\(\frac{7x}{2.5}+\frac{7x}{5.8}+\frac{7x}{8.11}+\frac{7x}{11.14}+\frac{7x}{14.17}+\frac{7x}{17.20}=\frac{21}{10}\)
tìm x
\(\frac{7x}{2.5}+\frac{7x}{5.8}+\frac{7x}{8.11}+\frac{7x}{11.14}+\frac{7x}{14.17}+\frac{7x}{17.20}=\frac{21}{10}\)tìm x zùm nha nếu không được thì chắc là đề sai đó
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)
\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
mk đầu tiên đó
\(=1\div3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\frac{9}{20}\)
\(=\frac{3}{20}\)
so sánh A với 1 , biếtA = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
A=...
<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}.\frac{9}{20}\)
\(A=\frac{3}{20}\)
Vì \(\frac{3}{20}< 1\Rightarrow A< 1\)
Tìm \(x\), biết :
\((x+\frac{7x}{9})\left(x+\frac{7x}{33}\right)(x+\frac{7x}{33})...\left(x+\frac{7x}{9200}\right)=\frac{186}{25}\)
Tớ biết làm đúng 100%:
\((x\cdot1+x\cdot\frac{7}{9})\left(x\cdot1+x\cdot\frac{7}{20}\right)...\left(x\cdot1+x\cdot\frac{7}{9200}\right)=\frac{186}{25}\)
\(x\cdot\left(1+\frac{7}{9}\right)\cdot x\left(1+\frac{7}{20}\right)\cdot...\cdot x\left(1+\frac{7}{9200}\right)=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)(\frac{16}{9}+\frac{27}{20}+...+\frac{9207}{9200})=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)\left(\frac{2\cdot8}{1\cdot9}+\frac{3\cdot9}{2\cdot10}+...+\frac{93\cdot99}{92\cdot100}\right)=\frac{186}{25}\)
\(x^{92}\cdot\frac{2\cdot8\cdot3\cdot9\cdot...\cdot93\cdot99}{1\cdot9\cdot2\cdot10\cdot...\cdot92\cdot100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{\left(2\cdot3\cdot...\cdot93\right)\cdot\left(8\cdot9\cdot...\cdot99\right)}{\left(1\cdot2\cdot...\cdot92\right)\cdot\left(9\cdot10\cdot...\cdot100\right)}=\frac{186}{25}\)
\(x^{92}\cdot\frac{93\cdot8}{100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{186}{25}=\frac{186}{25}\)
\(x^{92}=\frac{186}{25}:\frac{186}{25}\)
\(x^{92}=1\Rightarrow x=1\)
cô tớ giải rồi . x=1 (đúng 100%)
Nhanh +Đúng = Tick ( gắp )
\(A=\frac{1}{2.5}\frac{1}{5.8}\frac{1}{8.11}\frac{1}{11.14}\frac{1}{14.17}\frac{1}{17.20}\)
\(B=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
- A ở trên giữa các phân số là dấu " + " nha mấy bạn !
Tính : \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(A=\frac{1}{2}-\frac{1}{17}\)
\(A=\frac{15}{34}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)
\(4+\frac{x}{1+\frac{1}{2+\frac{1}{3}}}=\frac{x}{4+\frac{1}{3+\frac{1}{2}}}\)
\(\Leftrightarrow4+\frac{x}{1+\frac{1}{\frac{7}{3}}}=\frac{x}{4+\frac{1}{\frac{7}{2}}}\)
\(\Leftrightarrow4+\frac{x}{1+\frac{3}{7}}=\frac{x}{4+\frac{2}{7}}\)
\(\Leftrightarrow4+\frac{x}{\frac{10}{7}}=\frac{x}{\frac{30}{7}}\)
\(\Leftrightarrow4+\frac{7x}{10}=\frac{7x}{30}\)
\(\Leftrightarrow\frac{40+7x}{10}=\frac{7x}{30}\)
\(\Leftrightarrow120+21x=7x\)
\(\Rightarrow x=-\frac{60}{7}\)
P/S:Số khá xấu nên ko chắc đâu nha !
1 cách khác nó phức tạp và khó hơn "n" lần :)) Cơ mà nó chẳng khác của cậu là mấy :v
\(4+\frac{x}{1+\frac{1}{2+\frac{1}{3}}}=\frac{x}{4+\frac{1}{3+\frac{1}{2}}}\)
\(\Leftrightarrow4+\frac{x}{1+\frac{1}{\frac{7}{3}}}=\frac{x}{4+\frac{1}{\frac{7}{2}}}\)
\(\Leftrightarrow4+\frac{x}{1+\frac{3}{7}}=\frac{x}{4+\frac{2}{7}}\)
\(\Leftrightarrow4+\frac{x}{\frac{10}{7}}=\frac{x}{\frac{30}{7}}\)
\(\Leftrightarrow4+x.\frac{7}{10}=x.\frac{7}{30}\)
\(\Leftrightarrow4+\frac{7x}{10}=\frac{7x}{30}\)
\(\Leftrightarrow120+21x=7x\)
\(\Leftrightarrow120=7x-21\)
\(\Leftrightarrow120=-14x\)
\(\Leftrightarrow-\frac{120}{14}=-\frac{60}{7}=x\)
\(\Rightarrow x=-\frac{60}{7}\)
Tuấn Huỳnh cách của a có khác gì cách của e đâu.chỉ một bên chọn MSC còn a thì chuyển vế thôi mà
bài của Tuấn Huỳnh khác với bài của cậu chứ zZz Cool Kid zZz
Cho 3 số x , y , z , thỏa mãn :
\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{7x}{y+z}+\frac{7x}{z+x}+\frac{7x}{x+y}=\frac{133}{10}\)
Tính giá trị biểu thức : \(M=\left(x+y+z\right)^2\)
giải phương trình
a)\(\frac{7x+10}{x+1}\left(x^2-x-2\right)=\frac{7x+10}{x+1}\left(2x^2-3x-5\right)\)
b)\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
c)\(x^2+\frac{1}{x^2}+\frac{9x}{2}-\frac{9}{2x}+7=0\)