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a, 117 - \(x\) = 28 - (-7)
117 - \(x\) = 28 + 7
117 - \(x\) = 35
\(x\) = 117 - 35
\(x\) = 82
b, \(x\) - (-38 - 2\(x\)) = (-3) - 8 + 2\(x\)
\(x\) + 38 + 2\(x\) = - 11 + 2\(x\)
3\(x\) + 38 = - 11 + 2\(x\)
3\(x\) - 2\(x\) = - 11 - 38
\(x\) = - 49
a/ \(51-(-12+3x)=27\)
\(\Leftrightarrow51+12-27-3x=0\Leftrightarrow36=3x\Leftrightarrow x=\frac{36}{3}=12\)
KL:........
b/ $-x + 21=15+ 2x$
\(\Leftrightarrow2x+x=21-15\Leftrightarrow2x=6\Leftrightarrow x=3\)
KL: ...........
c) $7.(x-9)-5(6-x)=-6+11.x$
\(\Leftrightarrow7x-63-30+5x=-6+11x\Leftrightarrow7x+5x-11x=-6+63+30\Leftrightarrow x=87\)
KL:............
d) $(x-3).(x^2 + 2)=0$
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x\in\varnothing\end{matrix}\right.\)\(\Leftrightarrow x=3\)
e) $|2x-7|-22=-13$
\(\Leftrightarrow\left|2x-7\right|=9\Leftrightarrow\left[{}\begin{matrix}2x-7=9\\2x-7=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
KL: ...........
f) $(2x - 1)^3=-125$
\(\Leftrightarrow\left(2x-1\right)^3=\left(-5\right)^3\Leftrightarrow2x-1=-5\Leftrightarrow x=-2\)
KL: ...........
Bài 1L
a) \(\left(x-7\right)\left(x+3\right)< 0\)
TH1:
\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )
TH2:
\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )
Vậy \(-3< x< 7\)
Bài 2:
a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)
\(\Leftrightarrow5x+8-2x+15+21=2x-5\)
\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)
\(\Leftrightarrow x=-49\)
Vậy ...
a) 42 - x = 5
x = 42 - 5
x = 37
b) 15.(x + 4) = 75
x + 4 = 75 : 15
x + 4 = 5
x = 5 - 4
x = 1
d) x + 18 = 12
x = 12 - 18
x = -6
e) 2x - 15 = -19
2x = (-19) + 15
2x = -4
x = -4 : 2
x = -2
f) 36 : (x^3 - 12) = -3
x^3 - 12 = 36 : (-3)
x^3 - 12 = -12
x^3 = 0
=> x = 0
g)x + 15 = 35
x = 35 - 15
x = 20
h) 2x - 13 = 3^2
2x - 13 = 9
2x = 9 + 13
2x = 22
x = 22 : 2
x = 11
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
a) Ta có : 3x + 27 = 9 b) Ta có : 2x + 12 = 3 ( x - 7 ) c) Ta có : 2x^2 - 1 = 49
2x + 12 = 3x - 21 2x^2 = 49 + 1
3x = 9 - 27 2x + 12 + 21 = 3x 2x^2 = 50
3x = -18 33 = x x^2 = 50 : 2
x = ( - 18 ) : 3 x^2 = 25
x = - 6 x^2 = 5^ 2 = ( -5 ) ^ 2
Vậy x = 5 hoặc x = - 5
c) Ta có : /-9 - x / - 5 =12
/ -9 -x / = 12 + 5
/ - 9 - x / = 17
Suy ra : - 9 - x = 17
x = - 9 -17
x = - 26
Xong rồi đó olm duyệt nhanh lên giùm
a) 3x = -18
x= -6
b) 2x+12= 3x-21
12+21=3x-2x
33=x
c)2x^2=50
x^2=25
x=5
d) /-9-x/=17
-9-x =17 hoặc -9-x =-17
suy ra: x= -26 hoặc x= 8
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
a, =>2x-10=x-17
=>2x-x=-17+10
=>x=-7
vậy x=-7
b,=>3-5x-7=-2x-2x-3
=>-5x-4=-4x-3
=>5x-4x=3-4
=>x=-1
c,2x-9-15+27=21-12+x
=>2x+1=9+x
=>2x-x=9-1
=>x=8