\(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)

\(\Rightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)

\(\Rightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)

\(\Rightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

Mà \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}>0\Rightarrow x-124=0\Rightarrow x=124\)

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 #

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: $\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0$

$\Rightarrow x-357=0$

$\Rightarrow x=357$

6A=5.7.6+7.9.6+9.11.6+11.13.6+13.15.6+...+99.101.6

6A=5.7.(9-3)+7.9.(11-5)+9.11.(13-7)+11.13.(15-9)+13.15.(17-11)+...+99.101.(103-97)

6A=-3.5.7+5.7.9-5.7.9+7.9.11-7.9.11+9.11.13-9.11.13+11.13.15-11.13.15+13.15.17-...-97.99.101+99.101.103

6A=99.101.103-3.5.7 => A=(99.101.103-3.5.7)/6

\(2x^4+3x^3+8x^2+6x+5=0\)

\(\Leftrightarrow2x^4+2x^3+2x^2+x^3+x^2+x+5x^2+5x+5=0\)

\(\Leftrightarrow2x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+5\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(2x^2+x+5\right)=0\)

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

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