M(OH)₂ + H₂SO₄-----> MSO₄ + 2H₂O

d= MSO₄/M(OH)₂=\(\frac{M+96}{M+34}=2,069\)

1,069M=96-34.2,069

M=24=Mg

a+b=3600

a=4b+940  ( a>b>940)

=>4b+940+b=3600

=>5b=2660

b=532 loai

khong co 2 so nao nhu vậy

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

\(\left(x-2\right)\left(\frac{x+4}{x^2-2x+3}-\frac{x+1}{\sqrt{x+2}+2}\right)=0\)

+ x=2

+ chiu kho lam cai con lai

\(\frac{\overline{1a75}}{15}\in N\Rightarrow\overline{1a75}⋮3;5\)

hiển nhiên \(\overline{1a74}⋮5\)

\(\overline{1a74}⋮3\) => 1+a+7+5chia hết cho 3

1+a chia hết cho 3

a thuộc{2;5;8}

\(\overline{5ab}=26.\overline{ab}\)

500+\(\overline{ab}\)=26\(\overline{ab}\)

25\(\overline{ab}\)=500

\(\overline{ab}\)=20

\(z^2=\left(x+y\right)^2\le\left(1^2+1^2\right)\left(x^2+y^2\right)\)

\(\Rightarrow x^2+y^2\ge\frac{z^2}{2}\)

Min= \(\frac{z^2}{2}\) khi x=y=x/2

a)  \(AB^2+AC^2=BC^2\left(3^2+4^2=5^2\right)\)nên ABC Vuông tại A

b)\(tínhchấtpg:\frac{AE}{AB}=\frac{EC}{BC}=\frac{AE+EC}{AB+BC}=\frac{AC}{AB+BC}=\frac{4}{3+5}=\frac{1}{2}\)

=>AE=AB/2=3/2  ;EC=BC/2=5/2

+ Tam giac ABE có: tanAEB= AB/AE=2

=> góc AEB=63'26'

=> góc BEC=180-AEB=116'34'

c) không có diều gs như vậy.

a)A1+A2+A3=60+60+60=180=> MAN thảng hàng.

b)2tam giac ANB; ACM có: gócNAB =góc CAM=120

AN=AC; AB=AM(GT)

=> Tam giac ANB=tam giác ACM=> BN=CM

c) không rõ đề