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a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)
Tự làm nốt và kết luận
b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)
Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)
Vậy ....
Bài làm:
c) \(-\frac{2}{5}+\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=-\frac{7}{6}\)
\(\Leftrightarrow-\frac{2}{5}+\frac{5}{2}-\frac{4}{9}x=-\frac{7}{6}\)
\(\Leftrightarrow\frac{4}{9}x=-\frac{2}{5}+\frac{5}{2}+\frac{7}{6}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{49}{15}\)
\(\Leftrightarrow x=\frac{49}{15}\div\frac{4}{9}\)
\(\Rightarrow x=\frac{147}{20}\)
Vậy \(x=\frac{147}{20}\)
Bài 2:
a) Ta có: \(F=\frac{3x-2}{x+3}=\frac{\left(3x+9\right)-11}{x+3}=3-\frac{11}{x+3}\)
Để F nguyên \(\Rightarrow\frac{11}{x+3}\inℤ\Leftrightarrow x+3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Rightarrow x\in\left\{-14;-4;-2;8\right\}\)
Vậy \(x\in\left\{-14;-4;-2;8\right\}\)thì F nguyên
2b) Tách
\(G=\frac{x^2-2x+4}{x+1}=\frac{x^2+x-3x-3+7}{x+1}=\frac{x\left(x+1\right)-3\left(x+1\right)+7}{x+1}\)
\(=\frac{x\left(x+1\right)}{x+1}-\frac{3\left(x+1\right)}{x+1}+\frac{7}{x+1}=x-3+\frac{7}{x+1}\)
G là số nguyên <=> \(\frac{7}{x+1}\)là số nguyên <=> \(7⋮x+1\)<=> \(x+1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
<=> \(x\in\left\{0;-2;6;-8\right\}\)
a)x-3/x+5=5/7 suy ra 7.(x-3) = 5(x+5)
Tương đương : 7x - 21 = 5x + 25
7x - 5x = 25 + 21 = 46
2x = 46 suy ra : x = 46/2 = 23
Vậy x = 23
2.
a) Ta có:
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b) Ta có:
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)
Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)
Vậy, x = -2004