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1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
a) |x - 4| = 6
\(\Rightarrow\) \(\pm\)x = \(\pm\)10
\(\Rightarrow\) x = {10, -10}
b) \(\dfrac{x}{5}=\dfrac{9}{7}\)
\(\Rightarrow7x=45\)
\(\Rightarrow x=\dfrac{45}{7}\)
1.
2,5 : 7,5 = x : \(\dfrac{3}{5}\)
\(\dfrac{1}{3}\) = x : \(\dfrac{3}{5}\)
x = \(\dfrac{1}{3}.\dfrac{3}{5}\)
x = \(\dfrac{1}{5}\)
\(\dfrac{5}{6}:x=20:3\)
x = \(\dfrac{5}{6}:\dfrac{20}{3}\)
x = \(\dfrac{1}{8}\)
1,8 : 1,3 = -2,7 : 5x
\(\dfrac{18}{13}\) = -2,7 : 5x
-2,7 : 5x = \(\dfrac{18}{13}\)
5x = -2,7 : \(\dfrac{18}{13}\)
5x = -1,95
x = -0,39
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
a: =>x/3=-5/2
hay x=-15/2
b: \(\Leftrightarrow\dfrac{7}{3}:x=\dfrac{1}{5}-\dfrac{4}{9}=\dfrac{9-20}{45}=\dfrac{-11}{45}\)
\(\Leftrightarrow x=\dfrac{7}{3}:\dfrac{-11}{45}=\dfrac{7}{3}\cdot\dfrac{-45}{11}=\dfrac{-105}{11}\)
c: \(\Leftrightarrow x=\dfrac{-7}{2}\cdot2=-7\)
d: =>x/27=-1/3+2/9=2/9-3/9=-1/9=-3/27
=>x=-3
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
a, \(2,5:7,5=x:\dfrac{3}{5}\)
\(\Leftrightarrow x:\dfrac{3}{5}=2,5:7,5\)
=> \(x.7,5=\dfrac{3}{5}.2,5\) => \(x=\dfrac{1,5}{7,5}=\dfrac{1}{5}\)
b, \(2\dfrac{2}{3}:x=1\dfrac{7}{9}\)
=> \(x=2\dfrac{2}{3}:1\dfrac{7}{9}\)
=> \(x=\dfrac{3}{2}\)
c, \(\dfrac{5}{6}:x=20:3\)
=> \(x.20=\dfrac{5}{6}.3\) => \(x=\dfrac{2,5}{20}=\dfrac{1}{8}\)
3, Tìm x, biết:
a) 2,5 : 7,5 = x : \(\dfrac{3}{5}\)
=> \(x:\dfrac{3}{5}=\dfrac{1}{3}\)
=> \(x=\dfrac{1}{3}.\dfrac{3}{5}=>x=\dfrac{1}{5}\)
b) \(2\dfrac{2}{3}\) : x = \(1\dfrac{7}{9}\)
=> \(\dfrac{8}{3}:x=\dfrac{16}{9}\)
=> \(x=\dfrac{8}{3}:\dfrac{16}{9}=>x=\dfrac{3}{2}\)
c) \(\dfrac{5}{6}\) : x = 20 : 3
=> \(x=\dfrac{5}{6}:\dfrac{20}{3}=>x=\dfrac{1}{8}\)