Tìm Min
N=\(|x-2|+|5-x|\)
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A=\(\dfrac{a^2+b^2+2ab+ab}{\sqrt{ab}\left(a+b\right)}=\dfrac{\left(a+b\right)^2+ab}{\sqrt{ab}\left(a+b\right)}\) =\(\dfrac{a+b}{\sqrt{ab}}+\dfrac{\sqrt{ab}}{a+b}=\dfrac{a+b}{\sqrt{ab}}+\dfrac{4\sqrt{ab}}{a+b}-\dfrac{3\sqrt{ab}}{a+b}\)
\(\ge2\sqrt{\dfrac{a+b}{\sqrt{ab}}.\dfrac{4\sqrt{ab}}{a+b}}-\dfrac{3\sqrt{ab}}{a+b}\) =\(\ge4-\dfrac{3\left(a+b\right)}{2\left(a+b\right)}=4-\dfrac{3}{2}=\dfrac{5}{2}\)
dấu = xảy ra khi a=b
(x-42) - 17 = 127
=> x - 42 = 127 + 17 = 144
=> x = 144 + 42 = 186
23(x+1) = 69
=> x + 1 = 69 : 23 = 3
x = 3 - 1 = 2
2x + 5 = 120 : 2 = 60
=> 2x = 60 - 5 = 55
x = 55 : 2 = 27,5
5x - 2 = 613
=> 5x = 613 + 2 = 615
x = 615 : 5 = 123
a)(x-42)-17=127
(x-42)=127+17
(x-42)=144
x=144+42
x=186
b)23(x+1)=69
(x+1)=69:23
(x+1)=3
x=3-1
x=2
c)2.x+5=120:2
2.x+5=60
2.x=60-5
2.x=55
x=55:2
x=27,5
d)5.x-2=613
5.x=613+2
5.x=615
x=615:5
x=123
\((x-2)(x^2+2x+5+2x+4-5)=0<=> (x-2)(x^2+4x+4)=0 <=> (x-2)(x+2)^2 = 0 <=> x = 2 ; x = -2\)
\(\left(x-2\right)\left(x^2+2x+5\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+5+2x+4-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
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\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)
Vậy \(\left(x;y\right)=\left(5;-2\right)\)
Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
(3x+2).(x+1)=3x.(5+x)
\(\Rightarrow\)\(3x^2+3x+2x+2=15x+3x^2\)
\(\Rightarrow3x^2+5x+2=15x+3x^2\)
\(\Rightarrow5x-15x+2=3x^2-3x^2\)
\(\Rightarrow-10x+2=0\)
\(-10x=-2\)
\(x=\frac{1}{5}\)
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
Áp dụng BĐT /A/+/B/\(\ge\)/A+B/
\(N=\)/x-2/+/5-x/\(\ge\)/x-2+5-x/=3
dấu = xảy ra khi x=3