ta có

x+y+y+z+z+x=\(\frac{13}{12}\)

2(x+y+z)=\(\frac{13}{12}\)

=>x+y+z=\(\frac{13}{24}\)

z=(x+y+z)-(x+y)

y=y+z-z

x=x+Y-y

=\(\frac{96}{1000}\)

\(\frac{x+1}{97}\) + \(\frac{x+1}{98}\) - \(\frac{x+1}{99}\) - \(\frac{x+1}{100}\) \(\Leftrightarrow\) (x+1).(1/97 + 1/98 - 1/99 - 1/100) . Vì (1/97 = 1/ 98 - 1/99 - 1/100) \(\ne\) 0 \(\Rightarrow\) x+ 1= 0 \(\Leftrightarrow\) x= -1

\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=0\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)

\(\Leftrightarrow x+100=0\text{ (do }\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\ne0\text{)}\)

\(\Leftrightarrow x=-100\)

bn tự chép đề lại nha

từ đề bài suy ra  \(1+\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+3=1+1+1+0=3\)

suy ra \(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+3=3\)

suy ra \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=3-3=0\)

suy ra \(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

mà 1/99 +1/98+1/97 lớn hơn 0

từ 2 điều trên suy ra x+100=0 suy ra x=-100

(1-1/97),(1-1/98).....(1-1/1000)

=96/97.97/98.....999/1000

=(96.97...999)/(97.98...1000)

=96/1000=12/125

Dấu . Là dấu nhân nhé

=\(\frac{96}{97}\)x\(\frac{97}{98}\)x...x\(\frac{999}{1000}\)

triệt tiêu đi ta có:

=\(\frac{96}{1000}\)=\(\frac{12}{125}\)

tick nhé!!

Bài này chắc tác giả đánh sai tử thức của phân thức cuối cùng, biểu thức B phải là  \(B=\frac{1}{C}\)  trong đó \(C=\left(\frac{\sqrt{x}+3}{x+\sqrt{x}+1}-\frac{\sqrt{x}-3}{x\sqrt{x}-1}\right)\cdot\frac{x\sqrt{x}-\sqrt{x}+x^2-1}{\sqrt{x}}\)
Ta có \(C=\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\frac{\left(\sqrt{x}-3\right)}{x\sqrt{x}-1}\right)\cdot\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{x+\sqrt{x}}{x\sqrt{x}-1}\cdot\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\left(\sqrt{x}+1\right)^2\)

Thành thử ta được \(C=\left(\sqrt{x}+1\right)^2\)

 Ta có \(x=98+20\sqrt{6}=\left(5\sqrt{2}+4\sqrt{3}\right)^2\to\sqrt{x}=5\sqrt{2}+4\sqrt{3}\to\)

hay \(C=\left(\sqrt{x}+1\right)^2=x+2\sqrt{x}+1=99+20\sqrt{6}+2\left(5\sqrt{2}+4\sqrt{3}\right)\)

\(=99+20\sqrt{6}+10\sqrt{2}+8\sqrt{3}\to B=\frac{1}{C}=\frac{1}{99+20\sqrt{6}+10\sqrt{2}+8\sqrt{3}}\)

\(\left(1-\frac{1}{97}\right)x\left(1-\frac{1}{98}\right)x...x\left(1-\frac{1}{1000}\right)\)
\(\frac{96}{97}\cdot x\cdot\frac{97}{98}\cdot x\cdot...\cdot x\cdot\frac{999}{1000}\)
\(\frac{96}{97}\cdot\frac{97}{98}\cdot...\cdot\frac{999}{1000}\cdot x^{903}\)
\(\frac{96}{1000}\cdot x^{903}\)
\(\frac{12}{125}\cdot x^{903}\)

minh @gmail.com.vn

minh @ gmail.com.vn

nha

Vậy x+1;x+2;x+3;x+4=0

=> x=-1;-2;-3;-4

nha bạn        

\(\frac{x+1}{99}+1\frac{x+2}{98}+1\frac{x+3}{97}+1\frac{x+4}{96}+1=0\)

\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(x+100=0\)

\(x=-100\)