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a) \(\frac{y}{6}=\frac{2010}{15}\) c) \(x-\frac{1}{3}=\frac{1}{4}\) e)\(5y-1952=2500-1947\)
\(y=\frac{2010}{15}.6\) \(x=\frac{1}{4}+\frac{1}{3}\) \(5y-1952=553\)
\(y=804\) \(x=\frac{7}{12}\) \(5y=553+1952\)
\(5y=2505\)
\(y=2505:5=501\)
b) \(x+\frac{1}{2}=\frac{3}{4}\) c) \(3x+\frac{3}{8}=\frac{1}{2}\)
\(x=\frac{3}{4}-\frac{1}{2}\) \(3x=\frac{1}{2}-\frac{3}{8}\)
\(x=\frac{1}{4}\) \(3x=\frac{1}{8}\)
\(x=\frac{1}{8}:3\)
\(x=\frac{1}{24}\)
f)\(\left(8y-1942\right).1947=\left(240-194,2\right).19470\)
\(\left(8y-1942\right).1947=45,8.19470\)
\(\left(8y-1942\right)=45,8.19470:1947\)
\(8y-1942=45,8.10\)
\(8y-1942=458\)
\(8y=458+1942\)
\(8y=2400\)
\(y=2400:8\)
\(y=300\)
942-2567+2563-1942
=(2563-2567)+(942-1942)
= -4 + (-1000)
=-1004
942-2567+2563-1942=(942-1942)+(2563-2567)
=-1000+(-4)
=-1004
=[ 1942 - 942] + [ 2563+2567 ]
=1000 + 512
= 1512.
k mk nha mk kban
942-2567+2563-1942
= ( 942 - 1942 ) + ( 2567 - 2563 )
= - 1000 + 4
= -1004
Ta có: (8y-1942).1947=(2400-1942).19470
\(\Rightarrow\)8y-1942=458.10
\(\Rightarrow\)8y=4580+1942
\(\Rightarrow\)8y=6522
\(\Rightarrow\)y=6522:8=815,25
\(\left(8.y-1942\right).1947=\left(2400-1942\right).19470\)
\(\Leftrightarrow\left(8.y-1942\right).1947=458.19470\)
\(\Leftrightarrow\left(8.y-1942\right).1947=8917260\)
\(\Leftrightarrow8.y-1942=8917260:1947\)
\(\Leftrightarrow8.y-1942=4580\)
\(\Leftrightarrow8.y=4580+1942\)
\(\Leftrightarrow8.y=6522\)
\(\Leftrightarrow y=6522:8\)
\(\Leftrightarrow y=\frac{3261}{4}\)
~ Rất vui vì giúp đc bn ~