dấu chấm là dấu nhân nha e tại chị cấp 2quen viết dấu chấm nên hơi nhầm xíu e thông cảm

A=\(\dfrac{2020.2022+2020.2}{2024.2022-2024.2}=\dfrac{2020\left(2022+2\right)}{2024\left(2022+2\right)}=\dfrac{2020}{2024}=\dfrac{505}{506}\)

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

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

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

=\(\dfrac{2\sqrt{3}-4}{2-\sqrt{3}}=\dfrac{-2\left(2-\sqrt{3}\right)}{2-\sqrt{3}}=-2\)

a)trong \(\Delta ADB\) vuông tại A có

\(AD^2=DB.DO\Rightarrow DB=\dfrac{AD^2}{DO}=\dfrac{16^2}{12,8}\)=20

OB=DB-DO=20-12,8=7,2

b)trong \(\Delta ADO\) vuông tại O có

\(AO=\sqrt{DA^2-OD^2}=\sqrt{16^2-12,8^2}=9,6\)

trong \(\Delta ADC\) vuông tại D có

\(AD^2=AC.AO\Rightarrow AC=\dfrac{AD^2}{AO}=\dfrac{16^2}{9,6}=26,7\)

c.trong \(\Delta AOB\) vuông tại O có

\(AB=\sqrt{AO^2+OB^2}=\sqrt{9,6^2+7,2^2}=12\)

trong \(\Delta ADC\) vuông tại D có

\(DO.AC=AD.DC\Rightarrow DC=\dfrac{DO.AC}{AD}\)\(=\dfrac{12,8.26,7}{16}=21,36\)

diện tích hình thang ABCD là

\(\dfrac{\left(AB+DC\right)AD}{2}\)\(=\dfrac{\left(12+21,36\right)16}{2}=266,88\left(cm^2\right)\)

ĐKXĐ:2x+27\(\ge\)0\(\Rightarrow x\ge-13,5\)

\(\sqrt{2x+27}-6=6\Rightarrow\sqrt{2x+27}=12\)\(\Rightarrow|2x+27|=144\)

\(\Rightarrow\left[{}\begin{matrix}2x+27=144\\2x+27=-144\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=117\\2x=-171\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=58,5\left(TM\right)\\x=-85,5\left(KTM\right)\end{matrix}\right.\)

vậy:x=58,5

trong \(\Delta ABC\) vuông tại A

AB=AC.tanC=10.tan30=5,77

CB=\(\sqrt{AC^2+AB^2}=\sqrt{10^2+5,77^2}=11,55\)

\(AH.BC=AB.AC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{5,77.10}{11,55}=5\)

\(\widehat{B}=90-\widehat{C}=90-30=60\)

bài 2

a)\(R_{AB}=\dfrac{\left(R1+R2\right)R3}{R1+R2+R3}=\dfrac{\left(12+18\right)20}{12+18+20}=12\)

b)\(I1=\dfrac{U1}{R1}=\dfrac{24}{12}=2\)

mà:R1 nt R2 nên I1=I2=2

U2=I2.R2=2.18=36

U12=U1+U2=24+36=60

ta có:U12 song song U3 nên U=U3=U12=60

a.\(CaSO_3\)

b.HCl

c.\(H_2O\)

d.CaO

e.\(O_2\)

f.\(H_2O\)

g.CaO

h.\(H_2O\)

i.\(H_2SO_4;\)\(Na_2SO_4\)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{61,2}{102}=0,6\left(mol\right)\)

PTHH:\(3H_2SO_4+Al_2O_3\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

TPU:        0,6          0,6

PƯ:         0,6            0,2               0,2            0,2

SPU:         0              0,4              0,2             0,2

a)chất còn dư là \(Al_2O_3\)

\(m_{Al_2O_3dư}=n.M\)=0,4.102=40,8(g)

b)\(m_{Al_2\left(SO_4\right)_3}=n.M\)=0,2.342=68,4(g)

1.\(=\left|3-\sqrt{2}\right|=3-\sqrt{2}\)

2.\(=\left|1+\sqrt{5}\right|=1+\sqrt{5}\)

3.\(=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)

4.\(=\left(4-\sqrt{2}\right)^2\)

5.\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|=2+\sqrt{5}-\left(\sqrt{5}-2\right)\)

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

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

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

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

=9-4=5