Tìm x,y biết:
a. \(\dfrac{x}{9}=\dfrac{4}{x}\) b. \(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)
Tìm x, biết:
a) \(\dfrac{2}{5}\) + \(\dfrac{3}{4}\): x = \(\dfrac{-1}{2}\)
b) \(\dfrac{5}{7}\) - \(\dfrac{2}{3}\) . x = \(\dfrac{4}{5}\)
c) \(\dfrac{1}{2}\) x + \(\dfrac{2}{3}\) x = \(\dfrac{-2}{3}\)
d) \(\dfrac{4}{7}\)x - x= \(\dfrac{-9}{14}\)
a, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
c và d mình làm dược nhưng ko ghi được cái suy ra
Tìm x,y biết :
a) \(\left|3.x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}.y+\dfrac{3}{5}\right|\)= 0
b)\(\left|\dfrac{3}{2}.x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}.y-\dfrac{1}{2}\right|\le0\)
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
Tìm x, y, z, t ∈ Z biết:
a, \(\dfrac{5}{x}=\dfrac{-10}{12}\) b, \(\dfrac{4}{-6}=\dfrac{x+3}{9}\) c, \(\dfrac{x-1}{25}=\dfrac{4}{x-1}\) d, \(\dfrac{x+1}{y}=\dfrac{-3}{5}\)
e, \(\dfrac{-12}{6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{Z}{-17}=\dfrac{-t}{-9}\)
h, \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{Z^3}{-2}\)
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
Tìm x, biết:
a) \(\dfrac{2}{3}\)x - \(\dfrac{1}{2}\)x = \(\left(-\dfrac{7}{12}\right)\) . \(1\dfrac{2}{5}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2\) = \(\dfrac{9}{4}\)
c) (1,25 - \(\dfrac{4}{5}\)x)3 = -125
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
Tìm các số nguyên x,y biết :
a). \(\dfrac{x}{2}\)=\(\dfrac{-5}{y}\). b). \(\dfrac{3}{x}\)=\(\dfrac{y}{4}\), trong đó x > y > 0.
c). \(\dfrac{3}{x-1}\)= y+1. d). \(\dfrac{x+2}{5}\)=\(\dfrac{1}{y}\).
a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
x | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
y | -10 | 10 | -5 | 5 | -2 | 2 | -1 | 1 |
c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
x - 1 | 1 | -1 | 3 | -3 |
y + 1 | 3 | -3 | 1 | -1 |
x | 2 | 0 | 4 | -2 |
y | 2 | -4 | 0 | -2 |
b: =>xy=12
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
Cho \(A=\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\); \(B=\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\) \(\left(x\ge0;x\ne9\right)\). Biết \(C=\dfrac{B}{A}\). Tìm \(x\in Z\) để \(C< \dfrac{-1}{3}\).
\(C=\left(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}+4}\)
\(=\dfrac{-3}{2\sqrt{x}+4}\)
Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)
\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)
\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)
hay \(0\le x< \dfrac{25}{4}\)
làm đầy đủ theo các bước nhé
Tìm x biết :
a) \(^{\dfrac{4}{9}+x=\dfrac{5}{3}}\)
b)\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
c) \(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
d)\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)
a)
\(\frac{4}{9} + x = \frac{5}{3}\)
=> \(x = \frac{5}{3}-\frac{4}{9}\)
=> \(x = \) \(\frac{11}{9}\)
Vậy \(x = \dfrac{11}{9}\)
b)
\(\dfrac{3}{4} .x = \dfrac{-1}{2}\)
=> \(x = \dfrac{-1}{2} : \dfrac{3}{4}\)
=> \(x = \dfrac{-2}{3}\)
Vậy \(x = \dfrac{-2}{3}\)
c)
\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)
=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)
=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)
=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)
=> \(x = \dfrac{-15}{2}\)
Vậy \(x = \dfrac{-15}{2}\)
d)
\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)
=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)
=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)
=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)
=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)
=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)
=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)
=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)
Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }
Cho \(A=\dfrac{2\sqrt{x}+4}{\sqrt{x}-3}\) và \(B=\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\) (\(x\ge0;x\ne9\))
a, Rút gọn B.
b, Biết \(C=\dfrac{B}{A}\). Tìm \(x\in Z\) để \(C< -\dfrac{1}{3}\).
a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3}{\sqrt{x}-3}\)
Tìm x, biết :
\(a.3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)
\(b.\left(x+1\right):\dfrac{5}{6}=20:3\)
\(c.\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
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. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)