1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)

⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)

⇔-1\(x^2\) - 4x= 1- \(x^2\)

⇔ -1\(x^2\) -4x+ \(x^2\) = 1

⇔-4x=1

⇔ x = \(\dfrac{-1}{4}\)

Như thế này bn thấy rõ k

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

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

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

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

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

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

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

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

\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)

\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)

=56

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

\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)

\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)

\(=6\)

a)\((x^2- 4).(x^2 - 10) = 72 Đặt x^2 - 7 = a(1), ta có (a+3)(a-3)=72 a^2-9=72 a^2=81 a=+-9 xét 2 trường hợp a = 9 và -9 khi thay vào (1) ta có..... tự lm nốt nha \)

b) nhóm x+1 vs x+4 và x+2 vs x+3 ta sẽ có (x²+5x+4)(x²+5x+6)(x+5)=40

câu c dễ lắm... bạn có thể làm 2 cách 1 là xét 2 th khi vế trái bằng vế phải hay trái dấu vế phải

hoặc cách 2 đưa về hiệu 2 bình phương nhá.. cách này dễ hơn

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

\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)

\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)

\(\Leftrightarrow2x^2+4x+4=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)+2=0\)

\(\Leftrightarrow2\left(x+1\right)^2=-2\)

\(\Leftrightarrow\left(x+1\right)^2=-1\Rightarrow\) pt vô nghiệm

\(b,\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)

\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)-14x^2=0\)

\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)

\(\Leftrightarrow-8x-17=0\)

\(\Leftrightarrow-8x=17\)

\(\Leftrightarrow x=\dfrac{-17}{8}\)

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

\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(x^2+4x+4\right)-\dfrac{5}{2}=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(x+2\right)^2=\dfrac{5}{2}\)

\(\Rightarrow\left(x+2\right)^2=5\)

\(\Rightarrow\left[{}\begin{matrix}x+2=-\sqrt{5}\\x+2=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}-2\\x=\sqrt{5}-2\end{matrix}\right.\)

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

\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)

\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)

\(\Leftrightarrow2x^2+4x+4=0\Leftrightarrow\left(\sqrt{2}x\right)^2+2.\sqrt{2}.\sqrt{2}x+\left(\sqrt{2}\right)^2+2=0\) \(\Leftrightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+2=0\)

ta có : \(\left(\sqrt{2}x+\sqrt{2}\right)^2\ge0\Rightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+2\ge2>0\forall x\)

\(\Rightarrow\) phương trình vô nghiệm

vậy phương trình vô nghiệm

b) \(\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)

\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)=14x^2\)

\(\Leftrightarrow16x^2-9-2x^2-8x-8=14x^2\)

\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)

\(\Leftrightarrow-8x-17=0\Leftrightarrow-8x=17\Leftrightarrow x=\dfrac{-17}{8}\)

vậy \(x=\dfrac{-17}{8}\)

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

\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}x^2-x+\dfrac{1}{2}\)

\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)

\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x\right)^2+2.\sqrt{2}.\dfrac{\sqrt{2}}{2}x+\left(\sqrt{2}\right)^2-\dfrac{5}{2}=0\)

\(\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x+\sqrt{2}\right)^2-\dfrac{5}{2}=0\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x+\sqrt{2}\right)^2=\dfrac{5}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{2}}{2}x+\sqrt{2}=\sqrt{\dfrac{5}{2}}\\\dfrac{\sqrt{2}}{2}x+\sqrt{2}=-\sqrt{\dfrac{5}{2}}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{2}}{2}x=\sqrt{\dfrac{5}{2}}-\sqrt{2}=\dfrac{\sqrt{10}-2\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{2}x=-\sqrt{\dfrac{5}{2}}-\sqrt{2}=-\dfrac{\sqrt{10}+2\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2+\sqrt{5}\\x=-2-\sqrt{5}\end{matrix}\right.\)

vậy \(x=-2+\sqrt{5};x=-2-\sqrt{5}\)

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

\(\Rightarrow9x^2+12x+4-\left(4x^2-1\right)=5\left(x^2-4x+4\right)\)

\(\Rightarrow9x^2+12x+4-4x^2-1=5x^2-20x+20\)

\(\Rightarrow9x^2-4x^2-5x^2+12x+20x=20+1-4\)

\(\Rightarrow32x=17\Rightarrow x=\dfrac{17}{32}\)

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

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

\(\Rightarrow x^2+4x+4-x^2+x-3x+3-5x=0\)

\(\Rightarrow-3x=-3-4\Rightarrow-3x=-7\Rightarrow x=\dfrac{7}{3}\)

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

\(\Rightarrow3x^2-9x-x+3+x^2-4x+4=4x^2-20x+25\)

\(\Rightarrow3x^2+x^2-4x^2-9x-x-4x+20x=25-3-4\)

\(\Rightarrow6x=18\Rightarrow x=3\)

Chúc bạn học tốt!!!

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

4. 1\(\dfrac{1}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = 3\(\dfrac{1}{3}\).(1 - 2x) - \(\dfrac{5.\left(x+1\right)}{6}\)

<=> \(\dfrac{3}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = \(\dfrac{10}{3}\).(1 - 2x) - \(\dfrac{5x+5}{6}\)

<=> \(\dfrac{3}{2}x-3-\dfrac{x}{3}+\dfrac{5}{3}=\dfrac{10}{3}-\dfrac{20}{3}x-\dfrac{5x}{6}-\dfrac{5}{6}\)

<=> \(\dfrac{3}{2}x-\dfrac{x}{3}+\dfrac{20}{3}x-\dfrac{5x}{6}=\dfrac{10}{3}-\dfrac{5}{6}-3+\dfrac{5}{3}\)

<=> 7x = \(\dfrac{7}{6}\)

<=> x = \(\dfrac{1}{6}\)
@Nguyễn Hoàng Vũ