Tìm x:
a) Xx6+X /3 =1/2
b)X/24=5/9
Lưu ý: dấu / ở /3 là chia ba nha còn lại là phần
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Ta có:
\(\left(\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\right)+-\frac{1}{2}=\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\)\(-\frac{1}{2}\)
=\(\frac{6}{30}+\frac{10}{30}+\frac{9}{30}-\frac{15}{30}=\frac{6+10+9-15}{30}=\frac{10}{30}=\frac{1}{3}\)
a, Áp dụng t/c dtsbn:
\(5x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{y-x}{5-7}=\dfrac{2}{-2}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-7\\y=-5\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{y}=\dfrac{7}{2}\Rightarrow\dfrac{x}{7}=\dfrac{y}{2}=\dfrac{x+y}{7+2}=\dfrac{-27}{9}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-21\\y=-6\end{matrix}\right.\)
c, \(\dfrac{x}{32}=\dfrac{2}{x}\Rightarrow x^2=2\cdot32=64\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
d, \(\left|x+\dfrac{1}{3}\right|-2=\dfrac{1}{2}\Rightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{5}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{5}{2}\\x+\dfrac{1}{3}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=-\dfrac{17}{6}\end{matrix}\right.\)
\(A.12:\left(x+4\right)=2\)
\(\Rightarrow x+4=12:2\)
\(\Rightarrow x+4=6\)
\(\Rightarrow x=2\)
\(B,12:\left(x+4\right).3+18=24\)
\(12:\left(x+4\right).3=6\)
\(\Rightarrow12:\left(x+4\right)=2\)
\(\Rightarrow x+4=6\)
\(\Rightarrow x=2\)
a) 12 : ( x + 4 ) = 2
x + 4 = 12 : 2
x + 4 = 6
x = 6 - 4
x = 2
Vậy x = 2
b) 12 : ( x + 4 ) . 3 + 18 = 24
12 : ( x + 4) . 3 = 24 - 18
12 : ( x + 4 ) . 3 = 6
12 : ( x + 4 ) = 6 : 3
12 : ( x + 4 ) = 2
x + 4 = 12 : 2
x + 4 = 6
x = 6 - 4
x = 2
Vậy x = 2
Học tốt !!!
=))
1/2-(4/12+9/12)<x<1/24-(3/24-8/24)
1/2-13/12<x<1/24-(-5/24)
-7/12<x<1/4
=>x\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) E{0}
ta có:\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)=\frac{-1}{12}=-0,08333333\)
mà \(\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)=\frac{1}{4}=0.25\)
nên suy ra không có số nguyên x nào thỏa mãn đề bài.
a) Ta có: \(\frac{3+x}{5+y}=\frac{3}{5}\)
=> (3 + x).5 = 3(5 + y)
=> 15 + 5x = 15 + 3y
=> 5x = 3y
=> x = 3/5y
Mà x + y = 16
hay 3/5y + y = 16
=> (3/5 + 1).y = 16
=> 8/5.y = 16
=> y = 16 : 8/5
=> y = 10
=> x = 16 - 10 = 6
Vậy x = 6; y = 10
b) Ta có: \(\frac{x-7}{y-6}=\frac{7}{6}\)
=> (x - 7).6 = 7.(y - 6)
=> 6x - 42 = 7y - 42
=> 6x = 7y
=> x = 7/6y
Mà x - y = -4
hay 7/6y - y = -4
=> 1/6y = -4
=> y = -4 : 1/6
=> y = -24
=> x = -4 - 24 = -28
Vậy x = -28; y = -24
2 : \(buổi \) \(sáng\) \(bán\) \(dc :\)
\((360 - 142 : 2 =109 l\)
\(buổi\) \(chiều\) \(bán\) \(dc :\)
\(360 - 109 = 251 l\)
\(1,\)
\(a,x\times\dfrac{3}{9}=\dfrac{9}{15}\) \(b,x:\dfrac{1}{2}=\dfrac{5}{6}\)
\(x=\dfrac{9}{15}:\dfrac{3}{9}\) \(x=\dfrac{6}{5}\times\dfrac{1}{2}\)
\(x=\dfrac{81}{45}=\dfrac{9}{5}\) \(x=\dfrac{6}{10}=\dfrac{3}{5}\)
\(2,\) có bạn làm rồi nhé ;>
\(3,\)
\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\left(\dfrac{3}{4}\times\dfrac{2}{9}\right)=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{7+2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\left(\dfrac{4}{15}:\dfrac{2}{5}\right)=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{8-6}{9}=\dfrac{2}{9}\)