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\(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)
\(\Rightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)
\(\Rightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)
\(\Rightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Mà \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}>0\Rightarrow x-124=0\Rightarrow x=124\)
Ta có : \(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)
\(\Leftrightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)
\(\Leftrightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)
\(\Leftrightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Vì \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\)
Nên : 124 - x = 0
<=> x = 124
Vậy x = 124
Bài làm
\(\frac{x+19}{27}-\frac{x+17}{29}=\frac{x+15}{31}-\frac{x+13}{33}\)
\(\Leftrightarrow\left(\frac{x+19}{27}+1\right)-\left(\frac{x+17}{29}+1\right)=\left(\frac{x+15}{31}+1\right)-\left(\frac{x+13}{33}+1\right)\)
\(\Leftrightarrow\frac{x+46}{27}-\frac{x+46}{29}=\frac{x+46}{31}-\frac{x+46}{33}\)
\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}=\left(x+46\right).\frac{1}{31}-\left(x+46\right).\frac{1}{33}\)
\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}-\left(x+46\right).\frac{1}{31}+\left(x+46\right).\frac{1}{33}=0\)
\(\Leftrightarrow\left(x+46\right)\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)=0\)
Mà \(\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)>0\forall x\)
\(\Leftrightarrow x+46=0\)
\(\Leftrightarrow x=-46\)
Vậy phương trình trên có tập nghiệm S = { -46 }
# Học tốt #
Giải Phương Trình Sau:
\(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
Lời giải:
PT $\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0$
$\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0$
$(x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0$
Dễ thấy: $\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0$
$\Rightarrow x-357=0$
$\Rightarrow x=357$
\(2x^4+3x^3+8x^2+6x+5=0\)
\(\Leftrightarrow2x^4+2x^3+2x^2+x^3+x^2+x+5x^2+5x+5=0\)
\(\Leftrightarrow2x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+5\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(2x^2+x+5\right)=0\)
Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(2x^2+x+5=2\left[\left(x+\frac{1}{4}\right)^2+\frac{39}{16}\right]>0\forall x\)
Vậy tập nghiệm của pt là \(S=\varnothing\)
b, \(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)
\(\Leftrightarrow\left(\frac{x-342}{15}-1\right)+\left(\frac{x-323}{17}-2\right)+\left(\frac{x-300}{19}-3\right)+\left(\frac{x-273}{21}-4\right)=0\)
\(\Leftrightarrow\frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0\)
\(\Leftrightarrow\left(x-357\right)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0\)
\(\Leftrightarrow x-357=0\Leftrightarrow x=357\)
Vậy tập nghiệm của pt: \(S=\left\{357\right\}\)
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-163}{23}=10\)
\(\Leftrightarrow\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-166}{23}-4=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow x=258\)
Vậy \(x=258\)
Chúc bạn học tốt !!!
\(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)
\(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)
\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\) (do \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\))
\(\Leftrightarrow x=100\)
Ta có: \(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}=10\)
\(\Rightarrow\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}-10=0\)
\(\Leftrightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-170}{23}-4\right)=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow\left(x-258\right)=0\)
\(\Rightarrow x=258\)
Vậy x=258
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=-10\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=-10\)
\(.....................\)
đến đây thì dễ rồi :)
\(\dfrac{x-187}{13}+\dfrac{x-170}{15}+\dfrac{x-149}{17}+\dfrac{x-124}{19}=10\)
`<=>(x-187)/13+(x-170)/15+(x-149)/17+(x-124)/19-10=0`
`<=>(x-187)/13-1+(x-170)/15-2+(x-149)/17-3+(x-124)/19-4=0`
`<=>(x-200)/13+(x-200)/15+(x-200)/17+(x-200)/19=0`
`<=>(x-200)(1/13+1/15+1/17+1/19)=0`
`<=>x-200=0(1/13+1/15+1/17+1/19>0)`
`<=>x=200`
\(=>\left(\dfrac{x-187}{13}-1\right)+\left(\dfrac{x-170}{15}-2\right)+\left(\dfrac{x-149}{17}-3\right)+\left(\dfrac{x-124}{19}-4\right)=0\)\(< =>\left(\dfrac{x-187}{13}-\dfrac{13}{13}\right)+\left(\dfrac{x-170}{15}-\dfrac{30}{15}\right)+\left(\dfrac{x-149}{17}-\dfrac{51}{17}\right)+\left(\dfrac{x-124}{19}-\dfrac{76}{19}\right)=0\)
\(< =>\left(\dfrac{x-200}{13}\right)+\left(\dfrac{x-200}{15}\right)+\left(\dfrac{x-200}{17}\right)+\left(\dfrac{x-200}{19}\right)=0\)
\(< =>\left(x-200\right)\left(\dfrac{1}{13}+\dfrac{1}{15}+\dfrac{1}{17}+\dfrac{1}{19}\right)=0\)
\(< =>x-200=0\)
<=>x=200