pt\(\Leftrightarrow\sqrt{\left(3x\right)^2-2.3.2x+2^2}=2023\)

   \(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=2023\)

  \(\Rightarrow\)\(\left|3x-2\right|=2023\)

\(\Rightarrow\)Với x\(\ge\)\(\dfrac{2}{3}\)\(\Leftrightarrow\)3x = 2025 => x = \(\dfrac{2025}{3}\)=675

     Với x \(< \dfrac{2}{3}\)\(\Leftrightarrow\)2 - 3x = 2023 <=> 3x = -2021 <=> x = \(\dfrac{-2021}{3}\)

Vậy pt có tập nghiệm S = \(\left\{675;\dfrac{-2021}{3}\right\}\)

a) = 3\(\sqrt{2}\) - \(10\sqrt{2}\) + 6\(\sqrt{2}\) = -\(\sqrt{2}\)

b) = 7 - \(2\sqrt{21}\)+ 3 + 2\(\sqrt{21}\)=10

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

= \(\dfrac{4\sqrt{5}+4-4\sqrt{5}+4}{5-1}\) =\(\dfrac{8}{4}=2\)

d) = \(\sqrt{2}+5-\sqrt{2}\)  = 5

e)= 3\(\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}\) = 14\(\sqrt{2x}\)

f) =\(\sqrt[3]{3^3}+\sqrt[3]{5^3}-\sqrt[3]{4^3}\)= 3 + 5 - 4 = 4

b)Đkxđ: x \(\ge\)-9

<=> 3.\(\sqrt{x+3}\) \(-\dfrac{1}{4}.4.\sqrt{x+3}\)+ \(\sqrt{x+3}\)=6

<=>3\(\sqrt{x+3}\)=6

<=>\(\sqrt{x+3}\)=2 => x+3 = 4 => x=1

cau nao

?

a)PTHH: Fe + H2SO4 -> FeSO4 + H2

nFe = 5,6/56 = 0,1 mol

theo pthh => nH2 = 0,1 mol

-> Vh2 = 0,1 . 22,4 = 2,24 l

b) Theo pthh 

-> nH2SO4 = 0,1 mol

Đổi 200ml = 0,2 l

-> Ch2so4 = 0,1 / 0,2

= 0,5 M

c) thep PTHH -> nFeSO4 = 0,1 mol

-> mFeSO4 = 0,1 . (56 + 32 + 64)

                   = 0,1 . 152

=15,2 gam

b) \(x=\sqrt{2^2+2.2.\sqrt{3}+\sqrt{3^2}}\)

       \(=\sqrt{\left(2+\sqrt{3}\right)^2}\)

        \(=2+\sqrt{3}\)

\(\Rightarrow P=\dfrac{-1}{2+\sqrt{3}}\)