a)
<=> \(3x-12x^2+12x^2-6x=9\)
<=> \(-3x=9\)
<=> \(x=-3\)
b)
<=> \(6x-24x^2-12x+24x^2=6\)
<=> \(-6x=6\)
<=> \(x=-1\)
c)
<=> \(6x-4-3x+6=1\)
<=> \(3x+2=1\)
<=> \(x=-\frac{1}{3}\)
d)
<=> \(9-6x^2+6x^2-3x=9\)
<=> \(-3x=0\)
<=> \(x=0\)
e) KO HIỂU ĐỀ
f)
<=> \(4x^2-8x+3-\left(4x^2+9x+2\right)=8\)
<=> \(-17x+1=8\)
<=> \(x=-\frac{7}{17}\)
g)
<=> \(-6x^2+x+1+6x^2-3x=9\)
<=> \(-2x=8\)
<=> \(x=-4\)
h)
<=> \(x^2-x+2x^2+5x-3=4\)
<=> \(3x^2+4x=7\)
<=> \(\orbr{\begin{cases}x=1\\x=-\frac{7}{3}\end{cases}}\)
a. \(3x\left(1-4x\right)+6x\left(2x-1\right)=9\)
\(\Rightarrow3x-12x^2+12x^2-6x=9\)
\(\Rightarrow-3x=9\)
\(\Rightarrow x=-3\)
b. \(3x\left(2-8x\right)-12x\left(1-2x\right)=6\)
\(\Rightarrow6x-24x^2-12x+24x^2=6\)
\(\Rightarrow-6x=6\)
\(\Rightarrow x=-1\)
c. \(2\left(3x-2\right)-3\left(x-2\right)=1\)
\(\Rightarrow6x-4-3x+6=1\)
\(\Rightarrow3x+2=1\)
\(\Rightarrow3x=-1\)
\(\Rightarrow x=-\frac{1}{3}\)
Để mình làm nốt câu n,m,p,q
n, (x2 - 2x + 4)(x + 2) - x(x - 1)(x + 1) + 3 = 0
=> x2(x + 2) - 2x(x + 2) + 4(x + 2) - x(x2 - 1) + 3 = 0
=> x3 + 2x2 - 2x2 - 4x + 4x + 8 - x3 + x + 3 = 0
=> (x3 - x3) + (2x2 - 2x2) + (-4x + 4x + x) + (8 + 3) = 0
=> x + 11 = 0
=> x = -11
Vậy x = -11
m) (2x - 1)(x + 3) - (x - 4)(2x - 5) = 4x + 1
=> 2x(x + 3) - 1(x + 3) - x(2x - 5) + 4(2x - 5) = 4x + 1
=> 2x2 + 6x - x - 3 - 2x2 + 5x + 8x - 20 = 4x +1
=> (2x2 - 2x2) + (6x - x + 5x + 8x) + (-3 - 20) = 4x + 1
=> 18x - 23 = 4x + 1
=> 18x - 23 - 4x - 1 = 0
=> 14x + (-23 - 1) = 0
=> 14x - 24 = 0
=> 14x = 24
=> x = 12/7
Vậy x = 12/7
p) (2x - 1)(2x - 3) - (4x + 3)(x - 2) = 8 - x
=> 2x(2x - 3) - 1(2x - 3) - 4x(x - 2) - 3(x - 2) = 8 - x
=> 4x2 - 6x - 2x + 3 - 4x2 + 8x - 3x + 6 = 8 - x
=> (4x2 - 4x2) + (-6x - 2x + 8x - 3x) + (3 + 6) = 8 - x
=> -3x + 9 = 8 - x
=> -3x + 9 - 8 + x = 0
=> (-3x + x) + 1 = 0
=> -2x + 1 = 0
=> -2x = -1
=> x = 1/2
q, 6x2 - 2x(3x + 3/2) = 9
=> 6x2 - 6x2 - 3x = 9
=> -3x = 9
=> x = -3
