tim x biet -{x-3} +2.{3x+2}=x-4
la dau ngoac don
tim X biet
a(3x(X+2):7)x4=120
b2480-720:3+(200-(X+5))=1010
x là nhân
ngoac dung truoc la ngoac lon
giai chi tiết nha
a)[ 3 x ( x + 2 ) : 7 ] x 4 = 120
[ 3 x ( x + 2 ) : 7 ] = 120 : 4
[ 3 x ( x + 2 ) : 7 ] = 30
3 x ( x + 2 ) = 30 x 7.
3 x ( x + 2 ) = 210
( x + 2 ) = 210 : 3
( x + 2 ) = 70
=>x = 70 ‐ 2
x =68
b) 2480 - 720 : 3 + [200 - (x - 5)] = 1010
<=>2240 + 200 - (x-5) =1010
<=> -x+5 =-1430
<=> -x = -1435
<=> x =1435
NHỚ TK MK NHA
a)[ 3 x ( x + 2 ) : 7 ] x 4 = 120
[ 3 x ( x + 2 ) : 7 ] = 120 : 4
[ 3 x ( x + 2 ) : 7 ] = 30
3 x ( x + 2 ) = 30 x 7.
3 x ( x + 2 ) = 210
( x + 2 ) = 210 : 3
( x + 2 ) = 70
=>x = 70 ‐ 2
x =68
b) 2480 - 720 : 3 + [200 - (x - 5)] = 1010
<=>2240 + 200 - (x-5) =1010
<=> -x+5 =-1430
<=> -x = -1435
<=> x =1435
NHỚ TK MK NHA
tim x biet
x^2+3x0
b) x-3/x+2 <0
c) x{x-1}{x+1/3}0
ai nhanh mk tk
luu y : { } la dau ngoc don !
xin loi , cau a) la x^2 +3x <0
tim tong cac he so cua da thuc nhan duoc sau khi bo dau ngoac trong bieu thuc (3-4x+x^2)^2006.(3+4x+x^2)^2007
TIM TONG CAC HE SO CUA DA THUC NHAN DUOC SAU KHI BO DAU NGOAC TRONG BIEU THUC ;
A(X) = ( 3 - 4X + X2 )2004 . ( 3 + 4X + X2 )2005
Giá trị của đa thức sau khi bỏ dấu ngoặc tại x = 1
\(\Leftrightarrow A_{\left(1\right)}=\left(3-4.1+1^2\right)^{2004}\left(3-4.1+1^2\right)^{2005}=0\)
Tổng các hệ số của đa thức A(x) nhân được sau khi bỏ dấu ngoặc chính bằng A(1).
Ta có: \(A\left(1\right)=0^{2004}.8^{2005}\)
\(\Leftrightarrow A\left(1\right)=0\)
Chúc bạn học tốt ! truongthienvuong
bai 1:tim x(chu y dau * la dau nhan)
a)(x+1/4)+(3x-4)+2*(x-3)=1
b)2*(x-3)=3(x+2)-x+1
c)x*(x+3)+x(x-2)=2x*(x-1)
d)(x-1)*3x-2*(x+2)-2x=x(x-1)
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
tim x biet
|3x+9| + 3|x-1|=12-|x+1|20
giup mk voi mk danh dau cho
tim x biet (x-2)^3 -(x-3)(x^2+3x+9)+6(x+1)^2
tim x biet
|x+1|+|3x+2|=3x
|x-2|+4|x+1|=x-3
tim x biet x^3-3x^2+3x-1=0
x^3 - 3x^2 + 3x - 1 = 0
( x- 1)^3 = 0
=> x -1 = 0
=> x = 1