\(C=\frac{x+5}{2x+\left(x-2y\right)}+\frac{y-5}{2y-\left(x-2y\right)}\)

\(=\frac{x+5}{2x+10}+\frac{y-5}{2y-10}=\frac{x+5}{2\left(x+5\right)}+\frac{y-5}{2\left(y-5\right)}=\frac{1}{2}+\frac{1}{2}=1\left(x\ne-5,y\ne5\right)\)

Trả lời :

Ta có :

C = \(\frac{x+5}{3x-2y}+\frac{y-5}{4y-x}\)

C = \(\frac{2\left(x+5\right)}{2\left(3x-2y\right)}+\frac{2\left(y-5\right)}{2\left(4y-x\right)}\)

C = \(\frac{2x+10}{6x-4y}+\frac{2y-10}{8y-2x}\)

Thay x - 2y = 10 . Ta được :

C = \(\frac{2x+x-2y}{6x-4y}+\frac{2y-x-2y}{8y-2x}\)

C = \(\frac{x\left(2+1\right)-2y}{6x-4y}+\frac{y\left(2+2\right)-x}{8y-2x}\)

C = \(\frac{3x-2y}{6x-4y}+\frac{4y-x}{8y-2x}\)

C = \(\frac{1}{2}+\frac{1}{2}\)

C = \(1\)

Vậy C = 1

Hok tốt

a, xy + 2x - y = 9

=> x(y + 2) - y - 2 = 7

=> (x - 1)(y + 2) = 7

lap bang 

b, xy - 5x - y = 8

=> x(y - 5) - y + 5 = 13

=> (x - 1)(y - 5) = 13

c, xy - 5x + y = 8

=> x(y - 5) + y - 5 = 3

=> (x + 1)(y - 5) = 3

Ta có: 104²⁰¹⁵ + 2 < 104²⁰¹⁶ + 2.

=> A = \(\frac{104^{2015}+2}{104^{2016}+2}\)< 1 (1).

Ta có: 104²⁰¹⁶ + 2 > 104²⁰ + 2 > 104²⁰.

=> B = \(\frac{104^{2016}+2}{104^{20}}\) > 1 (2).

Từ (1) và (2) => A < 1 < B => A < B.

Chúc bạn học tốt nhé!

bạn ơi giúp mình sửa mẫu của B thành 104^2017+2 nhé

Thanks bạn nhìu

tu ve hinh :

a, xet tamgiac OCB va tamgiac OCA co : OC chung

goc OBC = goc OAC = 90 do BC | Oy va AC | Ox (GT)

OB = OA (gt)

=> tamgiac OCB = tamgiac OCA (ch - cgv)

=> goc BOC = goc AOC  (dn) ma OC nam giac Ox va Oy 

=> OC la phan giac cua goc xOy (dn)

b, xet tamgiac OBD va tamgiac OAE co : OB = OA (gt)

goc BOD = goc AOE (doi dinh)

goc OBD = goc OAE = 90 do BC | Oy va AC | Ox (GT)

=>  tamgiac OBD = tamgiac OAE  (cgv - gnk)

=> OD = OE (dn)

=> tamgiac ODE can tai O (dn)

c, tu nghi di cau c-g-c

tu  ve hinh :

cau b la vuong goc phai k

a, tamgiac ABC can tai A(gt) => AB = AC va goc ABC = goc ACB (dn)

goc ADB = goc ADC do AD | BC (GT)

=> tamgiac ADB = tamgiac ADC (ch - gn)

=> BD = DC (dn)

b, xet tamgiac BHD va tamgiac CKD co :  BD = DC (Cau a)

goc ABC = goc ACB (cau a)

goc BHD = goc DKC = 90 do HD | AB va HK | AC (gt)

=> tamgiac BHD = tamgiac CKD (ch - gn)

=> HD = DK (dn)

c, xet tamgiac AHD va tamgiac AKD co : AD chung

HD = DK (cau b) 

goc AHD = goc AKD = 90 do HD | AB va HK | AC (gt) 

=> tamgiac AHD = tamgiac AKD  (ch - cgv)

=> tamgiac AHK can tai A (dn)

=> goc AHK = (180 - goc BAC) : 2

tamgiac ABC can tai A (gt) => goc ABC = (180 - goc BAC) : 2

=> goc AHK = goc ABC  2 goc nay dong vi

=> HK // BC (tc)

d, tu ap dung py-ta-go 

bài 2 nữa ạ

tu ve hinh : 

a, AE | AB va AD | AC (gt) => goc DAC = goc BAE = 90 (dn)

goc DAB + goc BAC = goc DAC

goc EAC + goc CAB = goc BAE 

=> goc DAB = goc CAE 

xet tamgiac BDA va tamgiac ECA co : 

AD = AC (gt) va AB = AE (gt)

=> tamgiac BDA = tamgiac ECA  (c - g - c)

=> BD = CE (dn)