v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4c^3\right)\)

\(=-\dfrac{1}{2}\left(2x+6-4c^3\right)+3\left(2x+6-4c^3\right)\)

\(=-x^2-3x+2c^3x+6x+18-12c^3\)

\(=-x^2+3x+2c^3x+18-12c^3\)

f) \(\left(2x-5\right)\left(x^2-x+3\right)\)

\(=2x\left(x^2-x+3\right)-5\left(x^2-x+3\right)\)

\(=2x^3-2x^2+6x-5x^2+5x-15\)

\(=2x^3-7x^2+11x-15\)

w) \(\left(3x+1\right)\left(x^2-2x-5\right)\)

\(=3x\left(x^2-2x-5\right)+\left(x^2-2x-5\right)\)

\(=3x^3-6x^2-15x+x^2-2x-5\)

\(=3x^3-5x^2-17x-5\)

x) \(\left(6x-3\right)\left(x^2+x-1\right)\)

\(=6x\left(x^2+x-1\right)-3\left(x^2+x-1\right)\)

\(=6x^3+6x^2-6x-3x^2-3x+3\)

\(=6x^3+3x^2-9x+3\)

y) \(\left(5x-2\right)\left(3x+1-x^2\right)\)

\(=5x\left(3x+1-x^2\right)-2\left(3x+1-x^2\right)\)

\(=15x^2+5x-5x^3-6x-2+2x^2\)

\(=-5x^3+17x^2-x-2\)

z) \(\left(\dfrac{3}{4}x+1\right)\left(4x^2+4x+4\right)\)

\(=\dfrac{3}{4}x\left(4x^2+4x+4\right)+\left(4x^2+4x+4\right)\)

\(=3x^3+3x^2+3x+4x^2+4x+4\)

\(=3x^3+7x^2+7x+4\)

f: =2x^3-2x^2+6x-5x^2+5x-15

=2x^3-7x^2+11x-15

w: =3x^3-6x^2-15x+x^2-2x-5

=3x^3-5x^2-17x-5

x: =6x^3+6x^2-6x-3x^2-3x+3

=6x^3+3x^2-9x+3

y: =(5x-2)(-x^2+3x+1)

=-5x^3+15x^2+5x+2x^2-6x-2

=-5x^3+17x^2-x-2

z: =3x^3+3x^2+3x+4x^2+4x+4

=3x^3+7x^2+7x+4

\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a) 9x²-49=0

(3x)²-7²=0

<=> (3x-7)(3x+7)=0

th1: 3x-7=0

<=>3x=7

<=>x=\(\dfrac{7}{3}\)

th2: 3x+7=0

<=>3x=-7

<=>x=\(-\dfrac{7}{3}\)

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x² - 10x - 2x² + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x² + 2x + x + 2 - x² - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x² - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x² - 4x + 5 - 4x² + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x² = 4x - 5

<=> 2x² + x + 6x + 3 - 2x² - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0

<=> - 8x + 32 + 8x² + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

x=-15/6

p) \(\left(9-x\right)\left(x^2+2x-3\right)\)

\(=9\left(x^2+2x-3\right)-x\left(x^2+2x-3\right)\)

\(=9x^2+18x-27-x^3-2x^2+3x\)

\(=-x^3+7x^2+21x-27\)

n) \(\left(-x+3\right)\left(x^2+x+1\right)\)

\(=-x\left(x^2+x+1\right)+3\left(x^2+x+1\right)\)

\(=-x^3-x^2-x+3x^2+3x+3\)

\(=-x^2+2x^2+2x+3\)

o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)

\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)

\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)

\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)

q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)

\(=6x\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)

\(=6x^3-12x^2-18x+x^2-2x-3\)

\(=6x^3-11x^2-20x-3\)

r) \(\left(2x+1\right)\left(-x^2-3x+1\right)\)

\(=2x\left(-x^2-3x+1\right)+\left(-x^2-3x+1\right)\)

\(=-2x^3-6x^2+2x-x^2-3x+1\)

\(=-2x^3-7x^2-x+1\)

u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)

\(=2x\left(-x^2+x+6\right)-3\left(-x^2+x+6\right)\)

\(=-2x^3+2x^2+12x+3x^2-3x-18\)

\(=-2x^3+5x^2+9x-18\)

s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)

\(=-4x\left(x^2+3x-2\right)+5\left(x^2+3x-2\right)\)

\(=-4x^3-12x^2+8x+5x^2+15x-10\)

\(=-4x^3-7x^2+23x-10\)

v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)

\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)

\(=-x^2-3+2x^4+6x+18-12x^3\)

\(=2x^4-12x^3-x^2+6x+15\)

p: (-x+9)(x^2+2x-3)

=-x^3-2x^2+3x+9x^2+18x-27

=-x^3+7x^2+21x-27

n: (-x+3)(x^2+x+1)

=-x^3-x^2-x+3x^2+3x+3

=-x^3+2x^2+2x+3

o: (-6x+1/2)(x^2-4x+2)

=-6x^3+24x^2-12x+1/2x^2-2x+1

=-64x^3+49/2x^2-14x+1

q: (6x+1)(x^2-2x-3)

=6x^3-12x^2-18x+x^2-2x-3

=6x^3-11x^2-20x-3

r: (2x+1)(-x^2-3x+1)

=-2x^3-6x^2+2x-x^2-3x+1

=-2x^3-7x^2-x+1

u: =-2x^3+2x^2+12x+3x^2-3x-18

=-2x^3+5x^2+9x-18

s: =-4x^3-12x^2+8x+5x^2+15x-10

=-4x^3-7x^2+23x-10

1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2) ĐKXĐ: \(x\ge3\)

\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)

4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)