Làm sao 2 ẩn mà chỉ có 1 phương trình mà giải đc nhỉ ??

Thầy cho bọn tớ thế !

Ta có : \(\frac{6}{8}=\frac{3}{4}\)

\(\frac{x}{y}=\frac{3}{4}\)

\(\Rightarrow4x=3y\)

\(\Rightarrow x=3;y=4\)

A = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2

1/2^2 < 1/1*2

1/3^2 < 1/2*3

1/4^2 < 1/3*4

...

1/100^2 < 1/99*100

=> A < 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100

=> A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

=> A < 1 - 1/100

=> A < 1

minh deo can ban k dau :((

\(a,\frac{1}{2}x+\frac{3}{5}(x-2)=3\)

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

\(\Rightarrow\left[\frac{1}{2}+\frac{3}{5}\right]x=3+\frac{6}{5}\)

\(\Rightarrow\left[\frac{5}{10}+\frac{6}{10}\right]x=\frac{21}{5}\)

\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\)

\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}=\frac{21}{5}\cdot\frac{10}{11}=\frac{21}{1}\cdot\frac{2}{11}=\frac{42}{11}\)

Vậy x = 42/11

cậu cứ nhân 5 vào phương trình (2)

cộng 2 phương trình lại cậu sẽ ra được x+y-1=2

thế cái vừa tìm được vào 1 trong 2 phương trình thi sẽ ra thêm một phương trình 2x-y=-13

giải hệ rồi tìm được x và y

a )  

Ta có : 

\(\hept{\begin{cases}\frac{x}{5}=\frac{y}{6}\\\frac{y}{8}=\frac{z}{7}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{20}=\frac{y}{24}\\\frac{y}{24}=\frac{z}{21}\end{cases}}}\)

và \(x+y-z=69\)

ADTCDTSBN , ta có : 

\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{20}=3\\\frac{y}{24}=3\\\frac{z}{21}=3\end{cases}\Rightarrow\hept{\begin{cases}x=3.20=60\\y=3.24=72\\z=3.21=63\end{cases}}}\)

Vậy ...

b )  

Ta có : 

\(5y=72\Rightarrow y=\frac{72}{5}=14,4\)

\(\Rightarrow x=14,4.3:2=21,6\)

và \(3x+5y-7z=30\)

Thay vào làm tiếp : 

c ) 

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

\(=\frac{3\left(x-1\right)}{6}=\frac{4\left(y+3\right)}{16}=\frac{5\left(z-5\right)}{30}\)

\(=\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)

\(=\frac{5z-25-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)( ADTCDTSBN ) 

\(=\frac{5z-25-3x+3-4y-12}{8}=\frac{5z-3x-4y-34}{8}\)

\(=\frac{50-34}{8}=\frac{16}{8}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x-1}{2}=2\\\frac{y+3}{4}=2\\\frac{z-5}{6}=2\end{cases}\Rightarrow\hept{\begin{cases}x-1=2.2=4\\y+3=2.4=8\\z-5=2.6=12\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\y=5\\z=17\end{cases}}}\)

Vậy ...

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

ĐKXĐ: ..

Đặt \(\left\{{}\begin{matrix}\frac{1}{x+y-1}=u\\\frac{1}{2x-y+3}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4u-5v=-\frac{5}{2}\\3u+v=-\frac{7}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=-\frac{1}{2}\\v=\frac{1}{10}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y-1=-2\\2x-y+3=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\2x-y=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\frac{4}{x+y-1}-\frac{5}{2x-y+3}=\frac{-5}{2}\\\frac{3}{x+y-1}+\frac{1}{2x-y+3}=\frac{-7}{5}\end{matrix}\right.\)

đặt \(\frac{1}{x+y-1}=a\\ \frac{1}{2x-y+3}=b\)

ta có :

\(\left\{{}\begin{matrix}4a-5b=\frac{-5}{2}\\3a+b=\frac{-7}{5}\end{matrix}\right.\).......=>\(\left\{{}\begin{matrix}a=-\frac{1}{2}\\b=\frac{1}{10}\end{matrix}\right.\)

suy ra \(\left\{{}\begin{matrix}x+y-1=-2\\2x-y+3=10\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x+y=-1\\2x-y=7\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

#Mai.T.Loan

a) Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=a\\\frac{1}{y-1}=b\end{matrix}\right.\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}5a+b=10\\a-3b=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15a+3b=30\\a-3b=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-3b=18\\16a=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=3\\\frac{1}{y-1}=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=\frac{4}{5}\end{matrix}\right.\)

Vậy...

b) Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=a\\\frac{1}{\sqrt{y+6}}=b\end{matrix}\right.\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\frac{5}{2}\\5a+3b=\frac{13}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}31a-12b=\frac{15}{2}\\20a+12b=\frac{26}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7a-4b=\frac{5}{2}\\51a=\frac{97}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{97}{306}\\b=\frac{-43}{612}\end{matrix}\right.\)( loại vì \(a,b>0\) )

Vậy hệ vô nghiệm

Is that true .-.

Cho xin solve lại câu b)

hpt \(\Leftrightarrow\left\{{}\begin{matrix}21a-12b=\frac{15}{2}\\20a+12b=\frac{26}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5a+3b=\frac{13}{6}\\41a=\frac{97}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{97}{246}\\b=\frac{8}{123}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=\frac{97}{246}\\\frac{1}{\sqrt{y+6}}=\frac{8}{123}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{126379}{9409}\\y=\frac{14745}{64}\end{matrix}\right.\)

Vậy...