TÌM SỐ NGUYÊN X BIẾT :
a)|x+1|=3
b)|x|+5=7
c)3<x<6
d)|x|=12 và x>0
e)|x|=5 và x<0
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Bài 3:
a: \(x\in\left\{-5;-4;-3;-2;-1\right\}\)
b: \(x\in\left\{-3;-2;-1;0;1;2;3;4;5;6\right\}\)
a) \(\dfrac{3}{5}+x=\dfrac{4}{3}\)
\(\Leftrightarrow x=\dfrac{4}{3}-\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{20}{15}-\dfrac{9}{15}\)
\(\Leftrightarrow x=\dfrac{11}{15}\)
b) \(x+\dfrac{5}{6}=\dfrac{1}{7}\)
\(\Leftrightarrow x=\dfrac{1}{7}-\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{6}{42}-\dfrac{35}{42}\)
\(\Leftrightarrow x=-\dfrac{29}{42}\)
c) \(x\times\dfrac{3}{8}=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}:\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{4}\times\dfrac{8}{3}\)
\(\Leftrightarrow x=2\)
d) \(\dfrac{4}{5}\times x=\dfrac{3}{7}\)
\(\Leftrightarrow x=\dfrac{3}{7}:\dfrac{4}{5}\)
\(\Leftrightarrow x=\dfrac{3}{7}\times\dfrac{5}{4}\)
\(\Leftrightarrow x=\dfrac{15}{28}\)
a, \(\dfrac{3}{5}+x=\dfrac{4}{3}\Leftrightarrow x=\dfrac{4}{3}-\dfrac{3}{5}=\dfrac{20-9}{15}=\dfrac{11}{15}\)
b, \(x+\dfrac{5}{6}=\dfrac{1}{7}\Leftrightarrow x=\dfrac{1}{7}-\dfrac{5}{6}=\dfrac{6-35}{42}=\dfrac{-29}{42}\)
c, \(\dfrac{3x}{8}=\dfrac{3}{4}\Leftrightarrow\dfrac{3x}{8}=\dfrac{6}{8}\Rightarrow3x=6\Leftrightarrow x=2\)
d, \(\dfrac{4x}{5}=\dfrac{3}{7}\Leftrightarrow\dfrac{28x}{35}=\dfrac{15}{35}\Rightarrow x=\dfrac{15}{28}\)
a) Tìm x
\(6-\left(x-\frac{1}{3}\right)^2=\frac{2^{2013}}{\left(-2\right)^{2012}}\Rightarrow6-\left(x-\frac{1}{3}\right)^2=\frac{2^{2013}}{2^{2012}}=2^1=2\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=6-2=4=2^2\Rightarrow\hept{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{7}{3}\\x=\frac{-5}{3}\end{cases}}}\)
Vậy \(x\in\left\{\frac{7}{3};\frac{-5}{3}\right\}\)
b) Ta có : \(2a=3b\Rightarrow\frac{a}{3}=\frac{b}{2}\) và \(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{3}=\frac{b}{2}\Rightarrow\frac{a}{21}=\frac{b}{14}\\\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{14}=\frac{c}{10}\end{cases}}\Rightarrow\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{a+b-c}{21+14-10}=-\frac{50}{25}=-2\)
\(\Rightarrow a=\left(-2\right).21=-42\) \(b=\left(-2\right).14=-28\) \(c=\left(-2\right).5=-10\)
Vậy a = -42 ; b = -28 và c = -10
a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)
b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)
c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-11;1\right)\right\}\)
a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)
b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)
c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-1;11\right)\right\}\)
a, \(\left|x+1\right|=3\)
\(\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\)
\(\orbr{\begin{cases}\Rightarrow x=3-1=2\\\Rightarrow x=-3-1=-4\end{cases}}\)