\(\sqrt{4x^2+4x+5}+\sqrt{8x^2+8x+11}=4-4x^2-4x\)

<=> \(\sqrt{\left(2x+1\right)^2+4}+\sqrt{2\left(2x+1\right)^2+9}=5-\left(2x+1\right)^2\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{\left(2x+1\right)^2+4}\ge2\\\sqrt{2\left(2x+1\right)^2+9}\ge3\end{matrix}\right.\)

=> VT \(\ge\) 5 mà VP \(\le\) 5

Mà VT = VP

=> 2x + 1 = 0

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

Cách 2: bn hiểu cách nào thì lm cách đó nha

\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)

= \(\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

= \(\left(\dfrac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

= \(\dfrac{\left(b-a\right)\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\)

= b-a

A, C NẰM GIỮA

B, B NẰM GIỮA

C, A NẰM GIỮA

TICK NHA

\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}-\sqrt{3}-\sqrt{2}-\sqrt{5}+1\)

= \(\sqrt{2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}-\sqrt{3}-\sqrt{2}-\sqrt{5}+1\)

= \(\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{3}-\sqrt{2}-\sqrt{5}+1\)

= \(\sqrt{2}+\sqrt{3}+\sqrt{5}-\sqrt{3}-\sqrt{2}-\sqrt{5}+1\)

= 1

\(\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right).\sqrt{2}\)

= \(\sqrt{2}.\sqrt{2-\sqrt{3}}+\sqrt{2}.\sqrt{2+\sqrt{3}}\)

= \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

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

= \(\sqrt{\left(1-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)

= \(\sqrt{3}-1+1+\sqrt{3}\)

= 2\(\sqrt{3}\)

\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)

= \(\sqrt{2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)

= \(\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}\)

= \(\sqrt{2}+\sqrt{3}+\sqrt{5}\)

PTHH: 2Al + 6HCl ----> 2AlCl₃ + 2H₂\(\uparrow\) (1)

Mg + 2HCl ----> MgCl₂ +H₂\(\uparrow\) (2)

n_{\(H_2\)} = \(\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

- Gọi số mol của Al là a mol

- Gọi số mol của Mg là b mol

=> 27a + 24b = 18

Theo PTHH: (1) n_{\(H_2\)} = \(\dfrac{3}{2}n_{Al}\) = \(\dfrac{3}{2}.a\left(mol\right)\)

Theo PTHH (2) n_{\(H_2\)} = n_\(Mg\) = b (mol)

Từ đó ta có hệ:

\(\left\{{}\begin{matrix}27a+24b=18\\\dfrac{3}{2}a+b=0,9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,4\\b=0,3\end{matrix}\right.\)

=> m_\(Al\) = 0,4.27 = 10,8 (g)

=> m_\(Mg\) = 0,3.24 = 7,2 (g)

=> %m_Al = \(\dfrac{10,8}{18}.100\%=60\%\)

=> %m_Mg = \(\dfrac{7,2}{18}.100\%=40\%\)

a) PTHH: AgNO₃ + HCl ----> AgCl + HNO₃

m_{\(AgNO_3\)} = \(\dfrac{10\%.510}{100\%}=51\left(g\right)\)

=> n_{\(AgNO_3\)} = \(\dfrac{51}{170}=0,3\left(mol\right)\)

m_HCl = \(\dfrac{91,25.20\%}{100\%}=18,25\left(g\right)\)

b) n_HCl = \(\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

Ta có tỉ lệ: \(\dfrac{0,3}{1}< \dfrac{0,5}{1}\) => AgNO₃ phản ứng hết, HCl dư

Theo PTHH: n_HCl(p/ứ) = n_{\(AgNO_3\)} = 0,3 (mol)

=> n_HCl(dư) = 0,5-0,3 = 0,2 (mol)

=> m_HCl(dư) 0,2.36,5 = 7,3 (g)

c) Theo PTHH: n_\(AgCl\) = n_{\(AgNO_3\)} = 0,3 (mol)

=> m_\(AgCl\) = 0,3.143,5 = 43,05 (g)

d) m_ddsp/ứ = 510 + 91,25 - 43,05 = 558,2 (g)

C%_HCl(dư) = \(\dfrac{7,3}{558,2}.100\%=1,3\%\)

C%_{\(HNO_3\)} = \(\dfrac{18,9}{558,2}.100\%=3,39\%\)

a)PTHH: Mg + H₂SO₄ ----> MgSO₄ + H₂\(\uparrow\)

n_{\(H_2\)} = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PTHH: n_\(Mg\) = n_{\(H_2\)} = 0,3 (mol)

=> m_Mg = 0,3.24 = 7,2 (g)

=> a = 7,2 (g)

Theo PTHH: n_{\(H_2SO_4\)} = n_{\(H_2\)} = 0,3 (mol)

=> m_{\(H_2SO_4\)} = 0,3.98 = 29,4 (g)

=> m_{\(ddH_2SO_4\)} = \(\dfrac{29,4.100\%}{20\%}=147\left(g\right)\)

=> b = 147 (g)

b) m_{\(H_2\)} = 0,3.2 = 0,6 (g)

m_ddsp/ứ = 7,2 + 147 - 0,6 = 153,6 (g)

Theo PTHH: n_{\(MgSO_4\)} = n_{\(H_2\)} = 0,3 (mol)

=> m_{\(MgSO_4\)} = 0,3.120 = 36 (g)

=> C%_{\(MgSO_4\)} = \(\dfrac{36}{153,6}.100\%\approx23,44\%\)

a) PTHH: Fe + 2HCl ----> FeCl₂ + H₂\(\uparrow\)

n_\(Fe\) = \(\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PTHH: n_\(HCl\) = 2n_\(Fe\) = 2.0,1 = 0,2 (mol)

=> m_\(HCl\) = 0,2.36,5 = 7,3 (g)

=> C%_\(HCl\) = \(\dfrac{7,3}{100}.100\%=7,3\%\)

b) Theo PTHH: n_{\(H_2\)} = n_\(Fe\) = 0,1 (mol)

=> m_{\(H_2\)} = 0,1.2 = 0,2 (g)

m_ddsp/ứ = 5,6 + 100 - 0,2 = 105,4 (g)

Theo PTHH: n_{\(FeCl_2\)} = n_\(Fe\) = 0,1 (mol)

=> m_{\(FeCl_2\)} = 0,1.127 = 12,7 (g)

C%_{\(FeCl_2\)} = \(\dfrac{12,7}{105,4}.100\%\approx12,05\%\)