Đề bài - bài 116 trang 32 sbt toán 6 tập 2

Đề bài

Tìm \(y\), biết:

a) \(\displaystyle y + 30\%y = -1, 3\)

b) \(\displaystyle y - 25\% y = {1 \over 2}\)

c) \(\displaystyle 3{1 \over 3}y + 16{3 \over 4} = - 13,25\)

Lời giải chi tiết

a)\(\displaystyle y + 30\%y = -1, 3\)

\((1+30\%) y = -1, 3\)

\(\left(1+\dfrac{30}{100}\right)y=\dfrac{-13}{10}\)

\(\displaystyle \eqalign{
& {{130} \over {100}}y = \dfrac{-13}{10}\cr
& y = {{ - 13} \over {10}}:{{130} \over {100}} \cr
& y = {{ - 13} \over {10}}.{{100} \over {130}} \cr& y = {{ - 13} \over {10}}.{{10} \over {13}}\cr
& y = - 1 \cr} \)

b) \(\displaystyle y - 25\% y = {1 \over 2}\)

\(\displaystyle \left( {1 - 25\% } \right)y = {1 \over 2}\)

\(\left(1-\dfrac{25}{100}\right)y=\dfrac{1}{2}\)

\(\displaystyle \eqalign{
& {{75} \over {100}}y = {1 \over 2} \cr
& y = {1 \over 2}:{{75} \over {100}} \cr
& y = {1 \over 2}.{{100} \over {75}}\cr& y = {1 \over 2}.{{4} \over {3}}\cr
& y= {2 \over 3} \cr} \)

c)\(\displaystyle 3{1 \over 3}y + 16{3 \over 4} = - 13,25\)

\(\displaystyle {10 \over 3}y + {67 \over 4} = \dfrac{-53}{4}\)

\(\displaystyle \eqalign{
& {{10} \over 3}y = {-53 \over 4} - {67 \over 4} \cr& {{10} \over 3}y = {-120 \over 4} \cr
& {{10} \over 3}y = - 30 \cr
& \;y = - 30:{{10} \over 3} \cr
& y = - 30.{{3} \over 10}\cr
& y= - 9 \cr} \)