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\(a,\frac{x+15}{x}=\frac{4}{3}\Rightarrow4x=3x+45\Leftrightarrow x=45\)
\(b,\frac{7,5-x}{3,5+x}=\frac{5}{6}\Rightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Rightarrow x=2,5\)
\(c,\frac{x-20}{x-10}=\frac{x+40}{x+70}\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)
\(\Leftrightarrow x^2+50x-1400=x^2+30x-400\)
\(\Leftrightarrow20x=1000\)
\(\Rightarrow x=50\)
a. \(\frac{\left(x+15\right)}{x}=\frac{4}{3}\Leftrightarrow4x=3\left(x+15\right)\Leftrightarrow4x=3x+45\Leftrightarrow x=45\)
Vậy x=45
b. \(\frac{7,5-x}{3,5+x}=\frac{5}{6}\Leftrightarrow5\left(3,5+x\right)=6\left(7,5-x\right)\Leftrightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Leftrightarrow x=2,5\)
Vậy x=2,5
c. \(\frac{x+20}{x-10}=\frac{x+40}{x+70}\Leftrightarrow\left(x+40\right)\left(x-10\right)=\left(x+20\right)\left(x+70\right)\)
\(\Leftrightarrow x^2+30x-400=x^2+90x+1400\Leftrightarrow-60x=-30\Leftrightarrow x=-30\)
Vậy x=-30
a, \(\left(7,5-x\right):\left(3,5+x\right)=5:6\)
\(\Leftrightarrow\frac{7,5-x}{3,5+x}=\frac{5}{6}\)
\(\Leftrightarrow6.\left(7,5-x\right)=5.\left(3,5+x\right)\)
\(\Leftrightarrow45-6x=17,5+5x\)
\(\Leftrightarrow45-17,5=5x+6x\)
\(\Leftrightarrow27,5=11x\)
\(\Leftrightarrow x=\frac{27,5}{11}=2,5\)
Vậy : \(x=2,5\)
b) Tương tự như câu a.
Chúc bạn học tốt nhé !!
a) (x + 15) : x = 4 : 3
=> x : x + 15 : x = \(\frac{4}{3}\)
=> 1 + 15 : x = \(\frac{4}{3}\)
=> 15 : x = \(\frac{4}{3}\)- 1 = \(\frac{1}{3}\)
=> x = 15 : \(\frac{1}{3}\)
=> x = 45
\(\frac{x+15}{x}=\frac{4}{3}\)
\(\Rightarrow3\left(x+15\right)=4x\)
\(3x+45=4x\)
\(4x-3x=45\)
\(x=45\)
\(\frac{7,5-x}{3,5+x}=\frac{x+40}{x+70}\)
\(\Rightarrow\left(7,5-x\right)\left(x+70\right)=\left(3,5+x\right)\left(x+40\right)\)
\(7,5x+525-x^2-70x=3,5x+140+x^2+40x\)
\(4x+385-2x^2=110x\)
\(385-2x^2=106x\)
Đề sai òi bạn ~ nếu cái đề này thì x = 412372022 ...... số vô tỉ =.= bậc lên cũng cũng không được số hữa tỉ nào auto ...
a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
\(a,\left(x+15\right):x=4:3\)
=>\(1+\dfrac{15}{x}=\dfrac{4}{3}\)
=>\(\dfrac{15}{x}=\dfrac{1}{3}\)
=>\(x=3.15=45\)
Vậy x=45
b)\(\dfrac{7,5-x}{3,5+x}=\dfrac{5}{6}\)
\(6\left(7,5-x\right)=5\left(3,5+x\right)\)
=>\(45-6x=17,5+5x\)
=>\(-11x=-27,5\)
=>\(x=2,5\)
Vậy...
c)\(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
=>\(\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)
=>\(x^2+50x-140=x^2+30x-40\)
=>\(20x=100\)
=>\(x=50\)