=>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

 √(3 - √(5)) × (3 + √(5)) / √10+√2 = 2.86142715787

Ta có: \(\dfrac{\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(6+2\sqrt{5}\right)}{4\left(\sqrt{5}+1\right)}\)

\(=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)^2}{4\left(\sqrt{5}+1\right)}=1\)

Ta có: \(\dfrac{\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)

\(=\dfrac{\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)^2}{4\cdot\left(\sqrt{5}+1\right)}\)

\(=\dfrac{4}{4}=1\)

\(\dfrac{8}{3}\)=1\(\dfrac{5}{3}\)                         \(\dfrac{15}{6}\)=1\(\dfrac{9}{6}\)

\(\dfrac{85}{12}\)=7\(\dfrac{1}{12}\)                     \(\dfrac{5}{2}\)=1\(\dfrac{3}{2}\)

Chúc bạn học tốt.

\(\dfrac{8}{3}\rightarrow2\dfrac{2}{3}\)

\(\dfrac{15}{6}=\dfrac{5}{2}\rightarrow2\dfrac{1}{2}\)

\(\dfrac{85}{12}\rightarrow7\dfrac{1}{12}\)

\(\dfrac{5}{2}\rightarrow2\dfrac{1}{2}\)

\(\dfrac{8}{3}\Rightarrow2\dfrac{2}{3}\)        \(\dfrac{15}{6}=\dfrac{5}{2}\Rightarrow2\dfrac{1}{2}\)

\(\dfrac{85}{12}\Rightarrow7\dfrac{1}{12}\)     \(\dfrac{5}{2}\Rightarrow2\dfrac{1}{2}\)

nha bạn

1: Ta có: \(S_1=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2020\right)+2021\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(2019-2020\right)+2021\)

\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2021\)

\(=-1\cdot1010+2021\)

\(=-1010+2021=1011\)

2) Ta có: \(S_2=\left(-2\right)+4+\left(-6\right)+8+...+\left(-2014\right)+2016\)

\(=\left(-2+4\right)+\left(-6+8\right)+...+\left(-2014+2016\right)\)

\(=2+2+...+2\)

\(=2\cdot504=1008\)

Cho mình cảm ơn bạn NGUYỄN LÊ PHƯỚC THỊNH nhé 

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

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

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

=> 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

f(x)=3-x-a

nghiệm đa thức bằng 2 ⇒ x=2

⇒f(2)=3-2-a=0

        ⇒1-a=0

        ⇒a=1

Ta có: nghiệm đa thức bằng 2 thì f(x) = 0 

\(\Rightarrow\) f(2) = 3 - 2 - a = 0

f(2) = 1 - a  = 0

\(\Rightarrow\)a  = 1 - 0 = 1

Vậy a = 1 để nghiệm của đa thức f(x) = 3 - x - a có nghiệm là 2

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

\(= 143 , 8 × ( 2 , 25 + 7 , 75 )\)

\(= 143 , 8 × 10\)

\(=1438\)

\(\text{12 % × 5 + 0 , 4 × 2 + 2}\)

\(= 0 , 12 × 5 + 0 , 4 × 2 + 2\)

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

\(= 3 , 4\)

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

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

\(\left(x-4\right)^4=\left(x-4\right)^2\\ \Rightarrow\left(x-4\right)^2\left[\left(x-4\right)^2-1\right]=0\\ \Rightarrow\left(x-4\right)\left(x-4-1\right)\left(x-4+1\right)=0\\ \Rightarrow\left(x-4\right)\left(x-5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\\x=5\end{matrix}\right.\)