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a, \(\frac{5}{7}.3x-\frac{8}{5}=\frac{9}{35}\)
=> \(\frac{15}{7}x=\frac{9}{35}+\frac{8}{5}\)=> \(\frac{15}{7}x=\frac{9}{35}+\frac{56}{35}\)
=> \(\frac{15}{7}x=\frac{65}{35}=\frac{13}{7}\)=> \(x=\frac{13}{7}:\frac{15}{7}=\frac{13}{15}\)
vậy \(x=\frac{13}{15}\)
b, \(\frac{2}{9}.5x+\frac{1}{2}-\frac{1}{18}=\frac{5}{36}\)
=> \(\frac{10}{9}x+\frac{1}{2}=\frac{5}{36}+\frac{1}{18}\)=\(\frac{5}{36}+\frac{2}{36}=\frac{7}{36}\)
=> \(\frac{10}{9}x=\frac{7}{36}-\frac{1}{2}\)=\(\frac{7}{36}-\frac{18}{36}\)=\(\frac{-11}{36}\)=> \(x=\frac{-11}{36}:\frac{10}{9}\)=\(\frac{-11}{36}.\frac{9}{10}\)=\(\frac{-11}{40}\)
vậy x=\(\frac{-11}{40}\)
a, ta có : \(\frac{5}{7}.\frac{3x-8}{5}=\frac{9}{35}\Leftrightarrow\frac{3x-8}{5}=\frac{9}{35}:\frac{5}{7}\Leftrightarrow\frac{3x-8}{5}=\frac{9}{35}.\frac{7}{5}\)
\(\Leftrightarrow\frac{3x-8}{5}=\frac{9}{5.5}\Leftrightarrow3x-8=\frac{9}{25}.5\Leftrightarrow3x-8=\frac{9.5}{25}\)
\(\Leftrightarrow3x-8=\frac{9}{5}\Leftrightarrow3x-8=\frac{9}{5}+8\Leftrightarrow3x=\frac{9+8.5}{5}\)
\(\Leftrightarrow3x=\frac{49}{5}\Leftrightarrow x=\frac{49}{5}:3\Leftrightarrow x=\frac{49}{5}.\frac{1}{3}=\frac{49}{15}\)
~ Vậy, ta tìm được \(x=\frac{49}{15}\)
b, Ta có : \(\frac{2}{9}.\frac{5x+1}{2}-\frac{1}{18}=\frac{5}{36}\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{5}{36}+\frac{1}{18}\)
\(\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{5+2.1}{36}\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{7}{36}\)
\(\Leftrightarrow\frac{5x+1}{2}=\frac{7}{36}:\frac{2}{9}\Leftrightarrow\frac{5x+1}{2}=\frac{7}{36}.\frac{9}{2}\Leftrightarrow\frac{5x+1}{2}=\frac{7.9}{4.9.2}\)
\(\Leftrightarrow\frac{5x+1}{2}=\frac{7}{8}\Leftrightarrow5x+1=\frac{7}{8}.2\Leftrightarrow5x+1=\frac{7.2}{8}\)
\(\Leftrightarrow5x+1=\frac{7}{4}\Leftrightarrow5x=\frac{7}{4}-1\Leftrightarrow5x=\frac{7-1.4}{4}\)
\(\Leftrightarrow5x=\frac{3}{4}\Leftrightarrow x=\frac{3}{4}:5\Leftrightarrow x=\frac{3}{4}.\frac{1}{5}\Leftrightarrow x=\frac{3}{20}\)
~ Vậy, ta tìm được \(x=\frac{3}{20}\)
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8