\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(x+x+x+x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(4x+\frac{15}{16}=1\)

\(4x=1-\frac{15}{16}\)

\(4x=\frac{1}{16}\)

\(x=\frac{1}{16}\div4\)

\(x=\frac{1}{64}\)

ta coi x để ra mội bên ta có 1/2 + 1/4 + 1/8 + 1/6 = 25/24

x + 25/24 = 1

            x = 1 - 25/24

            x = -1/24

mk nha

Nhân lên 16 lần ta có đề bài mới: 

Tìm x biết: \(\left(x+8\right)+\left(x+4\right)+\left(x+2\right)+\left(x+1\right)=16\)

Bài giải

\(x\times4+1+2+4+8=16\)

\(x\times4=1\)

Vậy \(x=\frac{1}{4}\)

x=0

x=-3

(x+1)(x+2) = 2

x² + 2x + x + 2 = 2

            x² + 3x = 2 - 2

           x(x + 3) = 2

⇒ x = 0 hoặc x + 3 = 0

⇒ x = 0 hoặc x = -3

Vậy tập nghiệm S = {-3}

bay bài đó với

1) \(\Delta'=1^2-\left(m-1\right)=2-m\)

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)

Khi đó \(x_1=1+\sqrt{2-m};x_2=1-\sqrt{2-m}\)

TH1: \(2\left(1+\sqrt{2-m}\right)-\left(1-\sqrt{2-m}\right)=7\Leftrightarrow1+3\sqrt{2-m}=7\)

\(\Leftrightarrow\sqrt{2-m}=2\Leftrightarrow2-m=4\Rightarrow m=-2\left(tm\right)\)

TH2: \(2\left(1-\sqrt{2-m}\right)-\left(1+\sqrt{2-m}\right)=7\Leftrightarrow1-3\sqrt{2-m}=7\) (VÔ LÝ)

Vậy m = - 2.

2) \(P=\frac{x^4+3x^2+1}{x^2+1}=\frac{\left(x^4+2x^2+1\right)+\left(x^2+1\right)+2}{x^2+1}=\left(x^2+1\right)+\frac{2}{x^2+1}+1\)

Vì \(x^2+1\ge1\), áp dụng bđt Cô si ta có:

 \(\left(x^2+1\right)+\frac{2}{x^2+1}\ge2\sqrt{\left(x^2+1\right).\frac{2}{x^2+1}}=2\sqrt{2}\)

Vậy \(P\ge2\sqrt{2}+1\)

Dấu bằng xảy ra khi

 \(x^2+1=\frac{2}{x^2+1}\Leftrightarrow x^2+1=\sqrt{2}\Rightarrow x^2=\sqrt{2}-1\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\sqrt{2}-1}\\x=-\sqrt{\sqrt{2}-1}\end{cases}}\)

Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)

Đk:\(x\ge1\)

\(pt\Leftrightarrow3\left(x-2\right)\sqrt{x-1}\sqrt{x^2+x+1}+18\left(x-1\right)=x\left(x^2+x+1\right)\)

Chia 2 vế của pt cho \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)ta đc:

\(3\left(x-2\right)\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}+\frac{18\left(x-1\right)}{x^2+x+1}=x\)

Đặt \(y=\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}\left(y\ge0\right)\) pt trở thành

\(3\left(x-2\right)y+18y^2-x=0\)

\(\Leftrightarrow\left(3y-1\right)\left(6y+x\right)=0\)

\(\Leftrightarrow3y-1=0\left(y\ge0;x\ge1\Rightarrow6y+x\ge1\right)\)

\(\Leftrightarrow y=\frac{1}{3}\)\(\Leftrightarrow\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}=\frac{1}{3}\)

\(\Leftrightarrow9\left(x-1\right)=x^2+x+1\)

\(\Leftrightarrow x^2-8x+10=0\)

\(\Leftrightarrow x=4\pm\sqrt{6}\)

Vậy...