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1) \(\left(-27\right).\left(-28+128\right)=-27.100=-2700\)
2a)\(\left(x-3\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
b) \(\left(2x-1\right)^2=81\)
\(\sqrt{\left(2x-1\right)^2}=9\)
\(\left|2x-1\right|=9\)
\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
c) \(\left(2m+5\right)^3=-27\)
\(\sqrt[3]{\left(2m+5\right)^3}=-3\)
\(2m+5=-3\)
\(m=-4\)
d) \(\left(3x-2\right)^3=64\)
tương tự câu c

\(\left(2600+6400\right)-3\cdot x=2000\)
\(9000-3x=2000\)
\(3x=9000-2000\)
\(3x=7000\)
\(x=7000:3\)
\(x=\frac{7000}{3}\)


-2 . ( 3x+2 ) - 3 = 28
-6.2x - 4 = 31
-6 .2x = 35
2x = \(\dfrac{-35}{6}\)
x = \(\dfrac{-35}{6}.\dfrac{1}{2}\)
x = \(\dfrac{-35}{12}\)

\(-2\left(2x+2\right)-3=28\)
\(-2\left(2x+2\right)=28+3\)
\(-2\left(2x+2\right)=30\)
\(2x+2=30:\left(-2\right)\)
\(2x+2=15\)
\(2x=15-2\)
\(2x=13\)
\(x=13:2\)
\(x=\dfrac{13}{2}\)
Vậy .................

b ) -2(3x+2)-3 =28
-6x-4-3=28
-6x = 28 +4+3
-6x = 35
x =\(\dfrac{35}{-6}\)= \(\dfrac{-35}{6}\)= -5,8
vậy x= -5,8
c ) -(x+3)-5(x-7)=10x-20
-(x+3)-5x+35=10x-20
-x+3-5x+35=10x-20
-x-5x-10x=-20-3-35
-16x=-58
x=\(\dfrac{-58}{-16}\)=\(\dfrac{58}{16}\)=3,625
vậy x =3,625
d)2x+37=-6(3x-8)
2x+37=-18x+48
2x+18x=48-37
20x=11
x=\(\dfrac{11}{20}\)=0,55
vậy x=0,55

(x - 2)(2x - 6) = 0
=> x - 2 = 0 hoac 2x - 6 = 0
=> x = 2 hoac x = 3
vay_
(3x + 9)(1 - 3x) = 0
=> 3x + 9 = 0 hoac 1 - 3x = 0
=> x = -3 hoac x = 1/3
vay_
|2 - x| + 2 = x
=> |2 - x| = x - 2
=> 2 - x = x - 2 hoac 2 - x = 2 - x
=> -2x = -4 hoac x thuoc tap hop rong
=> x = 2
\(a,\left(x-2\right).\left(2x-6\right)=0\Leftrightarrow2\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(b,\left(3x+9\right).\left(1-3x\right)=0\Leftrightarrow3\left(x+3\right).\left(1-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\1-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
\(c,\left(x^2+1\right).\left(81-x^2\right)=0\Leftrightarrow\left(x^2+1\right).\left(9-x\right).\left(9+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9-x=0\\9+x=0\end{cases}}\) ( vì \(x^2+1\ne0\forall x\)) \(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
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