Gỉai phương trình
\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
Giải phương trình: \(a,\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)\(b,\frac{x-5}{x-5}+\frac{x-6}{x-5}+\frac{x-7}{x-5}+...+\frac{1}{x-5}=4\)
a, \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)(1)
ĐKXĐ: \(\hept{\begin{cases}x+9\ne0\\x+10\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-9\\x\ne-10\end{cases}}}\)
(1)\(\Leftrightarrow\frac{9.\left(x+9\right)}{90}+\frac{10.\left(x+10\right)}{90}=\frac{9.\left(x+9\right)}{\left(x+9\right)\left(x+10\right)}+\frac{10.\left(x+10\right)}{\left(x+9\right)\left(x+10\right)}\)
\(\Leftrightarrow9.\left(x+9\right)+10.\left(x+10\right)=9.\left(x+9\right)+10.\left(x+10\right)\)
\(\Leftrightarrow9x+81+10x+100=9x+81+10x+100\)
\(\Leftrightarrow9x+10x-9x-10x=81+100-81-100\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow x\in R\)trừ -9 và -10
\(\frac{X-6}{7}+\frac{X-7}{8}+\frac{X-8}{9}=\frac{X-9}{10}+\frac{X-10}{11}+\frac{X-11}{12}\)
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\)và \(5x+y-2z=28\)
Giải phương trình \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)
\(\Leftrightarrow\frac{9\left(X+9\right)\left(X+9\right)\left(X+10\right)+10\left(X+10\right)\left(X+10\right)\left(X+9\right)}{90\left(X+10\right)\left(X+9\right)}=\frac{9.90\left(X+9\right)+10.90\left(X+10\right)}{90\left(X+10\right)\left(X+9\right)}\)
\(\Rightarrow9\left(X+9\right)^2\left(X+10\right)+10\left(X+10\right)^2\left(X+9\right)=810\left(X+9\right)+900\left(X+10\right)\)
\(\Leftrightarrow\left(9X+90\right)\left(X^2+18X+81\right)+\left(10X+90\right)\left(X^2+20X+100\right)=810X+7290+900X+9000\)
\(\Leftrightarrow\)9X3+162X2+729X+90X2+1620X+7290+10X3+200X2+1000X+90X2+1800X+9000=1710X+16290
\(\Leftrightarrow\)19X3+542X2+5149X+16290=1710X+16290
\(\Leftrightarrow\)19X3+542X2=16290-16290+1710X-5149X
\(\Leftrightarrow\)19X3+542X2=-3439X
\(\Leftrightarrow\)19X3+542X2+3439X=0
RỒI GIẢI TIẾP
Mk nghĩ nên giải theo cách này thì hay hơn ( mk mớp 7 thui nên bài làm mang tính chất tham khảo nhé )
Ta có :
\(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)
\(\Leftrightarrow\)\(\left(\frac{x+9}{10}+1\right)+\left(\frac{x+10}{9}+1\right)=\left(\frac{9}{x+10}+1\right)+\left(\frac{10}{x+9}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+19}{10}+\frac{x+19}{9}=\frac{x+19}{x+10}+\frac{x+19}{x+9}\)
\(\Leftrightarrow\)\(\frac{x+19}{10}+\frac{x+19}{9}-\frac{x+19}{x+10}-\frac{x+19}{x+9}=0\)
\(\Leftrightarrow\)\(\left(x+19\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}\right)=0\)
Xét trường hợp \(x=0\)
\(\Rightarrow\)\(\left(x+19\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}\right)=\left(x+19\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{10}-\frac{1}{9}\right)=\left(x+19\right).0=0\)
( NHẬN )
\(\Rightarrow\) Nếu \(x\ne0\) thì \(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}\ne0\)
Xét trường hợp x nguyên dương ta có :
\(\frac{1}{10}>\frac{1}{x+10}\)
\(\frac{1}{9}>\frac{1}{x+9}\)
\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}>0\)
Xét trường hợp x nguyên âm ta có :
\(\frac{1}{10}< \frac{1}{x+10}\)
\(\frac{1}{9}< \frac{1}{x+9}\)
\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+9}-\frac{1}{x+10}< 0\)
Từ đó suy ra :
\(x+19=0\)
\(\Rightarrow\)\(x=-19\)
Vậy \(x=0\) hoặc \(x=-19\)
TIM X:\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
x = -1
ai tk mk
mk tk lại
hứa luôn
thank nhiều
Cho \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\). Tìm x.
Công mỗi phân số cho 1 .....................
mỗi hạng tử ở 2 vế cộng với 1 (có nghĩa là cộng 2 vế với 3 xong chia đều ra 3 hạng tử mỗi hạng tử cộng với 1)
Sau đó sẽ dẫn đến tất cả các hạng tử đều có chung tử số rồi nhóm tử ra ngoài là được
\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
Cộng mỗi p/s cho 1,ta đc:
\(\frac{x-6}{7}+1+\frac{x-7}{8}+1+\frac{x-8}{9}+1=\frac{x-9}{10}+1+\frac{x-10}{11}+1+\frac{x-11}{12}+1\)
\(\Leftrightarrow\frac{x-6+7}{7}+\frac{x-7+8}{8}+\frac{x-8+9}{9}=\frac{x-9+10}{10}+\frac{x-10+11}{11}+\frac{x-11+12}{12}\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\left(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\right)=0\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\)
=>x+1=0
=>x=-1
Giải phương trình: \(\frac{x+1}{4}-\frac{x+2}{5}+\frac{x+4}{7}-\frac{x+5}{8}+\frac{x+7}{10}-\frac{x+9}{12}=0\) = 0
Giải phương trình:
\(\frac{7}{10}-\frac{5}{8}\cdot x-\frac{7}{6}=\frac{5}{6}\cdot x\)
Ta có: \(\frac{7}{10}-\frac{5}{8}.x-\frac{7}{6}=\frac{5}{6}.x\)
\(\Rightarrow\frac{7}{10}-\frac{7}{6}=\frac{5}{6}x+\frac{5}{8}x\)
\(\Rightarrow\frac{21}{30}-\frac{35}{30}=\left(\frac{5}{6}+\frac{5}{8}\right)x\)
\(\Rightarrow-\frac{7}{15}=\left(\frac{15}{24}+\frac{20}{24}\right)x\)
\(\Rightarrow-\frac{7}{15}=\frac{35}{24}x\)
\(\Rightarrow x=-\frac{7}{15}:\frac{35}{24}\)
\(\Rightarrow x=-\frac{7}{15}.\frac{24}{35}\)
\(\Rightarrow x=-\frac{8}{25}\)
Vậy \(x=-\frac{8}{25}\)
Chuk bạn hok tốt!
\(\frac{x-7}{6}+\frac{x-11}{10}=\frac{x-8}{7}+\frac{x-10}{9}\). Tìm x=____
cho A là nghiệm của phương trình
\(\frac{x-7}{x-8}-\frac{x-8}{x-9}=\frac{x-10}{x-11}-\frac{x-11}{x-12}.\)
Tìm 6A
\(\frac{x-7}{x-8}-\frac{x-8}{x-9}=\frac{x-10}{x-11}-\frac{x-11}{x-12}\)
\(\frac{x-7}{x-8}-\frac{x-8}{x-9}-\frac{x-10}{x-11}+\frac{x-11}{x-12}=0\)
Rồi còn lại làm típ