\(450:\left(x-19\right)=15\)

\(=>x-19=450:15=30\)

\(=>x=30+19=49\)

\(\left(x-1\right)\left(y-2\right)=2=1.2=-1.-2\)

\(x-1=1=>x=2\)

\(y-2=1=>y=3\)

\(x-1=2=>x=3\)

\(y-2=2=>y=4\)

\(x-1=-1=>x=0\)

\(y-2=-1=>y=1\)

\(x-1=-2=>x=-1\)

\(y-2=-2=>y=0\)

Câu C làm tương tự câu b

Cho đường thẳng d đi qua M(2; 3) và tạo với chiều dương trục Ox một góc 45⁰. PTTQ của đường thẳng d là 

A. 2x - y - 1 = 0   B. x - y + 1 = 0   C. x + y - 5 0 =     D. -x + y - 1 = 0

= y x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 450

= y x 45 = 450

y = 10

Nho t ick cho mik

trình bày cách giải giúp mình ạ!

Cảm ơn bạn Nguyễn Công Danh nhiều. 

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7}{y}=-2\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\\dfrac{1}{x}=1+\dfrac{2}{7}=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\x=\dfrac{7}{9}\end{matrix}\right.\)

a) \(\dfrac{5}{7}-1\dfrac{4}{7}\left(450\%+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)

\(\dfrac{5}{7}-\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)

\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{5}{7}+\dfrac{1}{14}\)

\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{11}{14}\)

\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}.\dfrac{7}{11}\)

\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{1}{2}\)

\(\dfrac{2}{3}x=\dfrac{1}{2}-\dfrac{9}{2}=-4\)

\(x=-4:\dfrac{2}{3}=-4.\dfrac{3}{2}=-6\)

Vậy x = \(-6\)

b) \(100=6.7^{\left|x+2\right|}-194\)

\(100+194=6.7^{\left|x+2\right|}\)

\(294=6.7^{\left|x+2\right|}\)

\(294:6=49=7^{\left|x+2\right|}\)

\(\Rightarrow7^2=7^{\left|x+2\right|}\)

\(\Rightarrow2=\left|x+2\right|\Rightarrow\pm2=x+2\)

+ x + 2 = -2 \(\Rightarrow\) x = - 4

+ x + 2 = 2 \(\Rightarrow\) x = 0

Vậy x = - 4 hoặc 0

vd câu 1:
ta có x-y=4 =>x=4+y
ta có pt:
4+y/y-2=3/2
=>8+2y=3y-6
=>-y=-14
=>y=14
=>x=4+y=4+14=18
các bài khác cũng tương tự thôi bạn

dấu chéo có nghĩa là phân số hí

Vậy đề bài yêu cầu gì vậy bạn?

`a)TXĐ: R`

`b)TXĐ: R\\{0}`

`c)TXĐ: R\\{1}`

`d)TXĐ: (-oo;-1)uu(1;+oo)`

`e)TXĐ: (-oo;-1/2)uu(1/2;+oo)`

`f)TXĐ: (-oo;-\sqrt{2})uu(\sqrt{2};+oo)`

`h)TXĐ: (-oo;0) uu(2;+oo)`

`k)TXĐ: R\\{1/2}`

`l)ĐK: {(x^2-1 > 0),(x-2 > 0),(x-1 ne 0):}`

`<=>{([(x > 1),(x < -1):}),(x > 2),(x ne 1):}`

`<=>x > 2`

   `=>TXĐ: (2;+oo)`

\(\frac{1}{y\left(y+1\right)}\) + \(\frac{1}{\left(y+1\right)\left(y+2\right)}\) + \(\frac{1}{\left(y+2\right)\left(y+3\right)}\) + \(\frac{1}{\left(y+3\right)\left(y+4\right)}\)= \(\frac{1}{15}\)

\(\frac{1}{y}\) - \(\frac{1}{y+1}\) + \(\frac{1}{y+1}\) - \(\frac{1}{y+2}\) + \(\frac{1}{y+2}\) - \(\frac{1}{y+3}\) + \(\frac{1}{y+3}\) - \(\frac{1}{y+4}\) = \(\frac{1}{15}\) 

\(\frac{1}{y}\) + \(\frac{1}{y+1}\) - \(\frac{1}{y+1}\) + \(\frac{1}{y+2}\) - \(\frac{1}{y+2}\) + \(\frac{1}{y+3}\) - \(\frac{1}{y+3}\) - \(\frac{1}{y+4}\) = \(\frac{1}{15}\)

\(\frac{1}{y}\) - \(\frac{1}{y+4}\) = \(\frac{1}{15}\)

\(\frac{4}{y\left(y+4\right)}\) = \(\frac{1}{15}\) => \(\frac{4}{y\left(y+4\right)}\)= \(\frac{4}{60}\)

=> y(y+4)=60 Mà 60 = 1.60=2.30=3.20=4.15=5.12=6.10

Vậy y(y+4)=6.10 => y=6. Vậy y=6