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a) |x| = 4
\(\left[ {_{x = - 4}^{x = 4}} \right.\)
Vậy \(x \in \{ 4; - 4\} \)
b) |x| = \(\sqrt 7 \)
\(\left[ {_{x = - \sqrt 7 }^{x = \sqrt 7 }} \right.\)
Vậy \(x \in \{ \sqrt 7 ; - \sqrt 7 \} \)
c) ) |x+5| = 0
x+5 = 0
x = -5
Vậy x = -5
d) \(\left| {x - \sqrt 2 } \right|\) = 0
x - \(\sqrt 2 \) = 0
x = \(\sqrt 2 \)
Vậy x =\(\sqrt 2 \)
a) \(x+1^3=2^5-\left(-1^3\right)\)
\(\Rightarrow x+1=33\)
=> x = 32
b) \(3^7-x=1^4-\left(-3^5\right)\)
\(\Rightarrow2187-x=1+243=244\)
=> x = 1943
c: =>x+y-xy=-16
=>x+y-xy-1=-17
=>x(1-y)-(1-y)=-17
=>(1-y)(x-1)=-17
=>(x-1;y-1)=17
=>(x-1;y-1) thuộc {(1;17); (17;1); (-1;-17); (-17;-1)}
=>(x,y) thuộc {(2;18); (18;2); (0;-16); (-16;0)}
b: Tham khảo:
a)
\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{15}}\).
b)
\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)
Vậy \(x = 3\).
c)
\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)
Vậy \(x = \frac{8}{5}\)
d)
\(\begin{array}{l}3,2:x = - \frac{6}{{11}}\\\frac{{16}}{5}:x = - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 88}}{{15}}\).