\(a,=2\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =2\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{4+\sqrt{5}-1}\\ =2\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\\ =2\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\\ =2\left(\sqrt{5}-1\right)^2=2\left(6-2\sqrt{5}\right)=12-4\sqrt{5}\\ b,=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\\ =\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\\ =\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\\ =32-8\sqrt{15}+8\sqrt{15}-30=2\)

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{24\cdot12}{24+12}=8\Omega\)

\(I=\dfrac{U}{R}=\dfrac{12}{8}=1,5A\)

\(P=\dfrac{U^2}{R}=\dfrac{12^2}{8}=18W\)

\(Q_{tỏa1}=A_1=U_1\cdot I_1\cdot t=12\cdot\dfrac{12}{24}\cdot1\cdot3600=21600J\)

\(Q_{tỏa2}=A_2=U_2\cdot I_2\cdot t=12\cdot\dfrac{12}{12}\cdot1\cdot3600=43200J\)

nSO3=0,75(mol)

mH2SO4(bđ)= 600.15%=90(g) => nH2SO4(bđ)= 90/98=45/49(mol)

PTHH: SO3 + H2O -> H2SO4

0,75______0,75____0,75(mol)

=> mH2SO4(tổng)= 0,75.98+90= 163,5(g)

mddH2SO4= mSO3+mddH2SO4(bđ)= 0,75.80 + 600= 660(g)

=>C%ddH2SO4(sau)= (163,5/660).100=24,773%

VddH2SO4(sau)= 660/1,25= 528(ml)=0,528(l)

nH2SO4(tổng)= 45/49 + 0,75= 327/196(mol)

=> CMddH2SO4(sau)= 327/196 : 0,528=3,16(M)

1 C

2 D

3 A

4 D

5 D

1-c ,2-d ,3-a, 4-d ,5-d(cho mình cái đúng ,thanks)

1c,2d,3a,4d,5d