Tìm x biết (2x-6) x ( 3x-12)=0
Bài 1:a)Tìm x biết |3x-12|+4x=2x-2 b)tìm x,y biết |6+x|+(3+y)2=0
Bài 2:Tìm x để g(x)=0 biết rằng g(x)=3x2-3-8g(x)
b
\(\left|6+x\right|\ge0;\left(3+y\right)^2\ge0\Rightarrow\left|6+x\right|+\left(3+y\right)^2\ge0\)
Suy ra \(\left|6+x\right|+\left(3+y\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}6+x=0\\3+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)
a
Ta có:\(\left|3x-12\right|=3x-12\Leftrightarrow3x-12\ge0\Leftrightarrow3x\ge12\Leftrightarrow x\ge4\)
\(\left|3x-12\right|=12-3x\Leftrightarrow3x-12< 0\Leftrightarrow3x< 12\Leftrightarrow x< 4\)
Với \(x\ge4\) ta có:
\(3x-12+4x=2x-2\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\left(KTMĐK\right)\)
Với \(x< 4\) ta có:
\(12-3x+4x=2x-2\)
\(\Rightarrow10=x\left(KTMĐK\right)\)
Tìm số nguyên x, biết: a) 12 + (2x - 11) = 53; b) 21 - (-6+ 3x) = 9 c) - (2x + 4) + 11 = -27; d) 33 - (33 - x) = 0
Tìm x biết a) 3x^2+x)4-3x)=12 b)3x^2-2x-1=0
b: \(3x^2-2x-1=0\)
=>\(3x^2-3x+x-1=0\)
=>\(\left(x-1\right)\left(3x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a: Bạn ghi lại đề đi bạn
tìm x
a) 3x + 12 = 2x - 4
b) 14 - 3x = -x + 4
c) ( 2x - 8 ) ( x + 6 ) = 0
a) 3x + 12 = 2x - 4
=> 3x - 2x = -4 - 12
=> 1x = -16
=> x = -16
vậy___
b) 14 - 3x = -x + 4
=> 14-4 = -x+3x
=> 10 = 2x
=> x = 10 : 2
=> x = 5
vậy_____
\(c) ( 2x - 8 ) ( x + 6 ) = 0\)
\(\Rightarrow\orbr{\begin{cases}2x-8=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)
vậy_____
tìm x thuộc Z biết:
a) 8 . (7 - x) = 15 - 7x
b) ( x - 1) . ( -14 - 8) < 0
c) (x + 5). (3x - 12) > 0
d) 7 . ( -x + 10 ) - (3x + 7) . 6 = - 72
e) (x - 5) . ( 2x - 16) = 0
tìm số nguyên x biet
-2.(x+6)+6.(x-10)=8
x+5=20-(12-7)
(3x-9).(2x+6)=0
(2x+1).(3-x)=0
Tìm số nguyên x biết
a,3x+3y-2xy=7
b,xy+2x+y+11=0
c,xy+x-y=4
d,2x.(3y-2)+(3y-2)=12
e,3x+4y-xy=15
f,xy+3x-2y=11
g,xy+12=x+y
h,xy-2x-y=-6
i,xy+4x=25+5y
ii,2xy-6y+x=9
iii,xy-x+2y=3
k,2.x^2.y-x^2-2y-2=0
l,x^2.y-x+xy=6
Tìm x biết:
\(a) x^2+3x-10=0 \)
\(b) x^2-5x-6=0\)
\(c) 2x^2+3x-2=0\)
a: Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b: Ta có: \(x^2-5x-6=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
Tìm x biết
a) (x+2).(x+3) - (x-2).(x+5)=10
b) (3x+2). (2x+9) - (x+2). (8x+11)=(x+1).(3-2x)
c) 3.(2x-1).(3x-1)-(2x-3).(9x-1)=0
d) (5x-8).(4x-5)-(3x-4).(2x+12)=12
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3