\(\frac{-4}{7}\)=\(\frac{x-40}{31-5x}\)
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\(\frac{-4}{7}=\frac{x-40}{31-5x}\)
\(-4\left(31-5x\right)=7\left(x-40\right)\)
\(-124+20x=7x-280\)
\(20x-7x=124-280\)
\(13x=-156\)
\(x=-12\)
Vậy: \(x\in\left\{-12\right\}\)
=.= hk tốt!!
\(-\frac{4}{7}=\frac{x-40}{31-5x}\)
=> 7 ( x - 40 ) = - 4 ( 31 - 5x )
=> 7x - 7 . 40 = - 4 . 31 + 20x
=> 7x - 280 = - 124 + 20x
=> 124 - 280 = 20x - 7x
=> -156 = 13x
=> x = -12
\(\left\{{}\begin{matrix}\frac{5x-9}{x-2}+\frac{4y+31}{y+7}=11\\\frac{3x-4}{x-2}-\frac{2y+19}{y+7}=-11\end{matrix}\right.\) \(\left(x\ne2;y\ne-7\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{5\left(x-2\right)+1}{x-2}+\frac{4\left(y+7\right)+3}{y+7}=11\\\frac{3\left(x-2\right)+2}{x-2}-\frac{2\left(y+7\right)+5}{y+7}=-11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5+\frac{1}{x-2}+4+\frac{3}{y+7}=11\\3+\frac{2}{x-2}-2-\frac{5}{y+7}=-11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-2}+\frac{3}{y+7}=2\\\frac{2}{x-2}-\frac{5}{y+7}=-12\end{matrix}\right.\)
Đặt \(\frac{1}{x-2}=a;\frac{1}{y+7}=b\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}a+3b=2\\2a-5b=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2-3b\\2\left(2-3b\right)-5b=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{-26}{11}\\b=\frac{16}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-2}=\frac{-26}{11}\\\frac{1}{y+7}=\frac{16}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{41}{26}\\y=\frac{-101}{16}\end{matrix}\right.\)( thỏa mãn )
Vậy...
Vũ Minh TuấnNguyễn Ngọc Lộc HISINOMA KINIMADOTrên con đường thành công không có dấu chân của kẻ lười biếngPhạm Thị Diệu HuyềnNguyễn Thành TrươngNguyễn Lê Phước ThịnhNguyễn Trúc GiangNatsu Dragneel 2005Trần Thanh PhươngtthChiyuki Fujito
Bài 1 : Ta có:
\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)
= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)
= \(\frac{7}{9}\)
Bài 2 :
\(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)
=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)
=> 50x = 10
=> x = 10 : 50
=> x = 1/5
Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3
<=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng :
x + 3 | 1 | -1 | 3 | -3 |
x | -2 | -4 | 0 | -6 |
Vậy
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
\(\frac{-4}{7}=\frac{x-40}{31-5x}\)