a)2x+2-2x=192
b)(2x-1)2=(2x-1)5
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a: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)
\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)
\(\Leftrightarrow x=25\)
b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)
c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)
\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)
\(\Leftrightarrow-9x=1\)
hay \(x=-\dfrac{1}{9}\)
\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{2x+1}{-\left(2x+1\right)}-\dfrac{2x-1}{-\left(2x-1\right)}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{2x}{5x-5}\)
a: (2x+5)(4-3x)
=8x-6x^2+20-15x
=-6x^2-7x+20
b: (3xy+2x^2)*(-3x^2+xy)
=-9x^3y+3x^2y^2-6x^4+2x^3y
=-7x^3y+3x^2y^2-6x^4
c: (-1/2x^2y+6x)(1/2x^2y-2x)
=-1/4x^4y^2+x^3y+3x^3y-12x^2
=-1/4x^4y^2+4x^3y-12x^2
a) \(\left(2x+5\right)\left(4-3x\right)\)
\(=2x\left(4-3x\right)+5\left(4-3x\right)\)
\(=8x-6x^2+20-15x\)
\(=-6x^2+\left(8x-15x\right)+20\)
\(=-6x^2-7x+20\)
b) \(\left(3xy+2x^2\right)\left(-3x^2+xy\right)\)
\(=3xy\left(-3x^2+xy\right)+2x^2\left(-3x^2+xy\right)\)
\(=-9x^3y+3x^2y^2-6x^4+2x^3y\)
\(=3x^2y^2+\left(-9x^3y+2x^3y\right)-6x^4\)
\(=3x^2y^2-7x^3y-6x^4\)
c) \(\left(-\dfrac{1}{2}x^2y+6x\right)\left(-2x+\dfrac{1}{2}x^2y\right)\)
\(=-\dfrac{1}{2}x^2y\left(-2x+\dfrac{1}{2}x^2y\right)+6x\left(-2x+\dfrac{1}{2}x^2y\right)\)
\(=x^3y-\dfrac{1}{4}x^4y^2-12x^2+3x^3y\)
\(=\left(x^3y+3x^3y\right)-\dfrac{1}{2}x^4y^2-12x^2\)
\(=4x^3y-\dfrac{1}{2}x^4y^2-12x^2\)
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
^^
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
a) 2x+2 - 2x= 192
2x . 22 - 2x . 1 = 192
2x . ( 22 - 1) = 192
2x . 3 = 192
2x = 192 : 3
2x = 64 = 26
=> x = 6
bn có thể trả lời lại đc ko.?mk ko thấy câu trả lời