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a, 3 ( x + 1 ) - 2 ( 3 x - 4 ) = - 13
=> 3x + 3 - 6x + 8 = - 13
=> 6x - 3x = 3 + 8 + 13
=> 3x = 24
=> x = 8
b, 2 ( x - 3 ) - 4 ( 2 x - 1 ) = - 20
=> 2x - 6 - 8x + 4 = - 20
=> 8x - 2x = - 6 + 4 + 20
=> 6x = 18
=> x = 3
c, 2 x ( x + 3 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)
d, ( x - 1 ) ( 5 x - x ) = 0
=> \(\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)
e, ( x + 3 ) 2 ( 4 - x ) = 0
=> \(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x+3=0\\4-x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
a) \(3\left(x+1\right)-2\left(3x-4\right)=-13\)
\(\Leftrightarrow3x+3-6x+8=-13\)
\(\Leftrightarrow3x-6x=-13-3-8\)
\(\Leftrightarrow-3x=-24\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
b) \(2\left(x-3\right)-4\left(2x-1\right)=-20\)
\(\Leftrightarrow2x-6-8x+4=-20\)
\(\Leftrightarrow2x-8x=-20+6-4\)
\(\Leftrightarrow-6x=-18\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
c) \(2x\left(x+3\right)=0\)
\(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
d)\(\left(x-1\right)\left(5x-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
e)\(\left(x+3\right)^2\left(4-x\right)=0\)
\(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+3=0\\-x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
a) x+15 = 20-4x
=> x=1
b) 17-x=7-6x
=> x=-2
c) -12+x=5x-20
=> x=2
d) 4x-5=15-x
=> x=4
e) 9x-7=20-6x
=>x= \(\frac{9}{5}\)
g)2.(x-10)=3.(x-20)=x-4
=> x thuộc ∅
h) (x^2+2).(x-3) <0
=> x=3,...
\(\left(5x-55\right).\left(x+20\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-55=0\\x+20=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-20\end{cases}}\)
Vậy x = 11 hoặc x = -20 .
(5x-55).(x+20) =0
Ta có 2 trường hợp:
TH1: 5x-55 = 0
5x = 55
x = 11
TH2: x+20=0
x=-20
Vậy x= 11 hoặc x= -20.
Chúc bạn học giỏi!
a) Ta có: \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{2}\Rightarrow x=1\)
hoặc \(x-\frac{1}{2}=-\frac{1}{2}\Rightarrow x=-\frac{1}{2}+\frac{1}{2}=0\)
Vậy \(x\in\left\{1;0\right\}\)
b)\(20\%.x+\frac{2}{5}x=x-4\)
\(\Rightarrow\frac{1}{5}x+\frac{2}{5}x-x=-4\)
\(\Rightarrow\frac{-2}{5}x=-4\)
\(\Rightarrow x=-4:\left(-\frac{2}{5}\right)=10\)
Vậy x=10
c) 3(x+1)+4(x+3)-2(x+16) = 18
=> 3x+3+4x+12-2x-32 =18
=> (3x+4x-2x)+(3+12-32)=18
=> 5x - 17 = 18
=> 5x = 18 + 17 = 35
=> x = 35 : 5 = 7
Vậy x = 7
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\5x+20=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\5x=-20\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
#quankun^^
Bài giải
\(\left(x-3\right)\left(5x+20\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x+20=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\5x=-20\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{3\text{ ; }-4\right\}\)