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a) (4x – 2)(x + 5) = 0
⇔ \(\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}4x=2\\x=-5\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
Vậy ...
b) 2x – 9 = -8 – 9
⇔ 2x = - 8
⇔ x = - 4
Vậy x = - 4
c) 3.| x -1 | - 27 = 0
⇔ 3 . | x - 1| = 27
⇔ | x - 1| = 9
⇔ \(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
Vậy x ∈ { 10 ; - 8}
d) 5.(3x + 8) –7.(2x + 3) = 16
⇔ 15x + 40 - 14x - 21 = 16
⇔ x + 19 = 16
⇔ x = - 3
Vậy ..
a)\(\left(4x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
Vậy...
b)\(2x-9=-8-9\\ \Leftrightarrow2x-9=-17\\ \Leftrightarrow2x=-8\\ \Leftrightarrow x=-4\)
Vậy...
c)\(2\left|x-1\right|-27=0\\ \Leftrightarrow\left|x-1\right|=\frac{27}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{27}{2}\\x-1=-\frac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{2}\\x=-\frac{25}{2}\end{matrix}\right.\)
Vậy...
a,( x - 29 ) - ( 17 - 38 ) = -9
( x - 29 ) + 21 = - 9
x - 29 = - 9 - 21
x - 29 = - 30
x = -30 + 29
x = -1
c, ( 27 - x ) + ( 15 + x ) = - 24
27 - x + 15 + x = -24
- x + x = -24-27-15
Vô lí vì 0 ko bằng -66
Vậy \(x\in\varnothing\)
d, |2x - 7|- 9 =20
|2x-7|=11
* 2x-7=11 * 2x-7=-11
2x=11+7 2x=-11+7
2x=18 2x=-4
x=18:2 x=-4:2
x=9 x=-2
Vậy x=9 hoặc x=-2
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
Bài 1 : 12 - 12 + 11 + 10 - 9 + 8 - 7 + 5 - 4 + 3 + 2 - 1
= ( 12 - 12 ) + ( 11 - 1 ) + ( 10 - 9 ) + ( 8 - 7 ) + ( 5 - 4 ) + ( 3 + 2 )
= 0 + 10 + 1 + 1 + 1 + 5
= 18
Bài 2 :
3x + 27 = 9
3x = 9 - 27
3x = - 18
x = - 6
2x + 12 = 3( x - 7 )
2x + 12 = 3x - 21
3x - 2x = 12 + 21
x = 33
2x2 - 1 = 49
2x2 = 49 + 1
2x2 = 50
x2 = 50 : 2
x2 = 25
=> x = 5 hoặc x = - 5
- | 9 - x | - 5 = 12
- | 9 - x | = 12 + 5
- | 9 - x | = 17
TH1 : 9 - x >= 0 <=> x <= 9
=> - ( 9 - x ) = 17
=> x = 26 ( loại )
TH2 : 9 - x < 0 <=> x > 9
=> - ( 9 - x ) = -17
=> x = - 8 ( loại )
=> ko có giá trị nào thõa mãn
Bài 3 a,: A = ( - a - b + c ) - ( - a - b - c )
= - a - b + c + a + b + c
= 2c
b, thay c = - 2 vào biểu thức A = 2c
Ta được : A = 2 x ( -2 ) = - 4
a. 2x - 9 = -8 -9
=> 2x - 9 = -17
=> 2x = - 17 + 9
=> 2x = -8
=> x = -4
Vậy x = -4
b. 3 . | x - 1 | - 27 = 0
=> 3 . | x - 1 | = 27
=> | x - 1| = 9
=> x -1 =9 hoặc x-1 = -9
Với x - 1 =9
=> x = 10
Với x - 1 = -9
=> x = -8
Vậy .....
a)2x-9=-8-9
2x-9=-17
2x=-17+9
2x=-8
x=-8:2
x=-4
vậy x=-4
b)3.|x-1|-27=0
3.|x-1|=0+27
3.|x-1|=27
|x-1|=27:3
|x-1|=9
* x-1=9 * x-1=-9
x=9+1 x=-9+1
x=10 x=-8
vậy x=10 hoặc x=-8