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Câu a:
- \(\frac12\)(3 - 2\(x\)) - 7 = 5 - \(\frac13\)(\(x\) - \(\frac45\))
- \(\frac32\) + \(x\) - 7 = 5 - \(\frac13x\) + \(\frac{4}{15}\)
\(x\) + \(\frac13x\) = 5 + \(\frac{4}{15}\) + 7 + \(\frac32\)
(1 + \(\frac13\))\(x\) = \(\frac{150}{30}\) + \(\frac{8}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{158}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{368}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{413}{30}\)
\(x\) = \(\frac{413}{30}\) : \(\frac43\)
\(x\) = \(\frac{413}{40}\)
Vậy \(x=\frac{413}{40}\)
Câu b:
(5 - 3x/2) : - 1 3/8 = - 7 1/3
(5 - 3x/2) : (-11/8) = - 22/3
5 - 3x/2 = - 22/3 x (-11/8)
5 - 3x/2 = 121/12
3x/2 = 5 - 121/12
3x/2 = - 61/12
x = - 61/12 : 3/2
x = -61/18
Vậy x = - 61/18
Câu a:
\(\frac{-8}{3x-1}\) = \(\frac{4}{-7}\)
-8.(-7) = 4.(3\(x\) - 1)
56 = 12\(x\) - 4
12\(x\) = 56+ 4
12\(x\) = 60
\(x\) = 60 : 12
\(x\) = 5
Vậy \(x\) = 5
Câu b:
\(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(x^2\) = (-3)\(^2\)
\(\left[\begin{array}{l}x=-3\\ x=3\end{array}\right.\)
Vậy \(x\in\left\lbrace-3;3\right\rbrace\)
Câu c:
\(-\frac{4}{y}=\frac{x}{2}\)
-4.2 = \(x.y\)
\(xy=-8\)
Ư(8) = (-8; -4; -2; -1; 1; 2; 4; 8}
Vậy (\(x;y\)) = (-8; 1); (-4; 2); (-2; 4); (-1; 8); (1; -8); (2; -4); (4; -2); (8; -1)
Câu 2:
(\(x-1)\)(y + 2) = 7
Ư(7) = {-7; -1; 1; 7}
Lập bảng ta có:
\(x\)-1 | -7 | -1 | 1 | 7 |
\(x\) | -6 | 0 | 2 | 8 |
y+2 | -1 | -7 | 7 | 1 |
y | -3 | -9 | 5 | -1 |
\(x;y\in Z\) | tm | tm | tm | tm |
Theo bảng trên ta có:
(\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; - 1)
Vậy (\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; -1)
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
0,5x-2/3x=7/12
=>(1/2-2/3).x=7/12
=>-1/6x=7/12
=>x=7/12:1/6
=>x=7/2
b, x:4 1/3=-2,5
=>x:13/3=-2,5
=>x=-5/2x13/3
=>x=-13
c,5,5x=13/15
=>11/2x=13/15
=>x=13/15:11/2
=>x=26/165
d,(3x/7+1):-4=-1/28
=>3x/7+1=-1/28x(-4)
=>3x/7+1=-1/7
=>3x/7=-1/7-1
=>3x/7=-8/7
=>3x=-8/7x7
=>3x=8
=>x=8:3
=>x=8/3
\(0,5x-\frac{2}{3}x=\frac{7}{12}\)
\(\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\left(\frac{1}{2}-\frac{2}{3}\right)x=\frac{7}{12}\)
\(\frac{-1}{6}x=\frac{7}{12}\)
\(x=-21\)
b,c,d tương tự
a, \(\frac{17}{y}=\frac{-7}{11}\)
\(\Rightarrow17\cdot11=-7\cdot y\)
\(\Rightarrow187=-7\cdot y\)
\(\Rightarrow\frac{187}{-7}=y\)
b, \(\frac{-8}{3x-1}=\frac{4}{7}\)
\(\Rightarrow\frac{-8}{3x-1}=\frac{-8}{-14}\)
\(\Rightarrow3x-1=-14\)
\(\Rightarrow3x=-14+1\)
\(\Rightarrow3x=-13\)
\(\Rightarrow x=\frac{-13}{3}\)
c, \(\frac{x}{-3}=\frac{-3}{x}\)
\(\Rightarrow x\cdot x=-3\cdot\left(-3\right)\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x^2=\left(\pm3\right)^2\)
\(\Rightarrow x=\pm3\)
d, \(\frac{-4}{y}=\frac{x}{2}\)
\(\Rightarrow-4\cdot2=x\cdot y\)
\(\Rightarrow-8=x\cdot y\)
\(\Rightarrow x;y\inƯ\left(-8\right)=\left\{-1;1;-2;2;-4;4;-8;8\right\}\)
ta có bảng :
| x | -1 | -8 | -2 | -4 |
| y | 8 | 1 | 4 | 2 |
a)\(\frac{14}{y}\)\(=\) \(\frac{-7}{11}\)
\(\Rightarrow\)\(14\cdot11=y\cdot\left(-7\right)\)
\(y=\)\(\frac{14\cdot11}{-7}\)
\(y=22\)
c) \(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(\Rightarrow\) \(x\cdot x=\left(-3\right)\cdot\left(-3\right)\)
\(\Rightarrow\)\(x^2=9\)
\(\Rightarrow\)\(x^2=9\)hoặc \(x^2=-9\)
\(TH1:\) \(x^2=9\)
\(\Rightarrow\)\(x=3\)
\(TH2:\)\(x^2=-9\)
\(\Rightarrow\)\(x=-3\)
\(\left|x+\frac{7}{3}\right|=\left|x-\frac{7}{2}\right|\)
Vì \(x+\frac{7}{3}\ne x-\frac{7}{2}\)nên \(x+\frac{7}{3}=\frac{7}{2}-x\)
\(\Leftrightarrow x+x=\frac{7}{2}-\frac{7}{3}\)
\(\Leftrightarrow2x=\frac{21}{6}-\frac{14}{6}\)
\(\Leftrightarrow2x=\frac{7}{6}\)
\(\Leftrightarrow x=\frac{7}{6}\div2\)
\(\Leftrightarrow x=\frac{7}{12}\)
b) \(\left|2x-\frac{1}{3}\right|=\left|3x\left(-\frac{2}{3}\right)\right|\)
\(\Leftrightarrow\left|2x-\frac{1}{3}\right|=\left|-2x\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{3}=-2x\\2x-\frac{1}{3}=2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\x\in\left\{\varnothing\right\}\end{cases}}\Leftrightarrow x=\frac{1}{12}\)
Vậy \(x=\frac{1}{12}\)
\(\frac{11}{4}:\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{7}{2}\)
\(\Leftrightarrow\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{11}{4}:\frac{7}{2}\)
\(\Leftrightarrow\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{11}{14}\)
\(\Leftrightarrow\left|4x-\frac{1}{3}\right|=\frac{3}{2}:\frac{11}{14}\)
\(\Leftrightarrow\left|4x-\frac{1}{3}\right|=\frac{21}{11}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=\frac{21}{11}\\4x-\frac{1}{3}=-\frac{21}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{37}{66}\\x=-\frac{13}{33}\end{cases}}\)
Bài dưới tương tự
\(\frac{3}{x-5}=-\frac{4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=\left(-4\right)\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\)
\(\Leftrightarrow4\left(3x+1\right)=\left(x+2\right)7\)
\(\Leftrightarrow12+4=7x+14\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=2\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\)
\(\Leftrightarrow\left(-3\right)\left(2-2x\right)=\left(x+1\right)4\)
\(\Leftrightarrow-6+6x=4x+4\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
x=22/7;-20/7