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12 tháng 5 2022
\(a,\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}=\dfrac{2}{3}.\left(\dfrac{5}{4}-\dfrac{3}{4}\right)=\dfrac{2}{3}.\dfrac{2}{4}=\dfrac{1}{3}\)
\(b,2.\left(\dfrac{-3}{2}\right)-\dfrac{7}{2}=-6.\dfrac{1}{2}-7.\dfrac{1}{2}=\left(-6-7\right).\dfrac{1}{2}=-13.\dfrac{1}{2}=\dfrac{-13}{2}\)
\(c,-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}=-\dfrac{3}{4}.\left(\dfrac{68}{13}-\dfrac{36}{13}\right)=-\dfrac{3}{4}.\dfrac{32}{13}=-\dfrac{24}{13}\)
12 tháng 5 2022
a) \(\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}\)
\(=\dfrac{2}{3}.\left(\dfrac{5}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{2}{3}.\dfrac{2}{4}\)
\(=\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{3}\)
b) \(2.\left(\dfrac{-3}{2}\right)^2-\dfrac{7}{2}\)
\(=2.\dfrac{9}{4}-\dfrac{7}{2}\)
\(=\dfrac{9}{2}-\dfrac{7}{2}\)
\(=\dfrac{2}{2}=1\)
c) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
\(=-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
\(=\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
\(=\dfrac{3}{4}.\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)
\(=\dfrac{3}{4}.\dfrac{-104}{13}\)
\(=\dfrac{3}{4}.\left(-8\right)\)
\(=-6\)
\(a.\frac{x-3}{x+7}=\frac{1}{6}\) (đkxđ: \(x+7\neq0\Rightarrow x\neq-7\) )
\(\Rightarrow 6(x - 3) = 1(x + 7)\)
\(\Leftrightarrow 6x - 18 = x + 7\)
\(\Leftrightarrow 6x - x = 7 + 18\)
\(\Leftrightarrow5x=25\Rightarrow x=5\left(TM\right)\)
b. \(\frac{x+4}{x-3}=\frac{x-5}{x+1}\) (đkxđ: \(\begin{cases} x - 3 \neq 0 \\ x + 1 \neq 0 \end{cases} \Leftrightarrow \begin{cases} x \neq 3 \\ x \neq -1 \end{cases}\) )
\(\Rightarrow (x + 4)(x + 1) = (x - 5)(x - 3)\)
\(\Leftrightarrow x^2 + x + 4x + 4 = x^2 - 3x - 5x + 15\)
\(\Leftrightarrow x^2 + 5x + 4 = x^2 - 8x + 15\)
\(\Leftrightarrow x^2 + 5x - x^2 + 8x = 15 - 4\)
\(\Leftrightarrow13x=11\Rightarrow x=\frac{11}{13}\left(TM\right)\)
\(\frac{x-3}{x+7}=\frac{1}{6}\)ĐKXĐ: \(x + 7 \neq 0 \Leftrightarrow x \neq -7\)Ta có:
\(6\cdot (x-3)=1\cdot (x+7)\)
\(6x-18=x+7\)
\(6x-x=7+18\)
\(5x=25\)
\(x=5\text{\ (Tha\ mãn\ ĐKXĐ)}\)Vậy \(x = 5\).
Bài 2: Tìm \(x\), biết:
\(\frac{x+4}{x-3}=\frac{x-5}{x+1}\)ĐKXĐ: \(x \neq 3\) và \(x \neq -1\)Ta có:
\((x+4)(x+1)=(x-3)(x-5)\)
\(x^{2}+x+4x+4=x^{2}-5x-3x+15\)
\(x^{2}+5x+4=x^{2}-8x+15\)
\(x^{2}-x^{2}+5x+8x=15-4\)
\(13x=11\)
\(x=\frac{11}{13}\text{\ (Tha\ mãn\ ĐKXĐ)}\)Vậy \(x = \frac{11}{13}\).