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Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
Câu 6: Khôg có cau nào đúng
Câu 7: C
Câu 8: B
Câu 9: B
Câu 10: D
a, \(x=\dfrac{3}{4}-1=\dfrac{3}{4}-\dfrac{4}{4}=-\dfrac{1}{4}\)
b, \(x=\dfrac{1}{5}-4=\dfrac{1}{5}-\dfrac{20}{5}=-\dfrac{19}{5}\)
c, \(x=2+\dfrac{1}{5}=\dfrac{10}{5}+\dfrac{1}{5}=\dfrac{11}{5}\)
d, \(x=\dfrac{1}{81}-\dfrac{5}{3}=\dfrac{1}{81}-\dfrac{135}{81}=-\dfrac{134}{81}\)
a. x:(1/2+2/3)=6/5
=>x:7/6=6/5
=>x=6/5*7/6=>x=7/5
b.(x-1/2)-5(x-2/3)=3/2x
=>x-1/2-5x+10/3=3/2x
=>-4x+17/6=3/2x
=>17/6x=3/2x--4x
=>17/6=x(3/2+4)=>17/6=11/2x=>x=17/33
c.-5(x+1/5)-1/2(x-2/3)=x
=>-5x-1-1/2x+1/3=x=>-11/2x-2/3=x
=>-2/3=x+11/2x=>-2/3=x(1+11/2)=>-2/3=13/2x
=>x=-4/39
a) x : (1/2 + 2/3) = 6/5
=> x : 7/6 = 6/5
=> x = 6/5 x 7/6
x = 7/5
b) , c) ko bít hihi
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
a) \(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)
\(\frac{-1}{2}x+5=0\Rightarrow\frac{-1}{2}x=-5\Rightarrow x=10\)
b) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)
\(\Rightarrow3x-5x-x=\frac{1}{5}+\frac{3}{2}+3\)
\(-3x=\frac{47}{10}\)
\(x=\frac{-47}{30}\)
c) \(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}+1-\frac{1}{3}\)
\(-7x=\frac{-1}{6}\)
\(x=\frac{1}{42}\)
d) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(3x=\frac{1}{6}\)
\(x=\frac{1}{18}\)
Học tốt nhé bn!
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
