Lời giải:

ĐK: $x\geq 0; x\neq 1$

a) \(B=\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{(\sqrt{x}-1)(x+1)}\right):\left(\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(x+1)}+\frac{1}{x+1}\right)\)

\(=\frac{x+1-2\sqrt{x}}{(\sqrt{x}-1)(x+1)}:\frac{\sqrt{x}+1}{x+1}=\frac{(\sqrt{x}-1)^2}{(x+1)(\sqrt{x}-1)}.\frac{x+1}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b)

Với $B=\frac{\sqrt{x}-1}{\sqrt{x}+1}$ thôi thì $0< B< 2$ là không đúng. Bạn cho thử $x=0,5$ sẽ thấy.

A) \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}.3+\frac{1}{45}.3\)

                                                                                   \(< \frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)

B) \(\left(x-5\right).\left(x-y+1\right)=-23\)

=> x - 5 = 1; x - y + 1 = -23 hoặc x - 5 = -1; x - y + 1 = 23 hoặc x - 5 = 23; x - y + 1 = -1 hoặc x - 5 = -23; x - y + 1 = 1

+ Với x - 5 = 1; x - y + 1 = -23 

=> x = 6; x - y = -22

=> x = 6; y = 28

... Bn tự lm típ

Ủng hộ mk nha ^_-

a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)

\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)

\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)

\(\Rightarrow x=100\)

Vậy x = 100

b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)

\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)

\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)

\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

2) 

Ta có:

\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)

( vì \(75⋮25\) )

\(\Rightarrowđpcm\)

giup minh vs mai minh di hoc rui

a,\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

b,Áp dụng câu a:

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)

\(=\frac{1}{x}\)

a)

\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

b) S =\(\frac{1}{x}-\frac{1}{x+5}+\frac{1}{x+5}=\frac{1}{x}\)

Ta có :

x - y - z = 0

\(\Rightarrow\hept{\begin{cases}x=y+z\\y=x-z\\-z=y-x\end{cases}}\)

\(M=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)

\(M=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)

Thay các x , y, z vào đẳng thức M , ta sẽ có :

\(M=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}=\frac{y}{x}.\frac{-z}{y}.\frac{x}{z}=-\frac{z}{z}=-1\)

=> Với x - y - z = 0 (\(\forall x,y,z\ne0\)) thì M = -1