x(x+2)(x^2-3x+1) giúp e vs ạ T.T
Giúp e vs ạ Giải bất pt: a) 2x - x(3x + 1) < 15 - 3x(x + 2) b) 4(x - 3)² - (2x - 1)² ≥ 12x
a: =>2x-3x^2-x<15-3x^2-6x
=>x<-6x+15
=>7x<15
=>x<15/7
b: =>4x^2-24x+36-4x^2+4x-1>=12x
=>-20x+35>=12x
=>-32x>=-35
=>x<=35/32
\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)
Vậy S={-∞; 15/7}
\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)
Vậy S={-∞; 35/32]
X2- 3x +( x+1) ×(x-3) =0
Giúp e vs ạ
\(x^2-3x+\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
Tìm số tự nhiên x biết:
(x+2)-2=0
(x+3)+1=7
(3x-4)+4=12
(5x+4)-1=13
(4x-8)-3=5
8-(2x-4)=2
7+(5x+2)=14
5-(3x-11)=1
Giúp e vs ạ(Vui lòng trình bày ạ)
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
Giải = phương trình đặt ẩn phụ
a) C=(x2+x)2+3(x2+x)+2
b) E=(x2+x+1) (x2+3x+1)+x
LÀM ƠN GIÚP MK ĐI T.T
Giải = phương trình đặt ẩn phụ
a) C=(x2+x)2+3(x2+x)+2
b) E=(x2+x+1) (x2+3x+1)+x
LÀM ƠN GIÚP MK ĐI T.T
Phân tích đa thức thành nhân tử
a) A=( x^2+x)^2 + 4x^2 + 4x -12
b) B = ( x^2 + x + 1) (x^2 + x +2) -12
c) C= x^4y^4 + 4
Mn giúp mình vs ạ
Xíu nx mình đi học r T.T
a) (x-2)2 - (x+3)2 + (x+4).(x-4)
b). 2.(3x-2)2 - 3.(2x+5)2 -6.(x-1).(x+1)
Mn giúp e vs ạ !
a) (x-2)^2-(x+3)^2+(x+4)(x-4)
= (x^2-4x+4)-(x^2+6x+9)+(x^2-16)
= x^2-4x+4-x^2-6x-9+x^2-16
= x^2- 10x- 21
b) 2(3x-2)^2-3(2x+5)^2-6(x-1)(x+1)
= (6x^2-24x+8)-(6x^2+60x+75)-(6x^2-6)
=6x^2-24x+8-6x^2-60x-75-6x^2+6
=-6x^2-84x-61
Phân tích (3x-2)^2 sau đó nhân với 2: 2(3x^2+2*3x*2+2^2)=6x^-24x+8
A= 3x^3+6x^2-3x-x^3+1/2 tại x-1/3 Giúp mình vs ạ mình cần gấp Cảm ơn ạ
\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)
=13/54
Rút gọn phân thức : \(\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
Mọi ngườ giúp e vs ạ
\(\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
\(=\frac{2x}{x\left(x-3\right)}+\frac{2x}{x^2-3x-x+3}+\frac{x}{x-1}\)
\(=\frac{2}{x-3}+\frac{2x}{x\left(x-3\right)-\left(x-3\right)}+\frac{x}{x-1}\)
\(=\frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{2x}{\left(x-3\right)\left(x-1\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}\)
\(=\frac{2x-2+2x+x^2-3x}{\left(x-3\right)\left(x-1\right)}\)
\(=\frac{x^2+x-2}{\left(x-3\right)\left(x-1\right)}=\frac{x^2-x+2x-2}{\left(x-3\right)\left(x-1\right)}=\frac{x\left(x-1\right)+2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=\frac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(x-1\right)}=\frac{x+2}{x-3}\)