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Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
1+3+5+...+x=1600
=(x+1).[(x-1):2+1] /2 =1600
=(x+1).(x+1) /2 =1600
=(x+1)^2:2=40^2
=(x+1):2=40
=x+1=80
=x=79
a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)
b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)
\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)
c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)
\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)
e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)
f) \(\left|12-x\right|-7=5\)
th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)
th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)
i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)
k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)
a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)
b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)
Làm nốt nha
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
làm bài & thôi :
(x2 - 2x + 3) \(⋮\)(x - 1)
= x2 - 2x + 3
=) x2 - 2x + 3 - ( x - 1 )
=) x2 - 1
=) x2 - 1 - x( x - 1 )
=) 2 \(⋮\)x - 1
tự làm
a) Ta có: (x2 - 2x + 3) \(⋮\)(x - 1)
<=> [x(x - 1) - (x - 1) + 2] \(⋮\)(x - 1)
<=> [(x - 1)2 + 2] \(⋮\)(x - 1)
Do (x - 1)2 \(⋮\)(x - 1) => 2 \(⋮\)(x - 1)
=> (x - 1) \(\in\)Ư(2) = {1; -1; 2; -2}
Lập bảng :
| x - 1 | 1 | -1 | 2 | -2 |
| x | 2 | 0 | 3 | -1 |
Vậy ...
b) (3x - 1) \(⋮\)(x - 4)
<=> [3(x - 4) + 11] \(⋮\)(x - 4)
Do 3(x - 4) \(⋮\)(x - 4) => 11 \(⋮\)(x - 4)
=> (x - 4) \(\in\)Ư(11) = {1; -1; 11; -11}
Lập bảng:
| x - 4 | 1 | -1 | 11 | -11 |
| x | 5 | 3 | 15 | -7 |
vậy ...
c;d tương tự trên