=>15/4y=3/2

hay y=2/5

\(y\times3+\dfrac{y}{2}+\dfrac{y}{4}=1\dfrac{1}{2}\\ \Rightarrow y\times3+y\times\dfrac{1}{2}+y\times\dfrac{1}{4}=\dfrac{3}{2}\\ \Rightarrow y\times\left(3+\dfrac{1}{2}+\dfrac{1}{4}\right)=\dfrac{3}{2}\\ \Rightarrow y\times\dfrac{15}{4}=\dfrac{3}{2}\\ \Rightarrow y=\dfrac{3}{2}:\dfrac{15}{4}\\ \Rightarrow y=\dfrac{2}{5}\)

2/5

\(\left(x-\dfrac{1}{2}\right).\dfrac{5}{2}=\dfrac{7}{4}-\dfrac{1}{2}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right).\dfrac{5}{2}=\dfrac{5}{4}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{2}\)

=> x = 1

Ta có: \(\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{2}=\dfrac{7}{4}-\dfrac{1}{2}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{2}=\dfrac{5}{4}\)

\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{1}{2}\)

hay x=1

a,

143,8 x 2,25 + 143,8 x 7,75

= 143,8 x ( 2,25 + 7,75 )

= 143,8 x 10

= 1438

b, 12% x 5 + 0,4 x 2 + 2

= 0,12 x 5 + 0,4 x 2 + 2

= 0,6 + 0,8 + 2

= 0,6 + 3

= 3,6

#Giang

143,8 x 2,25 + 143,8 x 7,75

\(= 143 , 8 × ( 2 , 25 + 7 , 75 )\)

\(= 143 , 8 × 10\)

\(=1438\)

\(\text{12 % × 5 + 0 , 4 × 2 + 2}\)

\(= 0 , 12 × 5 + 0 , 4 × 2 + 2\)

\(0 , 6 + 0 , 8 + 2\)

\(= 3 , 4\)

\(\text{143,8 x 2,25 + 143,8 x 7,75 = 143,8 x ( 2,25 + 7,75 ) = 143,8 x 10 = 1438}\)

\(\text{12% x 5 + 0,4 x 2 + 2 = 0,12 x 5 + 0,4 x 2 + 2 = 0,6 + 0,8 + 2 = 0,6 + 3 = 3,4}\)

x-1=1/2

x   =1/2 + 1

x= 3/2

dấu / là phần nha.

ta có như sau

x-1=\(\dfrac{1}{2}\)

x   =\(\dfrac{1}{2}\) + 1

x= \(\dfrac{3}{2}\)

= 2,5 giờ x 7 + 2,5 giờ x 1 + 2,5 giờ x 2

= 2,5 giờ x (1 + 2 + 7)

= 2,5 giờ x 10

= 25 giờ 

HT

áp dụngBĐT cô si ta có

\(\frac{x^2}{y+1}\)+\(\frac{y+1}{4}\)\(\ge\)x

\(\frac{y^2}{z+1}\)+\(\frac{z+1}{4}\)\(\ge\)y

\(\frac{z^2}{x+1}\)+\(\frac{x+1}{4}\)\(\ge\)z

khi đó VT\(\ge\)x+y+z-\(\frac{x+y+z+3}{4}\)=\(\frac{3\left(x+y+z\right)-3}{4}\)

áp dụng BĐT cô si

x+y+z\(\ge\)\(3\sqrt[3]{xyz}\)=3

do đó VT\(\ge\)\(\frac{6}{4}\)=\(\frac{3}{2}\)  (đpcm)