x - 2x + 2²x - 2³x + 2⁴x -.....+ 2²⁰⁰⁶x - 2²⁰⁰⁷x = 2²⁰⁰⁸ - 1

x(1 - 2 + 2² - 2³ + 2⁴ -....+ 2²⁰⁰⁶ - 2²⁰⁰⁷) = 2²⁰⁰⁸ - 1

Đặt M = 1 - 2 + 2² - 2³ + 2⁴ -....+ 2²⁰⁰⁶ - 2²⁰⁰⁷

2M = 2 - 2² + 2³ - 2⁴ + 2⁵ -....+ 2²⁰⁰⁷ - 2²⁰⁰⁸

3M = 2M + M = 1 - 2²⁰⁰⁸

=> M = \(\frac{1-2^{2008}}{3}\)

=> x . \(\frac{1-2^{2008}}{3}\) = 2²⁰⁰⁸ - 1

=> x = (2²⁰⁰⁸ - 1)\(\frac{1-2^{2008}}{3}\)

Đến đây chịu

a, (3x - 2)(4x + 3) = (2 - 3x)(x - 1)

\(\Leftrightarrow\) (3x - 2)(4x + 3) - (2 - 3x)(x - 1) = 0

\(\Leftrightarrow\) (3x - 2)(4x + 3) + (3x - 2)(x - 1) = 0

\(\Leftrightarrow\) (3x - 2)(4x + 3 + x - 1) = 0

\(\Leftrightarrow\) (3x - 2)(5x + 2) = 0

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-2}{5}\end{matrix}\right.\)

Vậy S = {\(\frac{2}{3}\); \(\frac{-2}{5}\)}

b, x² + (x + 3)(5x - 7) = 9

\(\Leftrightarrow\) x² - 9 + (x + 3)(5x - 7) = 0

\(\Leftrightarrow\) (x - 3)(x + 3) + (x + 3)(5x - 7) = 0

\(\Leftrightarrow\) (x + 3)(x - 3 + 5x - 7) = 0

\(\Leftrightarrow\) (x + 3)(6x - 10) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{5}{3}\end{matrix}\right.\)

Vậy S = {-3; \(\frac{5}{3}\)}

c, 2x² + 5x + 3 = 0

\(\Leftrightarrow\) 2x² + 2x + 3x + 3 = 0

\(\Leftrightarrow\) 2x(x + 1) + 3(x + 1) = 0

\(\Leftrightarrow\) (x + 1)(2x + 3) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)

Vậy S = {-1; \(\frac{3}{2}\)}

d, \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}=\frac{3-2x}{2009}+\frac{3-2x}{2010}\)

\(\Leftrightarrow\) \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}-\frac{3-2x}{2009}-\frac{3-2x}{2010}=0\)

\(\Leftrightarrow\) (3 - 2x)\(\left(\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)\) = 0

\(\Leftrightarrow\) 3 - 2x = 0

\(\Leftrightarrow\) x = \(\frac{3}{2}\)

Vậy S = {\(\frac{3}{2}\)}

Chúc bn học tốt!!

Bạn ghi lại đề đi bạn

a) Đặt \(u=\sqrt{x^2+1}\left(u>0\right)\Rightarrow u^2-1=x^2\)

Phương trình trở thành :

\(2u^2+6x-\left(2x+6\right)t=0\)

\(\Rightarrow\Delta_t=\left(2x+6\right)^2-48x=\left(2x-6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2x+6-2x+6}{4}=3\\t=\dfrac{2x+6+2x-6}{4}=x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=3\\\sqrt{x^2+1}=x\end{matrix}\right.\)

đến đây thì ez rồi

c) Ta có :

\(2\sqrt{x^2-4x+5}=2\sqrt{\left(x-2\right)^2+1}\ge2\)

\(\sqrt{\dfrac{1}{4}x^2-x+1+4}=\sqrt{\left(\dfrac{1}{2}x-1\right)^2+4}\ge2\)

\(\Rightarrow2\sqrt{x^2-4x+5}+\sqrt{\dfrac{1}{4}x^2-x+5}\ge4\)

ta lại có: \(-4x^2+16x-12=-4\left(x^2-4x+4\right)+4\le4\)

\(\left\{{}\begin{matrix}VP\ge4\\VT\le4\end{matrix}\right.\)

Dấu bằng xảy ra khi x = 2

vậy x=2 là nghiệm của phương trình