( x-2) nhân (2x-8) =0
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a) \(\frac{8}{9}\cdot x-\frac{2}{3}=\frac{1}{3}\cdot x+1\frac{1}{3}\)
=> \(\frac{8x}{9}-\frac{2}{3}=\frac{x}{3}+\frac{4}{3}\)
=> \(\frac{8x}{9}-\frac{6}{9}=\frac{x+4}{3}\)
=> \(\frac{8x-6}{9}=\frac{x+4}{3}\)
=> \(3\left(8x-6\right)=9\left(x+4\right)\)
=> \(24x-18=9x+36\)
=> \(24x-18-9x=36\)
=> \(24x-9x=54\)
=> \(15x=54\)
=> \(5x=18\)
=> \(x=\frac{18}{5}\)
Vậy x = \(\frac{18}{5}\)
b) \(\left(x-\frac{1}{2}\right)\left(\frac{3}{2}-2x\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{1}{2}=0\\\frac{3}{2}-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=\frac{3}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}:2=\frac{3}{4}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{2};\frac{3}{4}\right\}\)
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 ...
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
Bài 1:
a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)
b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]
= 2xy.(x-y-1).(x+y+1)
c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2
= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)
Bài 2:
a) (x+2).(x^2-2x+4) - (x^3+2x) = 0
x^3 + 8 - x^3 - 2x = 0
8 - 2x = 0
x = 4
b) x^2 - 2x - 8 = 0
x^2 +2x - 4x - 8 = 0
x.(x+2) - 4.(x+2) = 0
(x+2).(x-4) = 0
...
bn tự làm tiếp nha
2x - (x + 5) = 300
=> 2x - x - 5 = 300
=> x - 5 = 300
=> x = 305
5x + (x - 40) = 8
=> 5x + x - 40 = 8
=> 6x - 40 = 8
=> 6x = 48
=> x = 8
(x+3).(x-2).2x = 0
<=> x + 3 = 0 => x = -3
<=> x - 2 = 0 => x = 2
<=> 2x = 0 => x = 0
5x - (2x +128) = 314 + 121
=> 5x - 2x - 128 = 435
=> 3x - 128 = 435
=> 3x = 563
=> x = 563/3
x^2 + x = 0
=> x^2 = -x
=> x = 0 hoặc x = -1.
mik ko bít
I don't now
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Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
(x-2) (2x-8)=0
\(\Rightarrow\)2x-8=0
2x=0+8
2x=8
x=8:2
x=4
Vậy x=4.
(x-2). (2x-8)=0
2x-8=0
2x= 0+8
2x=8
x= 8:2
x=4