a)Dat \(x^2-4x+3=a;x^2-7x+6=b \Rightarrow a+b=2x^2-11x+9\)

....

\(a,4x^2-\left(3x+1\right)\left(2x-1\right)=2\left(x-3\right)^2\)

\(\Leftrightarrow4x^2-\left(6x^2-3x+2x-1\right)=2\left(x^2-6x+9\right)\)

\(\Leftrightarrow4x^2-6x^2+x+1-2x^2+12x-18=0\)

\(\Leftrightarrow-4x^2+13x-17=0\)

\(\Leftrightarrow-4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)-\dfrac{103}{16}=0\)

\(\Leftrightarrow-4\left(x-\dfrac{13}{8}\right)^2=\dfrac{103}{16}\)

\(\Leftrightarrow\left(x-\dfrac{13}{8}\right)^2=\dfrac{-103}{64}\Rightarrow\) pt vô nghiệm

\(b,\left(5x-1\right)\left(x+1\right)-\left(2x-1\right)\left(2x+1\right)=x.\left(x+1\right)\)\(\Leftrightarrow5x^2+5x-x-1-\left(4x^2-1\right)=x^2+x\)

\(\Leftrightarrow5x^2+5x-x-1-4x^2+1-x^2-x=0\) \(\Leftrightarrow3x=0\Rightarrow x=0\)

\(c,7x^2-\left(2x-3\right)^2=1+3\left(x+2\right)^2\)

\(\Leftrightarrow7x^2-\left(4x^2-12x+9\right)=1+3\left(x^2+4x+4\right)\)

\(\Leftrightarrow7x^2-4x^2+12x-9=1+3x^2+12x+12\)\(\Leftrightarrow7x^2-4x^2+12x-9-1-3x^2-12x-12=0\)\(\Leftrightarrow-22=0\) ( vô lí)

Vậy phương trình vô nghiệm

cảm ơn nha

1) \(\left(x+1\right)^3-\left(x-1\right)^3=6.\left(x+2\right)^2-9\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)=6\left(x^2+4x+4\right)-9\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+24x+24-9\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2-24x-24+9=0\)

\(\Leftrightarrow-24x-13=0\Leftrightarrow-24x=13\Leftrightarrow x=\dfrac{-13}{24}\) vậy \(x=\dfrac{-13}{24}\)

2) \(\left(2x-1\right).\left(4x^2+2x+1\right)+\left(1-2x\right)^3=3.\left(2x+3\right)^2\)

\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3=3\left(.4x^2+12x+9\right)\)

\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3=12x^2+36x+27\)

\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3-12x^2-36x-27=0\)

\(\Leftrightarrow-42x-27=0\Leftrightarrow-42x=27\Leftrightarrow x=\dfrac{-27}{42}\) vậy \(x=\dfrac{-27}{42}\)

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

[ ( 2x )² - 3² ] - 4 . ( x² + 2.x.2 + 2²) = 6

4x² - 9 - 4 . ( x² + 4x + 4) = 6

4x² - 9 - 4x² - 16x - 16 = 6

-16x -25 = 6

x = \(-\dfrac{31}{16}\)

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

1: \(=x^3-6x^2+12x-8-8x^3-36x^2-54x-27+7\left(x-1\right)^3\)

\(=-7x^3-42x^2-42x-35+7x^3-21x^2+21x-7\)

\(=-63x^2-21x-42\)

2: \(=x^3+125-\left(x^3-8\right)=125+8=133\)

3: \(=8x^3-27-8x^3-12x^2-6x-1=-12x^2-6x-28\)

1. \(\left(x+5\right)\left(x^2-5x+25\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3+125-\left(x^3-8\right)=x^3+125-x^3+8=133\)

1,

\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x-2\right)\left(x^2+2x+4\right)\\ =\left(x^3+5^3\right)-\left(x^3-2^3\right)\\ =x^3+125-x^3+8\\ =\left(x^3-x^3\right)+\left(125+8\right)\\ =133\)

b,

\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+1\right)^3\\ =\left[\left(2x\right)^3-3^3\right]-\left[\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x+1+1\right]\\ =\left(8x^3-27\right)-\left(8x^3+12x^2+6x+1\right)\\ =8x^3-27-8x^3-12x^2-6x-1\\ =\left(8x^3-8x^3\right)-\left(12x^2+6x\right)-\left(27+1\right)\\ =-6x\left(2x+1\right)-28\\ =\left(-2\right)\left[3x\left(2x+1\right)+14\right]\)

1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ

2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ

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

\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)=2x-3\)

\(\Leftrightarrow x^2-25-x^2-6x-9=2x-3\)

\(\Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\)

\(\Leftrightarrow-8x-31=0\)

\(\Leftrightarrow x=\dfrac{-31}{8}\)

\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)

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

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

\(\Leftrightarrow\left(x^3-12x^2+48x-64\right)-x^3+5^3=x^3+2^3-\left(x^3+12x^2+48x+64\right)\)

\(\Leftrightarrow x^3-12x^2+48x-64-x^3+5^3=x^3+2^3-x^3-12x^2-48x-64\)

\(\Leftrightarrow-12x^2+48x-64+5^3=2^3-12x^2-48x-64\)

\(\Leftrightarrow-12x^2+48x-61=-12x^2-48x-56\)

\(\Leftrightarrow96x=-117\)

\(\Leftrightarrow x=\dfrac{-117}{96}=\dfrac{-39}{32}\)