giup mik vs
tim x biet : 25%x - 7/8 + x = 2/3x
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Ta có bảng
x+5 | 25 | 1 | -1 | -25 | 5 | -5 |
y-5 | 1 | 25 | -25 | -1 | 5 | -5 |
x | 22 | -4 | -6 | -30 | 0 | -10 |
y | 6 | 30 | -20 | 4 | 10 | 0 |
(x+3).(y-5)=25
Ta có:các số nhân nhau bằng 25 là:5x5
=>x+3=5 và y-5=5
x=5-3 y=5+5
x=2 y=10
Vậy:x=2;y=10 chúc bạn học tốt
25 – 2x = 16 – 3x
-2x + 3x = 16 - 25
x = -9
Vậy x = -9
x - (47 - 22 ) = 5 + (10-4x)
x - 47 + 22 = 5 + 10 - 4x
x + 4x = 5 + 10 - 22 + 47
5x = 40
x = 40 : 5
x = 8
Vậy x = 8.
~ HOK TỐT ~
\(\text{25 – 2x = 16 – 3x}\)
\(2x+3x=16-25\)
\(5x=-9\)
\(\Rightarrow x=\frac{-9}{5}\)
\(\text{d) x – (47 – 22) = 5+ (10 – 4x)}\)
\(x-47+22=5+10-4x\)
\(x+4x=5+10-22+47\)
\(5x=40\)
\(\Rightarrow x=8\)
học tốt
(x - 2)(y + 3) = 7
\(\Rightarrow\) x - 2 và y + 3 \(\in\) Ư(7)
Ư(7) = {1; -1; 7; -7}
Xét các TH:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=7\\y+3=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-7\\y+3=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=1\\y+3=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-1\\y+3=-7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-10\end{matrix}\right.\end{matrix}\right.\)
Vậy x \(\in\) {9; -5; 3; 1} thì y \(\in\) {-4; -2; 4; -10}
Mình làm mẫu phần này thì (x + 3)(y + 5) = -6 cũng vậy nha!
y + 3 chia hết cho x + 1 và 2x - 5 chia hết cho x + 4 mk làm sau nha!
Chúc bn học tốt
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=x+1-\left(x-9\right)\)
\(\Rightarrow6x^2+27x+4x+18-6x^2-x-12x-2=10\)
\(\Rightarrow18x+16=10\)
\(\Rightarrow18x=-6\)
\(\Rightarrow x=-\frac{6}{18}=-\frac{1}{3}\)
a,Để \(|2x+1|+|x-2|=0\Leftrightarrow\hept{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)(vô lý)
=> ko có x thỏa mãn
b,\(|x+5|=2x-1\Leftrightarrow1-2x< x+5< 2x-1\)
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
ta thấy \(2x^2+7>0\)
\(=>-3x-9< 0\)
\(=>-3x< 9\)
\(=>x>-3\)
vậy...
\(25\%x-\frac{7}{8}+x=\frac{2}{3}x\)
<=> \(\frac{1}{4}x-\frac{7}{8}+1x=\frac{2}{3}x\)
<=> \(\frac{1}{4}x+1x-\frac{2}{3}x=\frac{7}{8}\)
<=> \(x\left(\frac{1}{4}+1-\frac{2}{3}\right)=\frac{7}{8}\)
<=> \(x\cdot\frac{7}{12}=\frac{7}{8}\)
<=> \(x=\frac{12}{8}=\frac{3}{2}\)
Bài làm:
Ta có: \(25\%x-\frac{7}{8}+x=\frac{2}{3}x\)
\(\Leftrightarrow\frac{x}{4}+x-\frac{2}{3}x=\frac{7}{8}\)
\(\Leftrightarrow\frac{7}{12}x=\frac{7}{8}\)
\(\Leftrightarrow x=\frac{3}{2}\)