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\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4.5\right)=3.5\)
\(=>2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4.5-3.5=0\)
\(=>-6x-8=0\)
\(=>-2\left(3x+4\right)=0\)
\(=>3x+4=0\)(vì \(-2\ne0\))
\(=>x=\frac{-4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x\left(2x^2+10x-x-5\right)-\left(2x^3+9x^2+x+4.5\right)=3.5\)
\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4.5=3.5\)
=>-6x=8
hay x=-4/3
![](https://rs.olm.vn/images/avt/0.png?1311)
#)Giải :
\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-9x^2-x-4,5=3,5\)
\(\Leftrightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)
\(\Leftrightarrow-5x-4,5=3,5\)
\(\Leftrightarrow-5x=8\)
\(\Leftrightarrow x=-\frac{8}{5}\)
\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\)
\(\Leftrightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)
\(\Leftrightarrow-6x=8\)
\(\Leftrightarrow x=\frac{-8}{6}=\frac{-4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ => 4x2 - 4x + 1 + 4x2 + 4x + 1 = 16
=> 8x2 = 14
=> x2 = 14/8
=> x = \(\frac{\sqrt{7}}{2}\) hoặc x = \(-\frac{\sqrt{7}}{2}\)
b/ => 6x2 - (6x2 - 11x - 10) = 17
=> 6x2 - 6x2 + 11x + 10 = 17
=> 11x = 7
=> x = 7/11
c/ => 2x(x + 5) - x2 - 5x = 0
=> 2x(x + 5) - x(x + 5) = 0
=> (x + 5)(2x - x) = 0
=> x(x + 5) = 0
=> x = 0
hoặc x + 5 = 0 => x = -5
Vậy x = 0 ; x = -5
d/ \(x^2+\frac{1}{x^2}+2x+\frac{2}{x}=-3\)
đề là như vầy hả
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(\Rightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-9x^2-x-4,5=3,5\)
\(\Rightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)
\(\Rightarrow-5x-4,5=3,5\)
\(\Rightarrow-5x=8\)
\(\Rightarrow x=-\dfrac{8}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^2-3x\left(x-2\right)=36\\ \Rightarrow3x\left(x-x+2\right)=36\\ \Rightarrow6x=36\\ \Rightarrow x=6\)
\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\\ \Rightarrow3x^3-4x^2+2x-1+\left(4x^2-3x^3\right)=\dfrac{5}{2}\\ \Rightarrow2x-1=\dfrac{5}{2}\\ \Rightarrow x=\dfrac{7}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(^{\left(2x+1\right)^2-\left(x+2\right)^2-3x\left(x+2\right)=\left(2x+1\right)^2-\left(x+2\right)\left(x+2+3x\right)}\)
\(=\left(2x+1\right)^2-\left(x+2\right)\left(4x+2\right)=\left(2x+1\right)^2-2\left(x+2\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(1-2x-4\right)=\left(2x+1\right)\left(-3-2x\right)=-\left(2x+1\right)\left(3+2x\right)\)
\(\left(2x+1\right)^2-\left(x+2\right)^2-3x\left(x+2\right)\)
\(=4x^2+4x+1-\left(x^2+4x+4\right)-3x^2-6x\)
\(=4x^2+4x+1-x^2-4x-4-3x^2-6x\)
\(=-6x-3\)
\(=-3\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2
a) 4x(x-3)-3x+9
=4x(x-3)-3(x-3)
= (x-3)(4x-3)
b) x3+2x2-2x-4
=(x3+2x2)-(2x+4)
=x2(x+2)-2(x+2)
=(x+2)(x2-2)
c) 4x2-4y+4y-1
=4x2-1
=(2x-1)(2x+1)
d) x5-x
=x(x4-1)
=x(x2-1)(x2+1)
a) 4x(x-3)-3x+9
= 4x(x-3) - 3(x-3)
= (x-3)(4x-3)
b)x3 + 2x2 - 2x - 4
= x2(x + 2) - 2(x + 2)
= (x+2)(x2-2)
c) 4x2 - 4y +4y -1
= [(2x)2-12] + (-4y+4y)
= (2x+1)(2x-1)
d) x5-x
= x(x4 - 1)