Tìm x biết
7*x^2 -13*x + 6 = 0
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)390-(x-7)=169:13
<=>390-(x-7)=13
<=>x-7=390-13
<=>x-7=377
<=>x=377+7
<=>x=384
b)x-6:2-(48-24):2:6-3=0
<=>x-3-24:2:6-3=0
<=>x-6-12:6=0
<=>x-6-2=0
<=>x-8=0
<=>x=8
a) 390 - ( x - 7 ) = 169 : 13
390 - ( x - 7 ) = 13
x - 7 = 390 - 13
x - 7 = 377
x = 377 + 7
x = 384
b) x - 6 : 2 - ( 48 - 24 ) : 2 : 6 - 3 = 0
x - 3 - ( 48 - 24 ) : 2 : 6 = 0 + 3 = 3
x - 3 - ( 48 - 24 ) : 2 = 3 x 6 = 18
x - 3 - ( 48 - 24 ) = 18 x 2
x - 3 - 24 = 36
x - 3 = 36 + 24
x - 3 = 60
x = 60 + 3
x = 63
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>x=7-20=-13
b: =>x=-18+12=-6
c: =>x=9 hoặc x=-6
d: =>x=0 hoặc x=4
e: =>6-x=13-3+14=24
=>x=-18
Câu g và h đề thiếu rồi bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : ( x + 1 ).( 3 - x ) > 0
Th1 : \(\hept{\begin{cases}x+1>0\\3-x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x>3\end{cases}\Rightarrow}x>3}\)
Th2 : \(\hept{\begin{cases}x+1< 0\\3-x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x< 3\end{cases}\Rightarrow}x< -1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x εIN biết
a) 390 - (x-8) = 168:13
b) (x-140) : 7 = 27 - 24
c) x- 6 :2 - ( 48 - 24 ) :2 :6 - 3 = 0
d) x+5.2-(32+16.3:6-15)=0
b) \(\left(x-140\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(x-140=21\)
\(x=161\)
vay \(x=161\)
c) \(x-6:2-\left(48-24\right):2:6-3=0\)
\(x-3-24:2:6-3=0\)
\(x-3-2-3=0\)
\(x-8=0\)
\(x=8\)
vay \(x=8\)
d) \(x+5.2-\left(32+16.3:6-15\right)=0\)
\(x+10-\left(32+8-15\right)=0\)
\(x+10-25=0\)
\(x-15=0\)
\(x=15\)
vay \(x=15\)
a) \(390-\left(x-8\right)=168:13\)
\(390-x+8=\frac{168}{13}\)
\(x+8=390-\frac{168}{13}\)
\(x+8=\frac{5070}{13}-\frac{168}{13}\)
\(x+8=\frac{4902}{13}\)
\(x=\frac{4902}{13}-8\)
\(x=\frac{4798}{13}\)
vay \(x=\frac{4798}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
390-(x-7)=169:13
390-(x-7)=13
X-7=390-13
X-7=377
X=377+7
X=384
phần a bạn ở dưới làm r nhé
phần b chép sai đề
phần c :
x - 6 : 2 - ( 48 - 24 ) : 2 : 6 - 3 = 0
<=>x - 3 - 24 : 12 - 3 = 0
<=>x - 3 - 2 - 3 = 0
<=>x=8
Vậy x = 8
phần d :
x + 5 . 2 - ( 32 + 16 . 3 : 16 - 15 ) = 0
<=>x + 10 - ( 32 + 3 - 15 ) = 0
<=> x + 10 - 20 = 0
<=>x = 10
Vậy x = 10
Ta có :
\(7x^2-13x+6=0\)
=> \(7x^2-7x-13x+7x+6=0\)
=> \(7x\left(x-1\right)-6x+6=0\)
=>\(7x\left(x-1\right)-6\left(x-1\right)=0\)
=> \(\left(7x-6\right)\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}7x-6=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{6}{7}\\x=1\end{matrix}\right.\)
Vậy x \(\in\)\(\left\{\dfrac{6}{7};1\right\}\)
\(7x^2-13x+6=0\Leftrightarrow7x^2-7x-6x+6=0\)
\(\Leftrightarrow7x\left(x-1\right)-6\left(x-1\right)=0\Leftrightarrow\left(7x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\7x-6=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\7x=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=\dfrac{6}{7}\end{matrix}\right.\) vậy \(x=1;x=\dfrac{6}{7}\)