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1.
a, => 21-x+3 < 0
=> 24-x < 0
=> x < 24
b, => 7+x > 0
=> x > -7
c, => x-1 < 0 ; x+2 > 0 ( vì x-1 < x+2 )
=> x < 1 ; x > -2
=> -2 < x < 1
Tk mk nha
c) \(\left(x-7\right).\left(y+2\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-7=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0+7\\y=0-2\end{cases}}\Rightarrow\hept{\begin{cases}x=7\left(TM\right)\\y=-2\left(TM\right)\end{cases}}\)
Vậy \(\left(x;y\right)\in\left\{7;-2\right\}.\)
Chúc bạn học tốt!
a) \(\frac{x}{3}-\frac{10}{21}=-\frac{1}{7}\)
\(\Rightarrow\frac{x}{3}=-\frac{1}{7}+\frac{10}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{3}\)
\(\Rightarrow x=1\)
\(x-25\%=\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+\frac{1}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
c) \(-\frac{5}{6}+\frac{8}{3}+-\frac{29}{6}\le x\le-\frac{1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
b1
A,=\(\frac{35}{54}\) B,=4
b2
A,x=\(\frac{7}{15}\) B=\(\frac{-9}{7}\)
CHÚC BẠN HỌC GIỎI
Mình giải phần 1 ) thôi
\(1)\)
\(a)\frac{3}{2}x-\frac{1}{3}=1-x\)
\(\Rightarrow\frac{3}{2}x+x=1-\frac{1}{3}\)
\(\Rightarrow\frac{5}{2}x=\frac{2}{3}\)
\(\Rightarrow x=\frac{2}{3}:\frac{5}{2}\)
\(\Rightarrow x=\frac{2}{3}.\frac{2}{5}\)
\(\Rightarrow x=\frac{4}{15}\)
b ) \(\left(\frac{1}{3}+x\right)^3=27\)
\(\Rightarrow\frac{1}{3}+x=3\)
\(\Rightarrow x=3-\frac{1}{3}\)
\(\Rightarrow x=\frac{9}{3}-\frac{1}{3}\)
\(\Rightarrow x=\frac{8}{3}\)
Chúc bạn học tốt !!!
a, \(\frac{x}{5}-\frac{2}{3}+2x=\frac{1}{2}\)
\(\frac{6x}{30}-\frac{20}{30}+\frac{60x}{30}=\frac{15}{30}\)
\(\frac{66x-20}{30}=\frac{15}{30}\)
\(\Rightarrow\) 66x - 20 = 15
66x = 15 + 20
66x = 35
x = \(\frac{35}{66}\)
Vậy x = \(\frac{35}{66}\)
b, \(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{7}\)
\(\frac{5}{2}-1+3x=\frac{1}{7}\)
\(\frac{3}{2}+3x=\frac{1}{7}\)
3x = \(\frac{1}{7}-\frac{3}{2}\)
3x = \(\frac{-19}{14}\)
x = \(\frac{-19}{42}\)
Vậy x = \(\frac{-19}{42}\)
Phần c lỗi, chắc như thế này
c, \(4\cdot\left(\frac{1}{2}-x\right)+\frac{1}{2}=\frac{-5}{6}+x\)
\(2-4x+\frac{1}{2}=\frac{-5}{6}+x\)
\(\frac{5}{2}-4x=\frac{-5}{6}+x\)
\(-4x-x=\frac{-5}{6}-\frac{5}{2}\)
\(-5x=\frac{-10}{3}\)
x = \(\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
Chúc bn học tốt
a)\(\frac{5}{3}-\frac{2}{3}\times x=1\)
=>\(\frac{2}{3}\times x=\frac{5}{3}-1\)
=>\(\frac{2}{3}\times x=\frac{2}{3}\)
=>\(x=\frac{2}{3}:\frac{2}{3}\)
=>\(x=1\)
b)\(\frac{1}{2}+\frac{5}{7}:x=\frac{1}{6}\)
=>\(\frac{5}{7}:x=\frac{1}{6}-\frac{1}{2}\)
=>\(\frac{5}{7}:x=-\frac{1}{3}\)
=>\(x=-\frac{1}{3}\times\frac{5}{7}\)
=>\(x=-\frac{5}{21}\)
thank you!