\(\frac{x}{13}x\frac{15}{26}=\frac{5}{14}\)
tìm x
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= \(\frac{13}{45}x\left(\frac{15}{26}-1\frac{1}{26}+\frac{11}{26}\right)=\frac{13}{45}x\left(\frac{15}{26}-\frac{27}{26}+\frac{11}{26}\right)=\frac{13}{45}x\left(\frac{27}{26}-\frac{27}{26}\right)=\frac{13}{45}x0=0\)
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\) .Trừ 1 ở mỗi hạng tử của 2 vế ,ta có :
\(\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)\left(x-11\right)=\left(\frac{1}{14}+\frac{1}{15}\right)\left(x-11\right)\)
\(\Rightarrow\left[\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\right]\left(x-11\right)=0\)
\(\frac{1}{12}>\frac{1}{14};\frac{1}{13}>\frac{1}{15}\Rightarrow\frac{1}{12}+\frac{1}{13}>\frac{1}{14}+\frac{1}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\)
\(\Leftrightarrow\frac{x+1}{12}-1+\frac{x+2}{13}-1=\frac{x+3}{14}-1+\frac{x+4}{15}-1\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}-\frac{x-11}{14}-\frac{x-11}{15}=0\)
\(\Leftrightarrow\left(x-11\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Mà: \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\Leftrightarrow\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Mà \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
\(\frac{x-2}{11}+\frac{x-2}{12}+\frac{x-2}{13}-\frac{x-2}{14}-\frac{x-2}{15}=0\)= 0
\(\left(x-2\right).\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
\(\left(x-2\right).\frac{6791}{60060}=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy x = 2
Nhớ jk cho mình nhé! Thank you!!!
Ko cần phải làm vậy đâu nhá
ta có tất cả tử các ps trên đều giống nhau nhưng mẫu lại khác nhau
mà đề bài cho như vậy thì chắc chắn tử có giá trị = 0
từ đó suy ra x-2=0 suy ra x=2
tìm x
\(A=\frac{x+2}{10}+\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
A=\(\frac{x+2}{10}+\frac{x+2}{11}+\frac{x+2}{12}\)\(+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\frac{x+2}{10}+\frac{x+2}{11}+\frac{x+2}{12}\)\(+\frac{x+2}{13}\)-\(\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\left(x+2\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
x+2=0 vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\)khác\(0\)
x=0-2=-2
Vậy x=-2
\(\Leftrightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+324}{5}=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)\(Vì\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
=> x+329=0
=> x = -329
b) \(\Leftrightarrow\left(x+2\right)\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)\(Vì\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)\ne0\)
=> x+2 =0 => x =-2
\(\frac{x}{13}\times\frac{15}{26}=\frac{5}{14}\)
\(\frac{x}{13}=\frac{5}{14}\times\frac{26}{15}=\frac{13}{21}\)
\(\Rightarrow21\times x=13\times13\)
\(\Rightarrow x=169:21=\frac{169}{21}\)
Vậy, .....
\(\frac{x}{13}.\frac{15}{26}=\frac{5}{14}\)
\(\Rightarrow\frac{x}{13}=\frac{5}{14}:\frac{15}{26}\)
\(\Rightarrow\frac{x}{13}=\frac{5}{14}.\frac{26}{15}\)
\(\Rightarrow\frac{x}{13}=\frac{13}{21}\)
\(\Rightarrow x=\frac{13}{21}.13\)
\(\Rightarrow x=\frac{169}{21}\)
Vậy \(x=\frac{169}{21}\)