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Điều kiện \(\left\{{}\begin{matrix}x\ne-1\\x\ne0\\x\ne1\end{matrix}\right.\)
Đặt \(\dfrac{x+1}{x-1}=a\) thì pt trở thành
\(\dfrac{a-\dfrac{1}{a}}{1+a}=\dfrac{1}{2a}\)
\(\Leftrightarrow2a=3\)
\(\Leftrightarrow a=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}=\dfrac{3}{2}\)
\(\Leftrightarrow x=5\)
a) Đặt \(t=\left|2x-\dfrac{1}{x}\right|\Leftrightarrow t^2=\left(2x-\dfrac{1}{x}\right)^2=4x^2-4+\dfrac{1}{x^2}\Leftrightarrow t^2+4=4x^2+\dfrac{1}{x^2}\) ĐK \(t\ge0\)
từ có ta có pt theo biến t : \(t^2+4+t-6=0\)
\(\Leftrightarrow t^2+t-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|2x-\dfrac{1}{x}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{x}=1\\2x-\dfrac{1}{x}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x^2-x-1=0\\2x^2+x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
c: TH1: x>0
Pt sẽ là \(\dfrac{x^2-1}{x\left(x-2\right)}=2\)
=>2x^2-4x=x^2-1
=>x^2-4x+1=0
hay \(x=2\pm\sqrt{3}\)
TH2: x<0
Pt sẽ là \(\dfrac{x^2-1}{-x\left(x-2\right)}=2\)
=>-2x(x-2)=x^2-1
=>-2x^2+4x=x^2-1
=>-3x^2+4x+1=0
hay \(x=\dfrac{2-\sqrt{7}}{3}\)
b:
TH1: 2x^3-x>=0
\(4x^4+6x^2\left(2x^3-x\right)+1=0\)
=>4x^4+12x^5-6x^3+1=0
\(\Leftrightarrow x\simeq-0.95\left(loại\right)\)
TH2: 2x^3-x<0
Pt sẽ là \(4x^4+6x^2\left(x-2x^3\right)+1=0\)
=>4x^4+6x^3-12x^5+1=0
=>x=0,95(loại)
a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
a) đặc \(f\left(x\right)=y=\left|2x+1\right|+\left|2x-1\right|\)
\(D=R\) \(\Rightarrow\forall x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=\left|-2x+1\right|+\left|-2x-1\right|=\left|2x-1\right|+\left|2x+1\right|=f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm chẳn
b) đặc \(f\left(x\right)=y=\dfrac{\left|x+1\right|+\left|x-1\right|}{\left|x+1\right|-\left|x-1\right|}\)
\(D=R\backslash\left\{0\right\}\) \(\Rightarrow\forall x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=\dfrac{\left|-x+1\right|+\left|-x-1\right|}{\left|-x+1\right|-\left|-x-1\right|}=\dfrac{\left|x-1\right|+\left|x+1\right|}{\left|x-1\right|-\left|x+1\right|}\)
\(=-\dfrac{\left|x+1\right|+\left|x-1\right|}{\left|x+1\right|-\left|x-1\right|}=-f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm lẽ
What? Lớp 10? Mí bài nỳ dễ mak! Trên lp cs hc mak k giải đc thì thui lun!
e: =>-3<5x-12<3
=>9<5x<15
=>9/5<x<3
f: =>3x+15>=3 hoặc 3x+15<=-3
=>3x>=-12 hoặc 3x<=-18
=>x<=-6 hoặc x>=-4
b: =>(2x-7)(x-5)<=0
=>7/2<=x<=5
Giải bài này hơi dài, t ngại làm lắm :v you vào ib t chỉ cho =))
\(\Leftrightarrow\left(\dfrac{2}{5}\right)^n\left(\dfrac{2}{5}-1\right)=\dfrac{-12}{125}\)
\(\Leftrightarrow\left(\dfrac{2}{5}\right)^n\left(\dfrac{-3}{5}\right)=\dfrac{-12}{125}\)
\(\Leftrightarrow\left(\dfrac{2}{5}\right)^n=\dfrac{4}{25}=\left(\dfrac{2}{5}\right)^2\)
\(\Rightarrow n=2\)
Thanks