\(\frac{x-1}{3}=\frac{-2}{5}\)
ai giúp mình với
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Theo đề bài cho => \(4-\frac{5}{6-\frac{7}{8-\frac{9}{10}}}=4-\frac{5}{x}\)
=> \(x=6-\frac{7}{8-\frac{9}{10}}\)
Vậy x=...
Nếu thế mk đúng thì ủng hộ nha
Ta có:\(2-\frac{3}{4-\frac{5}{6-\frac{7}{8-\frac{9}{10}}}}=2-\frac{3}{4-\frac{5}{6-\frac{7}{\frac{71}{10}}}}=2-\frac{3}{4-\frac{5}{6-\frac{70}{71}}}\)
\(=2-\frac{3}{4-\frac{5}{\frac{356}{71}}}\)
=>x=\(\frac{356}{71}\)
\(\frac{2x+1}{3}=\frac{x-5}{2}\)
\(\Rightarrow2\left(2x+1\right)=3\left(x-5\right)\)
\(\Rightarrow4x+2=3x-15\)
\(\Rightarrow4x-3x=-15-2\)
\(\Rightarrow x=-17\)
Ta có:
\(\frac{2x+1}{3}=\frac{x-5}{2}\)
\(\Rightarrow\left(2x+1\right)2=\left(x-5\right)3\)
\(\Rightarrow4x+2=3x-15\)
\(\Rightarrow4x-3x=-15-2\)
\(\Rightarrow x=-17\)
Vậy x = -17
\(\frac{1}{x-3}=a,\frac{1}{y-4}=b\)
\(hpt\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{3}\\4a-3b=\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{13}{14}\\b=\frac{31}{42}\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{53}{13}\\y=\frac{166}{31}\end{cases}}\)
Đặt m = 1 / x - 3 và n = 1/y - 4
Khi đó ta có hệ m + n = 5/3
4 x x - 3 x n = 3/2
....Bạn tự giải tiếp nhé
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)
=>\(\frac{\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)}{99}=-4\)
=> (x+1)+(x+2)+(x+3)+(x+4)=-4.99=-396
=>4x+10=-396
4x=-406
x=-406:4=-101,5
\(\Leftrightarrow\frac{x^2+4}{8}-1+\frac{x^2+3}{7}-1+\frac{x^2+2}{6}-1=\frac{x^2+1}{5}-1+\frac{x^2}{4}-1+\frac{x^2-1}{3}-1\)
\(\Leftrightarrow\frac{x^2-4}{8}+\frac{x^2-4}{7}+\frac{x^2-4}{6}-\frac{x^2-4}{5}-\frac{x^2-4}{4}-\frac{x^2-4}{3}=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)
\(\Leftrightarrow x^2-4=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
a)\(-\frac{2}{5}+\frac{2}{3}x+\frac{1}{6}x=-\frac{4}{5}\Leftrightarrow\frac{5}{6}x=-\frac{2}{5}\Leftrightarrow x=-\frac{12}{25}\)
Vậy nghiệm là x = -12/25
b)\(\frac{3}{2}x-\frac{2}{5}-\frac{2}{3}x=-\frac{4}{15}\Leftrightarrow\frac{5}{6}x=\frac{2}{15}\Leftrightarrow x=\frac{4}{25}\)
Vậy nghiệm là x = 4/25
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)
Vậy nghiệm là x = -1
\(\frac{x-1}{3}=-\frac{2}{5}\)
\(\Rightarrow\frac{5\left(x-1\right)}{15}=\frac{-6}{15}\)
\(\Rightarrow5\left(x-1\right)=-6\)
\(\Rightarrow x-1=-\frac{6}{5}\)
\(\Rightarrow x=-\frac{6}{5}+1\)
\(\Rightarrow x=-\frac{1}{5}\)
Vậy \(x=-\frac{1}{5}\)
\(\frac{x-1}{3}=-\frac{2}{5}\)
\(\Leftrightarrow\left(x-1\right).5=3.-2\)
\(\Leftrightarrow5x-5=-6\)
\(\Leftrightarrow x=-\frac{1}{5}\)