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A) 3\7 + 5\9 +4\7+ 4\9=(3\7+4\7) + ( 5\9 +4\9) = 7\7+9\9=63\63+63\63=126\63=42\21
b) 1\5+4\11+4\5+ 7\11=(1\5+4\5)+ (4\11+7\11)=5\5+11\11=55\55+55\55=110\55=22\11
a) \(x-\frac{4}{9}=\frac{5}{6}\)
\(\Leftrightarrow\frac{18x}{18}-\frac{8}{18}=\frac{15}{18}\)
\(\Rightarrow18x-8=15\)
\(\Leftrightarrow18x=15+8\)
\(\Leftrightarrow18x=23\)
\(\Leftrightarrow x=\frac{23}{18}\)
Vậy \(x=\frac{23}{18}\)
b)\(x-7=\frac{1}{6}\)
\(\Leftrightarrow\frac{6x}{6}-\frac{42}{6}=\frac{1}{6}\)
\(\Rightarrow6x-42=1\)
\(\Leftrightarrow6x=42+1\)
\(\Leftrightarrow6x=43\)
\(\Leftrightarrow x=\frac{43}{6}\)
Vậy \(x=\frac{43}{6}\)
c)\(x-\frac{7}{8}=11\)
\(\Leftrightarrow\frac{8x}{8}-\frac{7}{8}=\frac{88}{8}\)
\(\Rightarrow8x-7=88\)
\(\Leftrightarrow8x=88+7\)
\(\Leftrightarrow8x=95\)
\(\Leftrightarrow x=\frac{95}{8}\)
Vậy \(x=\frac{95}{8}\)
#bn cs thể bỏ\(\Leftrightarrow;\Rightarrow\)đều đc nhé #
#hoktot<3#
a, x - 4/9 = 5/6
x = 5/6 + 4/9
x = 69/54
b, x - 7 = 1/6
x = 1/6 + 7
x = 43/6
c, x - 7/8 = 11
x = 11 + 7/8
x = 95/8
Bài 1:
a, \(\frac{1}{7}+3+\frac{5}{6}=\frac{1}{7}+\frac{3}{1}+\frac{5}{6}=\frac{6}{42}+\frac{126}{42}+\frac{35}{42}=\frac{167}{42}\)
b, \(\frac{4}{7}+\frac{5}{9}+4=\frac{4}{7}+\frac{5}{9}+\frac{4}{1}=\frac{36}{63}+\frac{35}{63}+\frac{252}{63}=\frac{323}{63}\)
a] 4/12 ; 5/12 ; 11/2 ; 1/4
b] 1 ; 9/12; 12/5 ; 11/3
c] 8/21; 6/11;8/7
d] 2/3 ;2/7 15/2
a]1/3 5/12 11/2 1/4
b]1 3/4 12/5 33/9
c]8/21 6/11 8/7
d]2/3 2/7 15/2
a) \(1678\cdot24+77\cdot1678-1678=1678\left(24+77-1\right)=167800\)
b) Ta có: \(\dfrac{2}{5}+\dfrac{4}{11}+\dfrac{3}{5}+\dfrac{18}{11}+\dfrac{1}{4}+\dfrac{1}{6}\)
\(=\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{4}{11}+\dfrac{18}{11}\right)+\dfrac{1}{4}+\dfrac{1}{6}\)
\(=3+\dfrac{1}{4}+\dfrac{1}{6}\)
\(=\dfrac{36}{12}+\dfrac{3}{12}+\dfrac{2}{12}=\dfrac{41}{12}\)
a,3/7+4/5+4/7.
= ( 3/7 + 4/7 ) + 4/5
= 1 + 4/5
= 9/5
b,6/11+1/3+5/11
= ( 6/11 + 5/11 ) + 1/3
= 1 + 1/3
= 4/3
#hien#
a,3/7+4/5+4/7.
= ( 3/7 + 4/7 ) + 4/5
= 1 + 4/5
= 9/5
b,6/11+1/3+5/11
= ( 6/11 + 5/11 ) + 1/3
= 1 + 1/3
= 4/3
*Ryeo*