3x+1-7=2
Giải giúp mình nhé!!!Mai nộp rồi@@@
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Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)
\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)
\(\Leftrightarrow15x^2-4x-4=0\)
\(\Leftrightarrow15x^2-10x+6x-4=0\)
Lỗi :vvvv
\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...
\(A=7^1+7^3+7^5+7^7+...+7^{1997}+7^{1999}\)
\(A=\left(7+7^3\right)+\left(7^5+7^7\right)+...+\left(7^{1997}+7^{1999}\right)\)
\(A=\left(7+7^3\right)+\left[\left(7+7^3\right)\cdot7^4\right]+...+\left[\left(7+7^3\right)\cdot7^{1996}\right]\)
\(A=\left(7+7^3\right)\cdot\left(1+7^4+...+7^{1996}\right)\)
\(A=350\cdot\left(1+7^4+...+7^{1996}\right)\)
Vì \(350⋮35\)nên \(A⋮35\left(đpcm\right)\)
(3\(x\) - 2)(\(x+4\)) - (1- \(x\))(2-\(x\)) =(\(x+1\))(\(x-2\))
3\(x^2\) + 12\(x\) - 2\(x\) - 8 - (\(x+1\))(\(x-2\)) - [-(\(x-2\))](1- \(x\)) = 0
3\(x^2\) + 10\(x\) - 8 - (\(x-2\))( \(x\) + 1 - 1 + \(x\)) = 0
3\(x^2\) + 10\(x\) - 8 - (\(x-2\)). 2\(x\) = 0
3\(x^2\) + 10\(x\) - 8 - 2\(x^2\) + 4\(x\) = 0
\(x^2\) + 14\(x\) - 8 = 0
\(x^2\) + 7\(x\) + 7\(x\) + 49 - 57 = 0
\(x\)( \(x\) + 7) + 7(\(x\) + 7) = 57
(\(x+7\))(\(x\) + 7) =57
(\(x+7\))2 = 57
\(\left[{}\begin{matrix}x+7=\sqrt{57}\\x+7=-\sqrt{57}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-7+\sqrt{57}\\x=-7-\sqrt{57}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -7 - \(\sqrt{57}\); - 7 + \(\sqrt{57}\)}
a) 1/2x - 2/3x = 7/12
-1/6x = 7/12
x = \(\frac{7}{12}\div\frac{-1}{6}\)
x = -7/2
b) x: 12 = -2,5
x = -2,5*12
x = -30
c) 11/2x = 13/15
x = 13/15:11/2
x = 26/165
d) 3x/7 + 1 = (-1/28)*(-4)
3x/7 + 1 = 1/7
3x/7 = 1/7 - 1
3x/7 = -6/7
3x = -6
x= -6/3
x= -2
a) x(0.5 - 2/3 ) = 7/12
(-1/6)x = 7/12
x = 7/12 : (-1/6)
x =-7/2
b) x : 12 = -2,5
x= -2.5 * 12 =-30
c) x = 13/15 : 5,5
x =26/125
d) 3x/7 + 1 = -1/28 * (-4)
3x/7 + 1 =1/7
3x/7 = 1/7 -1
3x/7 =-6/7
3x =-6/7 * 7
3x = -6
x =-2
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3x(x-1)=1-x
<=> 3x(x-1) +x-1=0
<=> (x-1)(3x+1)=0
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}}\)
Vậy...
3^x+1 - 7 = 2
3^x+1 = 2+7
3^x+1 = 9
3^x+1 = 3^2
=> x+1=2
x= 2-1
x=1
vậy.....
\(3^{x+1}-7=2\)
\(\Leftrightarrow3^{x+1}=2+7\)
\(\Leftrightarrow3^{x+1}=9\)
\(\Leftrightarrow3^{x+1}=3^2\)
\(\Leftrightarrow x+1=2\)
\(\Leftrightarrow x=2-1\)
\(\Leftrightarrow x=1\)