\(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{17+12\sqrt{2}}}\\ =\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\\ =\dfrac{\sqrt{5}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{5}+1}{3+2\sqrt{2}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{5}+1\right)\left(3-2\sqrt{2}\right)}{3^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{4\sqrt{10}-6}{9-8}=4\sqrt{10}-6\)

ĐKXĐ: x>=0

Ta có: \(\sqrt{5x}+\sqrt{125x}=30\)

=>\(\sqrt{5x}+5\cdot\sqrt{5x}=30\)

=>\(6\cdot\sqrt{5x}=30\)

=>\(\sqrt{5x}=5\)

=>5x=25

=>x=5

=>a=5

=>Chọn C

Gọi CTHH của muối là $R(NO_3)_n$

TH1 : Nếu chất rắn thu được là kim loại

$R(NO_3)_n \xrightarrow{t^o} R + nNO_2 + \dfrac{n}{2}O_2$

Theo PTHH : $n_{R(NO_3)_n} = n_R$
$\Rightarrow \dfrac{41,125}{R + 62n} = \dfrac{17,5}{R}$

$\Rightarrow R = 107,92n$(loại)

TH2 : Nếu chất rắn thu được là oxit

$2R(NO_3)_n \xrightarrow{t^o} R_2O_n + 2nNO_2 + \dfrac{n}{2}O_2$

Theo PTHH : $n_{R(NO_3)_n} = 2n_{R_2O_n}$

$\Rightarrow \dfrac{41,125}{R + 62n} = \dfrac{17,5}{2R + 16n}.2$

$\Rightarrow R = 32n$

Với n = 2 thì R = 64(Cu)

Vậy muối cần tìm là $Cu(NO_3)_2$

$2Cu(NO_3)_2 \xrightarrow{t^o} 2CuO + 4NO_2 + O_2$

$n_{NO_2} = 0,4375(mol); n_{O_2} = 0,109375(mol)$

Suy ra : $V_{khí} = 22,4.(0,4375 + 0,109375) = 12,25(lít)$

a: ĐKXĐ: x>=0; x<>1

Ta có: \(A=\left(\frac{\sqrt{x}+3}{1-x}-\frac{3}{1-\sqrt{x}}\right):\frac{\sqrt{x}+1}{2}\)

\(=\left(\frac{\sqrt{x}+3}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\frac{3}{1-\sqrt{x}}\right)\cdot\frac{2}{\sqrt{x}+1}\)

\(=\frac{\sqrt{x}+3-3\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\cdot\frac{2}{\sqrt{x}+1}=\frac{-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)^2}\cdot2=-\frac{4\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)^2}\)

\(=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)

b: Thay \(x=7+4\sqrt3=\left(2+\sqrt3\right)^2\) vào A, ta được

\(A=\frac{4\cdot\sqrt{\left(2+\sqrt3\right)^2}}{\left(\sqrt{\left(2+\sqrt3\right)^2}-1\right)\left(\sqrt{\left(2+\sqrt3\right)^2}+1\right)^2}=\frac{4\left(2+\sqrt3\right)}{\left(2+\sqrt3-1\right)\left(2+\sqrt3+1\right)^2}\)

\(=\frac{4\left(2+\sqrt3\right)}{\left(\sqrt3+1\right)\left(3+\sqrt3\right)^2}=\frac{2\left(4+2\sqrt3\right)}{\left(\sqrt3+1\right)^3\cdot3}=\frac{2\left(\sqrt3+1\right)^2}{3\left(\sqrt3+1\right)^3}\)

\(=\frac{2}{3\left(\sqrt3+1\right)}=\frac{2\left(\sqrt3-1\right)}{3\left(\sqrt3+1\right)\left(\sqrt3-1\right)}=\frac{\sqrt3-1}{3}\)

c: A<0

=>\(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}<0\)

=>\(\sqrt{x}-1<0\)

=>\(\sqrt{x}<1\)

=>0<=x<1

\(\dfrac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}+\dfrac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}=\dfrac{\left(\sqrt{7}-\sqrt{2}\right)^2+\left(\sqrt{7}+\sqrt{2}\right)^2}{7-2}\\ =\dfrac{7-2\sqrt{14}+2+7+2\sqrt{14}+2}{5}=\dfrac{18}{5}\)

a: ΔBHA vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HA^2=9^2-5,4^2=51,84=7,2^2\)

=>HA=7,2(cm)

Xét ΔHAB vuông tại H có

\(\sin A=\frac{BH}{BA}=\frac{5.4}{9}=\frac35\) =0,6

\(cosA=\frac{AH}{AB}=\frac{7.2}{9}=\frac45\) =0,8

tan A=\(\frac{BH}{AH}=\frac{5.4}{7.2}=\frac34\) =0,75

cot A=\(\frac{AH}{BH}=\frac{7.2}{5.4}=\frac43\) ≃1,33

Xét ΔABC vuông tại B có

\(cosC=\sin A=0,6\)

\(\sin C=cosA=0,8\)

\(\tan C=\cot A\) ≃1,33

cot C=tan A=0,75

b: ΔBHC vuông tại H

=>\(BH^2+HC^2=BC^2\)

=>\(BH=\sqrt{13^2-12^2}=5\left(\operatorname{cm}\right)\)

Xét ΔBHC vuông tại H có

sin C=\(\frac{BH}{BC}=\frac{5}{13}\) ≃0,38

\(cosC=\frac{CH}{CB}=\frac{12}{13}\) ≃0,92

tan C=\(\frac{BH}{CH}=\frac{5}{12}\) ≃0,42

cot C=\(\frac{CH}{BH}=\frac{12}{5}=2,4\)

Xét ΔBAC vuông tại B có

cosA=sin C≃0,38

sin A=cosC≃0,92

cot A=tan C≃0,42

tan A=cot C=2,4