Bmin=5 xay ra dau= khi va chi khi x=5

\(B=\sqrt{x^2-10x+34}+\sqrt{x^2-10x+29}\)

\(=\sqrt{\left(x-5\right)^2+9}+\sqrt{\left(x-5\right)^2+4}\)\(\ge\)\(\sqrt{9}+\sqrt{4}=5\)

Vậy Min \(B=5\)khi  \(x=5\)

a: =>x=5/4-5/14=25/28

b: =>x=6/7:2/3=6/7*3/2=18/14=9/7

a) \(\dfrac{35}{28}-x=\dfrac{5}{14}\)

\(\dfrac{35}{28}-x=\dfrac{10}{28}\)

\(x=\dfrac{35}{28}-\dfrac{10}{28}\)

\(x=\dfrac{25}{28}\)

b) \(\dfrac{6}{7}:x=\dfrac{2}{3}\)

\(x=\dfrac{6}{7}:\dfrac{2}{3}=\dfrac{6}{7}\times\dfrac{3}{2}\)

\(x=\dfrac{9}{7}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}=\dfrac{x^2-2y^2+z^2}{4-18+25}=\dfrac{44}{11}=4\\ \Leftrightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=20\end{matrix}\right.\)

| x - \(\frac{1}{2}\)| = 1

TH1:x - \(\frac{1}{2}\) = 1                                 TH2:x - \(\frac{1}{2}\) = -1

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

        x              = \(\frac{3}{2}\)                                 x             = \(\frac{-1}{2}\)

Vậy x thuộc \(\frac{3}{2}\)và \(\frac{-1}{2}\)

Trả lời:

\(a,\left|x-\frac{1}{2}\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=1\\x-\frac{1}{2}=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)

Vậy x = 3/2; x = - 1/2

g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

\(a.\)\(A=|x|+|2014-x|\ge|x+2014-x|=2014\)

Dấu '=' xảy ra khi\(x\left(2014-x\right)>0\)

TH1:\(\hept{\begin{cases}x>0\\2014-x>0\end{cases}\Leftrightarrow0< x< 2014\left(n\right)}\)

TH2:\(\hept{\begin{cases}x< 0\\2014-x< 0\end{cases}\left(l\right)}\)

Vậy \(A_{min}=2014\)khi\(0< x< 2014\)

\(b.\)\(|x^2+|x-1||=x^2+2\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+|x-1|=-x^2-2\\x^2+|x-1|=x^2+2\end{cases}\Leftrightarrow\orbr{\begin{cases}|x-1|=-2x^2-2\left(l\right)\\|x-1|=2\left(n\right)\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=-2\\x-1=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

V...

Nếu đề bài là giải phương trình thì :

\(\sqrt{x+3}=\sqrt{x-3}\)

Đk : \(x\ge3\)

Bình phương hai vế :

\(\Rightarrow x+3=x-3\)

\(x+3-x+3=0\)

\(0x=-6\)

\(\Rightarrow\)phương trình vô nghiệm

Để \(2x^3-4x^2+6x+a⋮x+2\)

\(\Leftrightarrow2x^3-4x^2+6x+a=\left(x+2\right)\cdot a\left(x\right)\)

Thay \(x=-2\)

\(\Leftrightarrow2\left(-2\right)^3-4\left(-2\right)^2+6\left(-2\right)+a=0\\ \Leftrightarrow-16-16-12+a=0\\ \Leftrightarrow-44+a=0\Leftrightarrow a=44\)