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Giải phương trình sau:
\(\dfrac{x}{50}\) +\(\dfrac{x_{ }-1}{49}\)+\(\dfrac{x-2}{48}\)+\(\dfrac{x-3}{47}\)+\(\dfrac{x-150}{25}\)= 0
⇔ \(\dfrac{\left(x-50\right)+50}{50}\)+\(\dfrac{\left(x-50\right)+49}{49}\)+\(\dfrac{\left(x-50\right)+48}{48}\)+\(\dfrac{\left(x-50\right)-100}{25}\)= 0
⇔\(\dfrac{x-50}{50}\)+ 1 + \(\dfrac{x-50}{49}\)+1+\(\dfrac{x-50}{48}\)+1+\(\dfrac{x-50}{47}\)+1+\(\dfrac{x-50}{25}\)-4 = 0
⇔\(\dfrac{x-50}{50}\)+\(\dfrac{x-50}{49}\)+\(\dfrac{x-50}{48}\)+\(\dfrac{x-50}{47}\)+\(\dfrac{x-50}{25}\)= 0
⇔ (x - 50 ) ( \(\dfrac{1}{50}\)+ \(\dfrac{1}{49}\)+\(\dfrac{1}{48}\)+\(\dfrac{1}{47}\)+\(\dfrac{1}{25}\)) = 0
⇔ x-50 =\(\dfrac{0}{\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}}\)
⇔ x- 50 = 0
⇔ x = 50
vậy S = \(\left\{50\right\}\)
c: =>|x-2|+3=-5 hoặc |x-2|+3=5
=>|x-2|=2
=>x-2=2 hoặc x-2=-2
=>x=4 hoặc x=0
\(\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+\dfrac{x+16}{4}=4\)
\(\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}=4-\dfrac{x+16}{4}\)
\(\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}\right)=-x\)
Mk giải đế đây rùi bạn tự giải nốt đi
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
a , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)
b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
a,(4x+1)2 e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)
b,(x-\(\dfrac{1}{2}\))2 g,\(\left(xy+1\right)^2\)
c,(\(x+\dfrac{3}{2}\))2 h,\(\left(x+5\right)^2\)
d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)
k,\(-\left(2x+3\right)^2\)
a: \(\Leftrightarrow5\left(x+1\right)\left(x-1\right)=2x-2-3x-3=-x-5\)
\(\Leftrightarrow5x^2-5+x+5=0\)
=>x(5x+1)=0
=>x=0 hoặc x=-1/5
b: \(\Leftrightarrow x^2-x-\left(2x-3\right)\left(x+1\right)=2x+3\)
\(\Leftrightarrow x^2-x-2x^2-2x+3x+3=2x+3\)
\(\Leftrightarrow-x^2+3=2x+3\)
=>-x(x+2)=0
=>x=0(nhận) hoặc x=-2(nhận)
c: \(\Leftrightarrow4x^2-25=0\)
=>(2x-5)(2x+5)=0
=>x=5/2 hoặc x=-5/2
\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x\left(1-\dfrac{x-1}{x+1}\right)\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x.\dfrac{x+1-\left(x-1\right)}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x.\dfrac{2}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{6x}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{6x}{x+1}=0\)
\(\Rightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-6x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{4x-6x^2+6x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{10x-6x^2}{\left(x-1\right)\left(x+1\right)=0}\)
\(\Rightarrow10x-6x^2=0\)
\(\Rightarrow x-6x^2=0\)
\(\Rightarrow2x\left(5-3x\right)=0\)
\(\Rightarrow x\left(5-3x\right)=0\)
\(\Rightarrow5-3x=0\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=\dfrac{5}{3}\)
a) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x\left(1-\dfrac{x-1}{x+1}\right)\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}3\left(1-\dfrac{x-1}{x+1}\right),\left(đk:x\ne1;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3-\dfrac{3\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3-\dfrac{3x-3}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{3x-3}{x+1}=3\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)\cdot\left(3x-3\right)}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{2\cdot2x+3x^2-3x-3x+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{4x+3x^2-3x-3x+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{-2x+3x^2+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow-2x+3x^2+3=3\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow-2x+3x^2+3=3\left(x^2-1\right)\)
\(\Leftrightarrow-2x+3x^2+3=3x^2-3\)
\(\Leftrightarrow-2x+3=-3\)
\(\Leftrightarrow-2x=-3-3\)
\(\Leftrightarrow-2x=-6\)
\(\Rightarrow x=3\left(đk:x\ne1,x\ne-1\right)\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
a)
2x-3=0 => x=3/2
b)
2x^2 +1 =0 => vô nghiệm
c) x^2 -25 =0 => x=5 loiaj
x=-5 nhân
d)
x^2 -25 =0 => x=5 loại
x=-5 loại
\(PT\Leftrightarrow\left(\dfrac{x-70}{130}-1\right)+\left(\dfrac{x-25}{175}-1\right)+\left(\dfrac{x-50}{150}-1\right)+\left(\dfrac{x-275}{25}+3\right)=0\)
\(\Leftrightarrow\left(x-200\right)\left(\dfrac{1}{130}+\dfrac{1}{175}+\dfrac{1}{150}+\dfrac{1}{25}\right)=0\Leftrightarrow x=200\).
Vậy...