Tìm x biết : a ) ( x - 2 ) . ( 3x - 9 ) = 0
b ) ( 3 - x ) . ( x + 5 ) = 0
c ) ( x + 1 ) . ( 4 - 2x ) = 0
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a) (x-2).(3x-9)=0
\(\Rightarrow\orbr{\begin{cases}x-2=0\\3x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\3x=9\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy x=2 hoặc x=3
Phần b,c tương tự
Ta có : (3 - x)(x + 5) = 0
\(\Leftrightarrow\orbr{\begin{cases}3-x=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
a) 2x2+3x-5=0
=> 2x2+5x-2x-5=0
=> x(2x+5)-(2x-5)=0
=> (2x-5)(x-1)=0
=> 2x-5=0, x-1=0
=> x=5/2; 1
\(2x^2+3x-5=0< =>2x^2-2+3x-3=0\)
\(< =>2\left(x+1\right)\left(x-1\right)-3\left(x-1\right)=0\)
\(< =>\left(x-1\right)\left(2x-1\right)=0< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 = 4
<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy S = { 5 ; 1 }
b) x2 - 9 = 0
<=> x2 = 9
<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy S = { 3 ; -3 }
c) x( x - 2x ) - x2 - 8 = 0
<=> x2 - 2x2 - x2 - 8 = 0
<=> -2x2 - 8 = 0
<=> -2x2 = 8
<=> x2 = -4 ( vô lí )
<=> x = \(\varnothing\)
Vậy S = { \(\varnothing\)}
d) 2x( x - 1 ) - 2x2 + x - 5 = 0
<=> 2x2 - 2x - 2x2 + x - 5 = 0
<=> -x - 5 = 0
<=> -x = 5
<=> x = -5
Vậy S = { -5 }
e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0
<=> x2 - 3x - ( x2 - x - 2 ) = 0
<=> x2 - 3x - x2 + x + 2 = 0
<=> - 2x + 2 = 0
<=> -2x = -2
<=> x = 1
Vậy S = { 1 }
f) x( 3x - 1 ) - 3x2 - 7x = 0
<=> 3x2 - x - 3x2 - 7x = 0
<=> -8x = 0
<=> x = 0
Vậy S = { 0 }
\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
a/ \(\orbr{\begin{cases}x-2=0\\2x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{5}{2}\end{cases}}\)
\(a,\left(x-2\right)\left(2x-5\right)=0.\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\2x=5\Leftrightarrow x=\frac{5}{2}\end{cases}}}\)
Vậy ....
\(b,\left(0,2x-3\right)\left(0,5x-8\right)=0\left(\text{Mạo muội sửa đề nha 0,5 thành 0,5x}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}0,2x-3=0\\0,5x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}0,2x=3\\0,5x=8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=15\\x=16\end{cases}}\)
Vậy ... ( có j sai thì bỏ qua cho)
\(c,2x\left(x-6\right)+3\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\2x=-3\Leftrightarrow x=-\frac{3}{2}\end{cases}}}\)
Vậy ...
\(d,\left(x-1\right)\left(2x-4\right)\left(3x-9\right)=0\)
\(\Leftrightarrow2.3\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
( ko có ngoặc vuông 3 cái nên mk trình bày kiểu này)
+ TH1:
x-1=0 <=> x= 1
+ TH2:
x-2=0 <=> x=2
+TH3:
x-3 = 0 <=> x = 3
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
a. x=0
b.x=1,7
c.x=5,3
G.X=7
h.x=6
Mk làm vậy thôi
hok tốt
Professor minhmama
a;\(x\left(x+0\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+0=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=0\end{cases}}}\)
\(b,\left(x-1\right)\left(7-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}}\)
\(c,\left(-x+5\right)\left(3-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)
\(d,\left(x+5\right)+\left(x-9\right)=13\)
\(\Rightarrow x+5+x-9=13\)
\(\Rightarrow2x=17\)
\(\Rightarrow x=\frac{17}{2}\)
\(e;\left(4+x\right)+\left(x-7\right)=x+2\)
\(\Rightarrow4+x+x-7=x+2\)
\(\Rightarrow x=5\)
\(f,\left(3x+5\right)-\left(2x-7\right)=4-x\)
\(\Rightarrow3x+5-2x+7=4-x\)
\(\Rightarrow2x=-8\Rightarrow x=-4\)
\(g,\left(x-1\right)^2=36\)
\(\Rightarrow\left(x-1\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=6\\x-1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=5\end{cases}}}\)
\(h,\left(3-x\right)^3=-27\)
\(\Rightarrow\left(3-x\right)^3=\left(-3\right)^3\)
\(\Rightarrow3-x=-3\)
\(\Rightarrow x=6\)
\(a,\left(x-2\right)\left(3x-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\3x=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(b,\left(3-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
\(c,\left(x+1\right)\left(4-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\4-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
a)(x-2 ).(3x-9)=0
x-2 =0
x=0+2
x=2
Hoặc :3x-9 =0
3x =0+9
3x =9
x =9\(\div\)3
x =3
Vậy : x bằng 2 hoặc bằng 3
b)(3-x).(x+5)=0
3-x =0
x =3-0
x =3
Hoặc : x+5 =0
x =0-5
x =-5
Vậy x bằng 3 hoặc bằng -5
c)(x+1).(4-2x)=0
x+1 =0-1
=-1
Hoặc : 4-2x =0
2x =4-0
2x =4
x =4\(\div\)2
x =2
Vậy x bằng -1 hoặc 2