\(\frac{x+29}{31}+\frac{x+27}{33}=\frac{x+17}{43}+\frac{x+15}{45}\)

\(\frac{x+29}{31}+1+\frac{x+27}{33}+1=\frac{x+17}{43}+1+\frac{x+15}{45}+1\)

\(\frac{x+60}{31}+\frac{x+60}{33}=\frac{x+60}{43}+\frac{x+60}{45}\)

\(\left(x+60\right)\left(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\right)=0\)

VÌ \(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\ne0\)

\(\Rightarrow x+60=0\)

\(\Rightarrow x=-60\)

Câu hỏi của honoka sonoka - Toán lớp 6 - Học toán với OnlineMath

ket qua x=-60

Cộng 3 vào cả 2 vế và chuyển vế xong đặt nhân tử x+50 ra ngoài ta được :

(x+50)(1/39+1/37+1/35-1/33-1/31-1/29)=0

vì (1/39+1/37+1/35-1/33-1/31-1/29) khác 0

=> x+50=0

=> x=-50

Nhớ k mình nhé

Chúc bạn học  tốt

x = -50 nha

x=-50 ok

Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.

Phương trình 4:
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Rightarrow\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}-15=0\)
\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
\(\Rightarrow\frac{x-90-10}{10}+\frac{x-76-24}{12}+\frac{x-58-42}{14}+\frac{x-36-64}{16}+\frac{x-15-85}{17}=0\)
\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

E = ( x - 29 ) / 1970 + ( x - 27 ) / 1972 + ( x - 25 ) / 1974 + ( x - 23 ) / 1976 + ( x - 21 ) / 1978 + ( x - 19 ) / 1980 = ( x - 1970 ) / 29 + ( x - 1972 ) / 27 + ( x - 1974 ) / 25 + ( x - 1976 ) / 23 + ( x - 1978 ) / 21 + ( x - 1980 ) / 19

( Trừ từng số hạng cho 1 ra như sau )

E = (x - 1999)/ 1970 + ( x - 1999 ) / 1972 + ( x - 1999) / 1974 + ( x - 1999)/ 1976 + ( x -1999) / 1978 + ( x - 1999)/ 1980 = ( x - 1999)/29 + ( x - 1999) / 27 + ( x - 1999 ) / 25 + ( x - 1999) / 23 + ( x - 1999)/21 + ( x - 1999) / 19

< = > ( x - 1999 ) / 1970 + ( x - 1999 ) / 1972 + ( x - 1999 ) / 1974 + ( x - 1999) / 1976 + ( x - 1999) / 1978 + ( x - 1999) / 1980 - ( x - 1999) / 29 - ( x - 1999)/ 27 - ( 1 - 1999) / 25 - ( x-1999) / 23 - ( x - 1999) / 21 - ( x - 1999) / 19 = 0 ( chuyển vế )

< = > ( x - 1999 ) ( 1/1970 + 1/ 1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 - 1/27 - 1/25 - 1/23 - 1/21 - 1/19) = 0

Vì ( 1/1970 + 1/1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 -1/27 - 1/25 - 123 - 1/21 - 1/19 ) khác 0 nên để đẳng thức bằng 0 thì bắt buộc x - 1999 = 0

< = > x = 0 + 1999 = 1999

Vậy tập nghiệm của phương trình là S = { 1999 }

b) Đặt x² + x + 1 = t > 0 (dễ c/m t > 0 rồi ha)

Khi đó, pt tương đương: \(t\left(t+1\right)=12\Leftrightarrow t^2+t-12=0\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-4\left(L\right)\end{matrix}\right.\)

t = 3 suy ra \(x^2+x+1=3\Leftrightarrow x^2+x-2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy...

c) Chị xem lại đề giúp em ạ.