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\(b.x^4+4x^2-5=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(c.x^3-19x-30=x^3-25x+6x-30\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)
=> (a+b)^2=(a-b)^2+4ab
- 2x – x2 + 2 – x – (3x2 + 6x + 5x +10) = – 4x2 + 2
- 2x – x2 + 2 – x – 3x2 – 6x – 5x – 10 = – 4x2 + 2 –10x = 10 x = – 1
- 2x2 – 6x + x – 3 = 0
(x – 3)(2x + 1) = 0
x = 3 hay x = -1/2
Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)
\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)
\(\Leftrightarrow15x^2-4x-4=0\)
\(\Leftrightarrow15x^2-10x+6x-4=0\)
Lỗi :vvvv
\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...
1.
a) \(\left(-2x^3\right)\)\(\left(x^2+5x-\frac{1}{2}\right)\) = \(-2x^5\)\(-10x^4\) \(+x^3\)
b) (\(6x^3-7x^2\)\(-x+2\))\(:\left(2x+1\right)\)=\(3x^2-5x+2\)
2.
a) 9x(3x-y) + 3y (y-3x)=9x(3x-y)-3y(3x-y)
= (9x-3y)(3x-y)
= 3(3x-y)(3x-y)
= 3(3x-y)^2
b) \(x^3-3x^2\)\(-9x+27\)= \(\left(x^3-3x^2\right)\)\(-\left(9x-27\right)\)
= \(x^2\left(x-3\right)\)\(-9\left(x-3\right)\)
= \(\left(x^2-9\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)^2\)
Bài 1 ) a ) \(\left(-2x^3\right)\left(x^2+5x-\frac{1}{2}\right)\)
\(=-2x^5-10x^4+x^3\)
b ) \(\left(6x^3-7x^2+x+2\right):\left(2x+1\right)\)
\(=3x^2-5x+2\)
2 ) a ) \(9x\left(3x-y\right)+3y\left(y-3x\right)\)
\(=9x\left(3x-y\right)-3y\left(3x-y\right)\)
\(=\left(3x-y\right)\left(9x-3y\right)\)
\(=3\left(3x-y\right)\left(x-y\right)\)
b ) \(x^3-3x^2-9x+27\)
\(=\left(x^3-3x^2\right)-\left(9x-27\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x^2-9\right)\left(x-3\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x-3\right)\)
(3x+2)2-(3x-2)2=5x+38
⇒(3x+2-3x+2)(3x+2+3x-2)=5x+38
⇒4.6x=5x+38
⇒19x=38
⇒x=2
Vậy...
\(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)
\(\Leftrightarrow\left(9x^2+12x+4\right)-\left(9x^2-12x+4\right)=5x+38\)
\(\Leftrightarrow24x=5x+38\Leftrightarrow19x=38\Leftrightarrow x=\frac{38}{19}=2\)
Vậy $x=2$