x-6/7+x-7/8+x-8/9=x-9/10+x-10/11+x-11/12
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\(\Leftrightarrow\frac{x-6}{7}+1+\frac{x-7}{8}+1+\frac{x-8}{9}+1=\frac{x-9}{10}+1+\frac{x-10}{11}+1\)\(+\frac{x-11}{12}+1\) ( cộng 2 vế với 3 )
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\right)\)
\(\Leftrightarrow x=-1\)
\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\)và \(5x+y-2z=28\)
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
\(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)
=>\(\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)=\left(\frac{x+8}{11}+1\right)+\left(\frac{x+10}{9}+1\right)+\left(\frac{x+12}{7}+1\right)\)
=> \(\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}=\frac{x+19}{11}+\frac{x+19}{9}+\frac{x+19}{7}\)
=> \(\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)
\(\Rightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)
=> x + 19 = 0 Vì \(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\ne0\)
=> x = - 19
Bài làm:
Ta có: \(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)
\(\Leftrightarrow\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)-\left(\frac{x+8}{11}+1\right)-\left(\frac{x+10}{9}+1\right)-\left(\frac{x+12}{7}+1\right)=0\)
\(\Leftrightarrow\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)
\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}\right)=0\)
Mà \(\hept{\begin{cases}\frac{1}{15}< \frac{1}{11}\\\frac{1}{13}< \frac{1}{9}\end{cases}\Rightarrow}\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}< 0\)
\(\Rightarrow x+19=0\)
\(\Rightarrow x=-19\)
X = ( 1+10 ) * 10 / 2 = 11 * 10 / 2 = 110 / 2 = 55
X= (2+20) * 10 /2 = 110
X= (1+19)*10/2= 100
1+2+3+4+5+6+7+8+9+10= (10+1)X10 : 2 = 55
2+4+6+8+10+12+14+16+18+20 = (20+2)x10 : 2 = 110
1+3+5+7+9+11+13+15+17+19 = (19+1) x 10 : 2 = 100
\(x-\frac{6}{7}+x-\frac{7}{8}+x-\frac{8}{9}=x-\frac{9}{10}+x-\frac{10}{11}+x-\frac{11}{12}\)
\(x+x+x-x-x-x=\frac{6}{7}+\frac{7}{8}+\frac{8}{9}-\frac{9}{10}-\frac{10}{11}-\frac{11}{12}\)
\(0=\frac{6}{7}+\frac{7}{8}+\frac{8}{9}-\frac{9}{10}-\frac{10}{11}-\frac{11}{12}\)
X triệt tiêu hết ròi! Vậy đề bài yêu cầu tìm gì vậy. Nhưng mà...giá trị của 2 vế ko bằng nhau.
\(\Leftrightarrow\left(\frac{x+1}{7}-1\right)+\left(\frac{x+1}{8}-1\right)+\left(\frac{x+1}{9}-1\right)=\left(\frac{x+1}{10}-1\right)+\left(\frac{x+1}{11}-1\right)+\left(\frac{x+1}{12}-1\right)\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=0\)
\(\text{Vì}\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\ne\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\)\(\Rightarrow\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)