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\(3x\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)
\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)
\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)
\(=\frac{6}{7}\)
Tìm x
\(a,3x(2x+1)=0\)
\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)
Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)
\(b.\frac{2}{3}-\frac{1}{3}(x-\frac{3}{2})-\frac{1}{2}(2x+1)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-x(\frac{1}{3}+1)=5\)
\(\frac{4}{3}x=\frac{2}{3}-5\)
\(\frac{4}{3}x=\frac{-13}{3}\)
\(x=\frac{-13}{3}\div\frac{4}{3}\)
\(x=\frac{-13}{4}\)
Chúc ban học tốt
1 )
(-10 + 5) -(4-x)=12 -(5-6)
<=> -10 + 5 - 4 + x = 12 - 5 + 6
<=> x = 12 - 5 + 6 + 10 - 5 + 4
<=> x = 22
2 )
14-(5-8+3-x)=|-7+10|
<=> 14 - 5 + 8 - 3 + x = | 3 |
<=> x = | 3 | -14 + 5 - 8 + 3
<=> x = 3 - 14 + 5 - 8 + 3
<=> x = -11
3 )
-19-(3 +x-5)=|4-15+2|
<=> -19 - 3 - x + 5 = | -9 |
<=> -x = | -9 | - 5 + 19 + 3
<=> -x = 9 - 5 + 19 + 3
<=> -x = 26
<=> x = -26
4 )
45-(x+45-3)=|-7+3|
<=> 45 - x - 45 + 3 = | -4 |
<=> -x + 3 = | -4 |
<=> -x = 4 - 3
<=> -x = 1
<=> x =-1
10 )
-(-12 +4-10)-(4-x)=-(-3)
<=> 12 - 4 + 10 - 5 + x = 3
<=> x = 3 -12 + 4 - 10 + 5
<=> x = -10
Mình là Siêu Phẩm Hacker , rất mong được thi đấu một lần vs Edokawa conan
1) (-10 + 5) - (4 - x) = 12 - (5 - 6)
=> -5 - 4 + x = 12 + 1
=> -5 - 4 + x = 13
=> 4 + x = -5 - 13
=> 4 + x = - 18
=> x = -18 - 4
=> x = -22
2) 14 - (5 - 8 + 3 - x) = |-7 + 10|
=> 14 + 5 + 8 - 3 + x = |3|
=> 24 + x = 3
=> x = 3 - 24
=> x = -21
3) -19 - (3 + x - 5) = |4 - 15 + 2|
=> -19 - 3 - x + 5 = |-9|
=> -17 - x = 9
=> x = -17 - 9
=> x = -26
4) 45 - (x + 45 - 3) = |-7 + 3|
=> 45 - x - 45 + 3 = |-4|
=> 3 - x = 4
=> x = 3 - 4
=> x = -1
5) -(-12 + 4 - 10) - (4 - x) = -(-3)
=> -(-29) - (4 - x) = 3
=> 29 - (4 - x) = 3
=> 4 - x = 29 - 3
=> 4 - x = 26
=> x = 4 - 26
=> x = -22
a) \(2xy+2x-y=8\)
\(\Rightarrow\ 2x\left(y+1\right)-\left(y+1\right)=7\)
\(\Leftrightarrow\left(2x-1\right)\left(y+1\right)=7\)
\(\Rightarrow\left[\begin{matrix}\begin{cases}2x-1=-7\\y+1=-1\end{cases}\\\begin{cases}2x-1=-1\\y+1=-7\end{cases}\end{matrix}\right.\left[\begin{matrix}\begin{cases}2x-1=7\\y+1=1\end{cases}\\\begin{cases}2x-1=1\\y+1=7\end{cases}\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=4\\y=0\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=1\\y=6\end{cases}\\\left[\begin{matrix}\begin{cases}x=-3\\y=-2\end{cases}\\\begin{cases}x=0\\y=-8\end{cases}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
c)\(x^2+xy+x+y=2\)
\(\Leftrightarrow x\left(x+1\right)+y\left(x+1\right)=2\)
\(\Leftrightarrow\left(x+y\right)\left(x+1\right)=2\)
\(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x+y=2\\x+1=1\end{cases}\\\begin{cases}x+y=1\\x+1=2\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x+y=-2\\x+1=-1\end{cases}\\\begin{cases}x+y=-1\\x+1=-2\end{cases}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=0\\y=2\end{cases}\\\begin{cases}x=1\\y=0\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=-2\\y=0\end{cases}\\\begin{cases}x=-3\\y=2\end{cases}\end{matrix}\right.\end{matrix}\right.\)
a) \(\frac{18}{45}-\frac{4}{12}-\frac{6}{9}-\frac{21}{35}=\frac{2}{5}-\frac{1}{3}-\frac{2}{3}-\frac{3}{5}\)
\(=\left(\frac{2}{5}-\frac{3}{5}\right)-\left(\frac{1}{3}+\frac{2}{3}\right)\)
\(=\frac{-1}{3}-1\)
\(=\frac{-4}{3}\)
b) \(\frac{4}{3}+\frac{3}{5}+\frac{7}{3}+\frac{2}{5}+\frac{1}{3}=\left(\frac{4}{3}+\frac{7}{3}+\frac{1}{3}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=4+1=5\)
c) \(\frac{1}{3}.\frac{4}{5}.\frac{1}{3}.\frac{6}{5}=\frac{8}{75}\)
d) \(\frac{6}{7}.\frac{8}{13}+\frac{6}{7}.\frac{9}{13}-\frac{3}{13}.\frac{6}{7}\)
\(=\frac{6}{7}.\left(\frac{8}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{6}{7}.\frac{14}{13}\)
\(=\frac{12}{13}\)
1a) \(\frac{x-3}{x+7}=\frac{-5}{-6}\)
=> \(\frac{x-3}{x+7}=\frac{5}{6}\)
=> (x - 3).6 = 5.(x + 7)
=> 6x - 18 = 5x + 35
=> 6x - 5x = 35 + 18
=> x = 53
b) \(\frac{x-7}{x+3}=\frac{4}{3}\)
=> (x - 7). 3 = (x + 3). 4
=> 3x - 21 = 4x + 12
=> 3x - 4x = 12 + 21
=> -x = 33
=> x = -33
c) \(\frac{x-10}{6}=-\frac{5}{18}\)
=> (x - 10) . 18 = -5 . 6
=> 18x - 180 = -30
=> 18x = -30 + 180
=> 18x = 150
=> x = 150 : 18 = 25/3
d) \(\frac{x-2}{4}=\frac{25}{x-2}\)
=> (x - 2)(x - 2) = 25 . 4
=> (x - 2)2 = 100
=> (x - 2)2 = 102
=> \(\orbr{\begin{cases}x-2=10\\x-2=-10\end{cases}}\)
=> \(\orbr{\begin{cases}x=12\\x=-8\end{cases}}\)
e) \(\frac{7}{x}=\frac{x}{28}\)
=> 7 . 28 = x . x
=> 196 = x2
=> x2 = 142
=> \(\orbr{\begin{cases}x=14\\x=-14\end{cases}}\)
f) \(\frac{40+x}{77-x}=\frac{6}{7}\)
=> (40 + x) . 7 = (77 - x).6
=> 280 + 7x = 462 - 6x
=> 280 - 462 = -6x + 7x
=> -182 = x
=> x = -182
1)\(\left(x+1\right).\left(y-2\right)=0\) \(\left(x,y\inℤ\right)\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
2)\(\left(x-5\right).\left(y-7\right)=1\)
x-5 | 1 | -1 |
y-7 | 1 | -1 |
x | 6 | 4 |
y | 8 | 6 |
3)\(\left(x+4\right).\left(y-2\right)=2\)
x+4 | 1 | 2 | -1 | -2 |
y-2 | 2 | 1 | -2 | -1 |
x | -3 | -2 | -5 | -6 |
y | 4 | 3 | 0 | 1 |
4)\(\left(x-4\right).\left(y+3\right)=-3\)
x-4 | 1 | -1 | 3 | -3 |
y+3 | -3 | 3 | -1 | 1 |
x | 5 | 3 | 7 | 1 |
y | -6 | 0 | -4 | -2 |
5)\(\left(x+3\right).\left(y-6\right)=-4\)
x+3 | -1 | 1 | -4 | 4 | 2 | -2 |
y-6 | 4 | -4 | 1 | -1 | -2 | 2 |
x | -4 | -2 | -7 | 1 | -1 | -5 |
y | 10 | 2 | 7 | 5 | 4 | 8 |
6)\(\left(x-8\right).\left(y+7\right)=5\)
x-8 | 1 | 5 | -1 | -5 |
y+7 | 5 | 1 | -5 | -1 |
x | 9 | 13 | 7 | 3 |
y | -2 | -6 | -12 | -8 |
7)\(\left(x+7\right).\left(y-3\right)=-6\)
x+7 | -1 | 1 | -6 | 6 | -2 | 2 | -3 | 3 |
y-3 | 6 | -6 | 1 | -1 | 3 | -3 | 2 | -2 |
x | -8 | -6 | -13 | -1 | -9 | -5 | -10 | -4 |
y | 9 | -3 | 4 | 2 | 6 | 0 | 5 | 1 |
8)\(\left(x-6\right).\left(y+2\right)=7\)
x-6 | 1 | 7 | -1 | -7 |
y+2 | 7 | 1 | -7 | -1 |
x | 7 | 13 | 5 | -1 |
y | 5 | -1 | -9 | -3 |
ok :)
=X/46 +7/25=36,78/80