a,-2.(x-6)+6.(x-10)=8
b,-4.(2x+9)-(8x+3)-(x+13)=0
c,7x.(2+x)-7x.(x+3)=14
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+ -2X-12+6X-60=8
<=>4X=80]
<=> X = 20
V.....
+ -8X-36 +8X -3 - X - 13 =0
<=> -X = 52
<=> X = -52
V.....
+ 14X +7\(X^2\)-7\(X^2\)-21X=14
<=> -7X =14
<=> X= -2
V .......
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
a, - 2 .( x + 6 ) + 6 . ( x - 10 ) = 8
- 2x - 12 + 6x - 60 = 8
4x - 72 = 8
4x = 8 + 72
4x = 80
x = 20
b, - 4 . ( 2x + 9 ) - ( - 8x + 3 ) - ( x + 13 ) = 0
- 8x - 36 + 8x - 3 - x - 13 = 0
- x - 52 = 0
x = - 52
c, 7x . ( 2 + x ) - 7x . ( x + 3 ) = 14
7x . ( 2 + x - x - 3 ) = 14
7x . ( - 1 ) = 14
7x = 14 : ( - 1 )
7x = - 14
x = - 2
d, 2 . ( 5 + 3x ) + x = 31
10 + 6x + x = 31
10 + 7x = 31
7x = 31 - 10
7x = 21
x = 3
a)-2(x+6)+6(x-10)=8
-2x+-12+6x+-60=8
4x+-72=8
4x=80
x=80:4
x=20
b)-4(2x+9)-(-8x+3)-(x+13)=0
-8x+-36+8x-3+x-13=0
(-8x+8x)+-36+-3+x-13=0
0+-52+x=0
x=0-(-52)
x=-52
c)7x(2+x)-7x(x+3)=14
14x+7x2
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
a) -2. (x+ 6) + 6. (x - 10) =8
<=> -2x - 12 + 6x - 60 = 8
<=> 4x = 36
<=> x = 9
b) -4. (2x + 9) - (-8x + 3) - (x + 13) = 0
<=> -8x - 36 + 8x - 3 - x - 13 = 0
<=> x = 52
c) 7x. (2 + x) - 7x. (x + 3) = 14
<=> 14x + 7x2 - 7x2 - 21x = 14
<=> -7x = 14
<=> x = -2
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Trả lời
Bài 1:
a)(45-25).(-11)+29.(-3-17)
=20.(-11)+29.(-20)
=20.(-11)+(-29).20
=20.[(-11)+(-29)]
=20.(-30)
=-600
b)(36-6).(-5)+21.(-17-3)
=30.(-5)+21.(-20)
=(-150)+(-420)
=-570
trả lời:
a,(45-25).(-11)+29.(-3-17)
=20.(-11)+29.(-20)
=20.(-11)+(-29).20
=20.[(-11)+(-29).20
=20.(-30)
=-600
học tốt
a, =>-2x+12+6x-60=8
=>4x+48=8
=>4x=56
=>x=14
b,=>-8x-36-8x-3-x-13=0
=>-17x-52=0
=>-17x=52
=>x=-52/17
c,=>14x+7x^2-7x^2-21x=14
=>-7x=14
=>x=-2
a <=> -2x+12 +6x-60-8=0
<=>4x-56=0
<=>x=56/4=14
Vậy S=(4)