tam giác ABC có : 

góc ABC + góc ACB + góc BAC = 180 (Đl)

góc BAC = 60 (gt); góc ACB = 50 (gt)

=> góc ABC = 180 - 60 - 50 = 70

BD là phân giác của góc ABC (gt) => góc ABD = góc ABC : 2 (đl)

=> góc ABD = 70 : 2 = 35 

xét tam giác ABD có : 

góc A + Góc ABD + góc ADB = 180 (Đl)

có góc A = 60 (gt); góc ABD = 35 (Cmt)

=> góc ADB = 180 - 60 - 35 = 95

góc ADB + góc CDB = 180 (kb)

=> góc CDB = 180 - 95 = 85 

giúp ik mn

\(\frac{2x+3}{3x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(3x+2\right)\left(4x+5\right)\)

\(\Leftrightarrow20x^2+34x+6=12x^2+23x+10\)

\(\Leftrightarrow8x^2+11x-4=0\)

Ta có: \(\Delta=11^2+4.4.8=249,\sqrt{\Delta}=\sqrt{249}\)

Vậy pt có 2 nghiệm:

\(x_1=\frac{-11+\sqrt{249}}{16}\);\(x_2=\frac{-11-\sqrt{249}}{16}\)

a) \(A=\frac{5}{\sqrt{x}+1}\)

A nguyên\(\Leftrightarrow\frac{5}{\sqrt{x}+1}\)nguyên\(\Leftrightarrow5⋮\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Mà \(\sqrt{x}+1\ge1\)nên \(\sqrt{x}+1\in\left\{1;5\right\}\)

\(TH1:\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

\(TH2:\sqrt{x}+1=5\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)

b) \(B=\frac{7}{\sqrt{x}-3}\)

A nguyên \(\Leftrightarrow\frac{7}{\sqrt{x}-3}\)nguyên\(\Leftrightarrow7⋮\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow\sqrt{x}-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

c) \(C=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}\)

\(=1+\frac{4}{\sqrt{x}-3}\)

C nguyên\(\Leftrightarrow\frac{4}{\sqrt{x}-3}\in Z\Leftrightarrow4⋮\sqrt{x}-3\)

Tương tự hai câu a,b

d) \(D=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{\sqrt{x}-1+3}{\sqrt{x}-1}\)

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

D nguyên\(\Leftrightarrow\frac{3}{\sqrt{x}-1}\)nguyên

Tương tự