7: \(2x\left(3-x\right)-2\left(x+5\right)^2=12\)
=>\(6x-2x^2-2\left(x^2+10x+25\right)=12\)
=>\(6x-2x^2-2x^2-10x-50-12=0\)
=>\(-4x^2-4x-62=0\)
=>\(4x^2+4x+1+61=0\)
=>\(\left(2x+1\right)^2+61=0\)(vô lý)
=>\(x\in\varnothing\)
8: \(\left(1-2x\right)\left(1+2x\right)+\left(4x-1\right)\left(x+2\right)=3\)
=>\(1-4x^2+4x^2+8x-x-2=3\)
=>7x-1=3
=>7x=4
=>\(x=\dfrac{4}{7}\)
9: \(3x\left(x-2\right)-3\left(x+1\right)^2=1\)
=>\(3x^2-6x-3\left(x^2+2x+1\right)=1\)
=>\(3x^2-6x-3x^2-6x-3=1\)
=>-12x=4
=>\(x=\dfrac{4}{-12}=-\dfrac{1}{3}\)
Bài 21.7:
Ta có:
\[
2x(3 - x) = 6x - 2x^2
\]
\[
(x + 5)^2 = x^2 + 10x + 25 \Rightarrow 2(x + 5)^2 = 2(x^2 + 10x + 25) = 2x^2 + 20x + 50
\]
\[
6x - 2x^2 - (2x^2 + 20x + 50) = 12
\]
\[
6x - 2x^2 - 2x^2 - 20x - 50 = 12
\]
\[
-4x^2 - 14x - 50 = 12
\]
\[
-4x^2 - 14x - 62 = 0
\]
\[
4x^2 + 14x + 62 = 0
\]
\[
D = b^2 - 4ac = 14^2 - 4 \cdot 4 \cdot 62 = 196 - 992 = -796
\]
Vì \( D < 0 \), nên phương trình không có nghiệm thực.
Bài 21.8:
Ta có:
\[
(1 - 2x)(1 + 2x) = 1 - 4x^2
\]
\[
(4x - 1)(x + 2) = 4x^2 + 8x - x - 2 = 4x^2 + 7x - 2
\]
\[
1 - 4x^2 + 4x^2 + 7x - 2 = 3
\]
\[
1 - 2 + 7x = 3
\]
\[
7x - 1 = 3
\]
\[
7x = 4 \Rightarrow x = \frac{4}{7}
\]
Bài 21.9:
Ta có:
\[
3x(x - 2) = 3x^2 - 6x
\]
\[
(x + 1)^2 = x^2 + 2x + 1 \Rightarrow 3(x + 1)^2 = 3(x^2 + 2x + 1) = 3x^2 + 6x + 3
\]
\[
3x^2 - 6x - (3x^2 + 6x + 3) = 1
\]
\[
3x^2 - 6x - 3x^2 - 6x - 3 = 1
\]
\[
-12x - 3 = 1
\]
\[
-12x - 4 = 0
\]
\[
-12x = 4 \Rightarrow x = -\frac{1}{3}
\]
