{1}{t}\right)^{2015}=2^{2016}\)(1)

Áp dụng Côsi ta có: 

\(1+t\ge2\sqrt{t}\Rightarrow\left(1+t\right)^{2015}\ge2^{2015}.\sqrt{t^{2015}}\)

\(1+\frac{1}{t}\ge\frac{2}{\sqrt{t}}\Rightarrow\left(1+\frac{1}{t}\right)^{2015}\ge\frac{2^{2015}}{\sqrt{t^{2015}}}\)

\(\Rightarrow\left(1+t\right)^{2015}+\left(1+\frac{1}{t}\right)^{2015}\ge2^{2015}\left(\sqrt{t^{2015}}+\frac{1}{\sqrt{t^{2015}}}\right)\)

\(\ge2^{2015}.2\sqrt{\sqrt{t^{2015}}.\frac{1}{\sqrt{t^{2015}}}}=2^{2016}\)

Dấu "=" xảy ra khi và chỉ khi t = 1.

Do đó, từ (1) => \(t=\frac{1}{x+\sqrt{x^2-1}}=1\Rightarrow x+\sqrt{x^2-1}=1\)

\(\Rightarrow1-x=\sqrt{x^2-1}\Rightarrow\left(1-x\right)^2=x^2-1\Leftrightarrow2-2x=0\Leftrightarrow x=1\)

Vậy: \(x=1\text{ là nghiệm (nguyên) duy nhất của phương trình.}\)

Ta có :

\(2006\left|x-1\right|+\left(x-1\right)^2=2005\left|1-x\right|\)

\(\Rightarrow2006\left|x-1\right|+\left(x-1\right)^2=2005\left|x-1\right|\)

\(\Rightarrow2006\left|x-1\right|+\left(x-1\right)^2-2005\left|x-1\right|=0\)

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

Vì \(\begin{cases}\left|x-1\right|\ge0\\\left(x-1\right)^2\ge0\end{cases}\)\(\forall x\)

\(\Rightarrow\begin{cases}x-1=0\\x-1=0\end{cases}\)

=> x = 1

Vậy x = 1

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005

2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005

 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2

1/2 - 1/x+1 = 2003/4010

1/x+1 = 1/2 - 2003/4010

1/x+1 = 2005/4010 - 2003/4010

1/x+1 = 1/2005

=> x+1 = 2005

=> x = 2004

Vậy x = 2004

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2005}\)

\(\Leftrightarrow x+1=2005\)

\(\Leftrightarrow x=2004\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2005}\)

\(\Leftrightarrow x=2004\)

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005

2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005

 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2

1/2 - 1/x+1 = 2003/4010

1/x+1 = 1/2 - 2003/4010

1/x+1 = 2005/4010 - 2003/4010

1/x+1 = 1/2005

=> x+1 = 2005

=> x = 2004

Vậy x = 2004

 ai tích mk tích lại cho 

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005

2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005

 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2

1/2 - 1/x+1 = 2003/4010

1/x+1 = 1/2 - 2003/4010

1/x+1 = 2005/4010 - 2003/4010

1/x+1 = 1/2005

=> x+1 = 2005

=> x = 2004

Vậy x = 2004

 ai tích mk tích lại cho 

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005

2 x (1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005

 1/2x3 + 1/3x4 + 1/4x5 + ... + 1/x(x+1) = 2003/2005 : 2

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 x 1/2

1/2 - 1/x+1 = 2003/4010

1/x+1 = 1/2 - 2003/4010

1/x+1 = 2005/4010 - 2003/4010

1/x+1 = 1/2005

=> x+1 = 2005

=> x=2004

Vậy x = 2004