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Thêm 2 vào pt có :
\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\) (1)
\(\Leftrightarrow\frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1\)
\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\) (2)
\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\ne0\)
\(\Leftrightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
a, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Leftrightarrow\left(\dfrac{59-x}{49}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{45}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right).\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
Mà \(\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
b, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-5x+3=-7x+6x^2\)
\(\Leftrightarrow6x^2-5x+3+7x-6x^2=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
Vậy \(S=\left\{\dfrac{-3}{2}\right\}\)
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 = \(\left\{50\right\}\)
Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
đúng là toán 8 khó thật nhìn mà hoa cả mắt *_* T_T
duyệt đi
chẳng hoa j cả
áp dụng tỉ lệ thức ta có :
\(\Leftrightarrow\frac{96x+1634}{2303}=\frac{x-25}{45}\Rightarrow\left(96x+1634\right)45=2303\left(x-25\right)\)
tự giải tiếp ra
=>x=-65
b)\(\dfrac{x+14}{86}+\dfrac{x+15}{85}+\dfrac{x+16}{84}+\dfrac{x+17}{83}+\dfrac{x+116}{4}=0\)
\(\Leftrightarrow\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+1+\dfrac{x+116}{4}-4=0\)
\(\Leftrightarrow\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}+\dfrac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x+100=0\).Do \(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\ne0\)
\(\Leftrightarrow x=-100\)
c)\(\dfrac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\dfrac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\dfrac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\left(x^2+1\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+3\right)}+...+\dfrac{1}{\left(x^2+4\right)\left(x^2+5\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+2}+\dfrac{1}{x^2+2}-\dfrac{1}{x^2+3}+...+\dfrac{1}{x^2+4}-\dfrac{1}{x^2+5}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+5}=-1\)\(\Leftrightarrow\dfrac{4}{x^4+6x^2+5}=-1\)
\(\Leftrightarrow\dfrac{x^4+6x^2+9}{x^4+6x^2+5}=0\Leftrightarrow x^4+6x^2+9=0\)
\(\Leftrightarrow\left(x^2+3\right)^2>0\forall x\) (vô nghiệm)
h.
\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)
\(\Leftrightarrow\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
Vì: \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\)
Suy ra: 2004 - x = 0
Vậy x = 2004
\(a,\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
\(\Leftrightarrow\dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)
\(\Leftrightarrow x-23=0\) ( vì \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\) )
\(\Leftrightarrow x=23\)
Vậy pt có tập nghiệm S = { 23 }
\(b,\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy pt có tập nghiệm S = { 100 }
\(c,\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
Vậy pt có tập nghiệm S = { 2005 }
\(d,\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Leftrightarrow\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0\)
\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
\(\Leftrightarrow300-x=0\)
\(\Leftrightarrow x=300\)
Vậy pt có tập nghiệm S = { 300 }
\(e,\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
\(\Leftrightarrow\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-45-55}{55}+\dfrac{x-47-53}{53}-\dfrac{x-55-45}{45}-\dfrac{x-53-47}{47}=0\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
Vậy pt có tập nghiệm S = { 100 }
\(f,\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)
\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
Vậy pt có tập nghiệm S = { 10 }
\(h,\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1=\dfrac{1-x}{2003}+1+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2-x+2002}{2002}-\dfrac{1-x+2003}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
\(\Leftrightarrow2004-x=0\)
\(\Leftrightarrow x=2004\)
Vậy pt có tập nghiệm S = { 2004 }
\(g,\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
\(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy pt có tập nghiệm S = { -100 }
b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100
Vậy x = 100
pt \(\Leftrightarrow\left(\dfrac{x+16}{49}+1\right)+\left(\dfrac{x+18}{47}+1\right)=\left(\dfrac{x+20}{45}+1\right)\)
\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}-\dfrac{x+65}{45}=0\)
\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)
\(\Leftrightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
\(\dfrac{x+16}{49}+\dfrac{x+18}{47}=\dfrac{x+20}{45}+1\left(1\right)\)
Thêm (2) vào vế của p.trình (1) ta có :
\(\Leftrightarrow\dfrac{x+16}{49}+1+\dfrac{x+18}{47}+1=\dfrac{x+20}{45}+1\)
\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}-\dfrac{x+65}{45}=0\) (2)
\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)
\(\Leftrightarrow x+65=0\) ( vì \(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\ne0\))
\(\Leftrightarrow x=-65\)