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

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

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

=> 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

các bạn giúp mink với mink cầm gấp

bạn ghi + - x : nhá

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

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

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

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

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= \(\dfrac{3}{2}\)

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

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

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

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

\(\text{a ) 15 : ( - 5 ) . ( - 3 ) - 8}\)

\(=\left(-3\right).\left(-3\right)-8\)

\(=9-8\)

\(=1\)

\(\text{b ) ( 6 . 8 - 10 : 5 ) + 3 . ( -7)}\)

\(=\left(48-2\right)+\left(-21\right)\)

\(=46+\left(-21\right)\)

\(=25\)

học tốt

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