\(a,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{100}{100}-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(b,\frac{x}{y}=\frac{3}{5}\)

\(\Leftrightarrow\frac{x}{3}=\frac{y}{5}\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\)

\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{18}{8}=\frac{9}{4}\)

\(\Rightarrow\frac{x}{3}=\frac{9}{4}\Rightarrow x=\frac{27}{4}\)

\(\frac{y}{5}=\frac{9}{4}\Rightarrow y=\frac{45}{4}\)

Cô gái vô sinh

\(X=2\)

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

làm ơn k mk 1 cái

ĐKXĐ : \(\hept{\begin{cases}x^2+x-6\ne0\\x^2+4x+3\ne0\\2x-1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(x-2\right)\ne0\\\left(x+1\right)\left(x+3\right)\ne0\\x\ne\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x\ne2;-3\\x\ne-1;-3\\x\ne\frac{1}{2}\end{cases}}}}\)

TXĐ : \(x\ne\left\{-3;-1;\frac{1}{2};2\right\}\)

\(pt\Leftrightarrow\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)

\(\Leftrightarrow\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)

\(\Leftrightarrow\frac{3x+9}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)

\(\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{-3}{2x-1}\)

\(\Leftrightarrow\frac{1}{x^2-x-2}=\frac{1}{1-2x}\)

\(\Leftrightarrow x^2-x-2-1+2x=0\)

\(\Leftrightarrow x^2+x-3=0\)

\(\Leftrightarrow\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{13}{4}=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{1-\sqrt{13}}{2}\right)\left(x+\frac{1+\sqrt{13}}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{2}\\x=\frac{-\sqrt{13}-1}{2}\end{cases}}\)

\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4+3}=-\frac{3}{2x-1}\)

<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{3x+9}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{1}{x-2}=-\frac{1}{2x-1}\)

<=> x-2=1-2x <=> 3x=3

=> x=1

Đáp số: x=1

Ta có:\(\frac{x+y}{2}=\frac{y-5}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:\(\frac{x+y}{2}=\frac{y-5}{3}=\frac{x+y+y-5}{2+3}=\frac{x+2y-5}{5}\)

\(\Rightarrow\frac{x+2y-5}{5}=\frac{x+2y-5}{y-1}\)\(\Rightarrow y-1=5\Rightarrow y=6\)

\(\Rightarrow\frac{x+6}{2}=\frac{6-5}{3}\)\(\Rightarrow\frac{x+6}{2}=\frac{1}{3}\)

\(\Rightarrow3\cdot\left(x+6\right)=2\)

\(\Rightarrow3x+18=2\)

\(\Rightarrow3x=-16\Rightarrow x=\frac{-16}{3}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

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

\(=\frac{x+2y-5}{y-1}\) (theo đề bài)

=> y - 1 = 5

=> y = 5 + 1 = 6

Thay y = 6 vào đề bài ta có: \(\frac{x+6}{2}=\frac{7-6}{3}=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}.2-6=\frac{-16}{3}\)

Vậy \(x=\frac{-16}{3};y=6\)

Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+y}{2}=\frac{y-5}{3}=\frac{x+y+y-5}{2+3}=\frac{x+2y-5}{5}=\frac{x+2y-5}{y-1}\)

\(\Rightarrow y-1=5\)

\(\Rightarrow y=6\)

Ta có: \(\frac{x+y}{2}=\frac{y+5}{3}\)

\(\Rightarrow\frac{x+6}{2}=\frac{6-5}{3}\)

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

\(\Rightarrow3\left(x+6\right)=2\)

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

\(\Rightarrow x=\frac{-16}{3}\)

Vậy \(x=\frac{-16}{3},y=6\)

để\(\frac{2x-1}{3+x}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}2x-1< 0\\3+x>0\end{cases}}\\\hept{\begin{cases}2x-1>0\\3+x< 0\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x< \frac{1}{2}\\x>-3\end{cases}\left(ktm\right)}\\\hept{\begin{cases}x>\frac{1}{2}\\x< -3\end{cases}\left(tm\right)}\end{cases}}\)

Vậy -3<x<1/2

Để 2x-1/3+x<0 

TH1 : 2x-1<0<=>2x<1<=>x<1/2

    và 3+x>0<=>x> -3 ( ktm)

TH2: 2x-1>0<=> 2x>1<=>x>1/2

    và 3+x<0<=>x< -3  (tm)

vậy -3<x<1/2

mk viết thường nên có chỗ nào ko hiểu thì ib cho mk nha nhớ đó

Ta có: \(A=\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right):\frac{4xy}{y^2-x^2}\)

\(=\left[\frac{1}{\left(x+y\right)^2}-\frac{1}{\left(x+y\right)\left(x-y\right)}\right].\frac{\left(y+x\right)\left(y-x\right)}{4xy}\)

\(=\frac{1}{x+y}\left(\frac{1}{x+y}-\frac{1}{x-y}\right).\frac{\left(x+y\right)\left(y-x\right)}{4xy}\)

\(=\frac{-2y}{\left(x+y\right)\left(x-y\right)}.\frac{x-y}{-4xy}\)

\(=\frac{1}{\left(x+y\right).2x}\)

Kb với mình nha mn!

\(\left|x+\frac{9}{2}\right|\ge0\)

\(\left|y+\frac{3}{4}\right|\ge0\)

\(\left|z+\frac{7}{2}\right|\ge0\)

Mà \(\left|x+\frac{9}{2}\right|+\left|y+\frac{3}{4}\right|+\left|z+\frac{7}{2}\right|\le0\)

\(\Rightarrow\hept{\begin{cases}x+\frac{9}{2}=0\\y+\frac{3}{4}=0\\z+\frac{7}{2}=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{9}{2}\\y=-\frac{3}{4}\\z=-\frac{7}{2}\end{cases}}}\)

Ta có: p + e + n = 58

Mà p = e, nên: 2p + n = 58

=> n = 58 - 2p

Ta có: \(p\le n\le1,5p\)

Ta lại có: \(\left\{{}\begin{matrix}p\le58-2p\\58-2p\le1,5p\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3p\le58\\58\le3,5p\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p\le19\left(3\right)\\p\ge16,6\end{matrix}\right.\)

=> \(p=\left\{17;18;19\right\}\)

Biện luận: 

=> p = 19 hạt. 

Dựa vào bảng hóa trị, suy ra: 

Y là kali (K)