\(\Leftrightarrow2x^2-x-2x^2-6x=8\\ \Leftrightarrow-7x=8\Leftrightarrow x=-\dfrac{8}{7}\)

`x(2x-1)-2x(x+3)=8`

`=>3x^2-x-2x^2-6x=8`

`=>-7x=8`

`=>x=8:(-7)`

`=>x=-8/7`

\(a,=\left(4x^2-1\right)\left(2x-5\right)=8x^3-20x^2-2x+5\\ b,=\left[x^2+\left(x-3\right)\right]\left[x^2-\left(x-3\right)\right]=x^4-\left(x-3\right)^2\\ =x^4-x^2+6x-9\)

𝑎)2𝑥−1𝑥−3+4=−1𝑥−3

⇔2x-1x+1x=-3+3-4

⇔2x=-4

⇔x=-2

𝑏)3𝑥−22𝑥+5=6𝑥+14𝑥−3

⇔5+3=6x+14x-3x+22x

⇔8=39x

⇔x=\(\frac{8}{39}\)

𝑐)𝑥+3𝑥+1+𝑥−2𝑥=2

⇔x+3x+x-2x=2-1

⇔3x=1

⇔x=\(\frac{1}{3}\)

𝑑)x+1−2𝑥−3𝑥−1=2𝑥+3𝑥²−1

⇔3x²+2x+2x+3x-x-1-1+1=0

⇔3x²+6x-1=0

⇔3x²+3x+3x+3-4=0

⇔3x(x+1)+3(x+1)-4=0

⇔3(x+1)(x+1)-4=0

⇔3(x+1)²-4=0

⇔(x+1)²=\(\frac{4}{3}\)

⇔\(\left[{}\begin{matrix}x+1=\frac{4}{3}\\x+1=-\frac{4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}-1\\x=-\frac{4}{3}-1\end{matrix}\right.\)

Vậy ...

a, 2x - x - 3 + 4 = -x - 3

\(\Leftrightarrow\) x + 1 = -x - 3

\(\Leftrightarrow\) x + x = -3 - 1

\(\Leftrightarrow\) 2x = -4

\(\Leftrightarrow\) x = -2

Vậy S = {-2}

b, 3x - 22x + 5 = 6x + 14x - 3

\(\Leftrightarrow\) -19x + 5 = 20x - 3

\(\Leftrightarrow\) -19x - 20x = -3 - 5

\(\Leftrightarrow\) -39x = -8

\(\Leftrightarrow\) x = \(\frac{8}{39}\)

Vậy S = {\(\frac{8}{39}\)}

c, x + 3x + 1 + x - 2x = 2

\(\Leftrightarrow\) 3x + 1 = 2

\(\Leftrightarrow\) 3x = 2 - 1

\(\Leftrightarrow\) 3x = 1

\(\Leftrightarrow\) x = \(\frac{1}{3}\)

Vậy S = {\(\frac{1}{3}\)}

Phần d mình ko hiểu, bạn viết rõ được ko!

Chúc bn học tốt!!

d, x + 1 - 2x - 3x - 1 = 2x + 3x² - 1

\(\Leftrightarrow\) x + 1 - 2x - 3x - 1 - 2x - 3x² + 1 = 0

\(\Leftrightarrow\) -3x² - 6x + 1 = 0

\(\Leftrightarrow\) -(3x² + 6x - 1) = 0

\(\Leftrightarrow\) 3x² + 6x - 1 = 0

\(\Leftrightarrow\) 3x² + 3x + 3x + 3 - 4 = 0

\(\Leftrightarrow\) 3x(x + 1) + 3(x + 1) - 4 = 0

\(\Leftrightarrow\) 3(x + 1)(x + 1) - 4 = 0

\(\Leftrightarrow\) 3(x + 1)² - 4 = 0

\(\Leftrightarrow\) (x + 1)² = \(\frac{4}{3}\)

\(\Leftrightarrow\) x + 1 = \(\sqrt{\frac{4}{3}}\) hoặc x + 1 = \(-\sqrt{\frac{4}{3}}\)

\(\Leftrightarrow\) x = \(\sqrt{\frac{4}{3}}\) - 1 và x = \(-\sqrt{\frac{4}{3}}\) - 1

\(\Leftrightarrow\) x = \(\frac{2\sqrt{3}-3}{3}\) và x = \(\frac{-2\sqrt{3}-3}{3}\)

Vậy S = {\(\frac{2\sqrt{3}-3}{3}\); \(\frac{-2\sqrt{3}-3}{3}\)}

Chúc bn học tốt!!

Nguyễn Thị Anh Thư cái này bạn gửi một lần r mà!

\(b,\Leftrightarrow x+7=38\Leftrightarrow x=31\\ c,\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\\ d,\Leftrightarrow2x=160-49=111\Leftrightarrow x=\dfrac{111}{2}\\ e,\Leftrightarrow x-8=20\Leftrightarrow x=28\\ f,\Leftrightarrow x-3=\dfrac{59}{4}\Leftrightarrow x=\dfrac{71}{4}\\ g,\Leftrightarrow x=3\\ h,\Leftrightarrow2x+1=5\Leftrightarrow2x=4\Leftrightarrow x=2\)

Câu 1: A

Câu 21: A

\(16,A\\ 17,C\\ 18,A\\ 19,C\\ 20,A\\ 21,A\)

1a

5) a) 2x(x^2 - 9) = 0

<=> 2x(x - 3)(x + 3) = 0

<=> x = 0 hoặc x = 3 hoặc x = -3

b) 2x(x - 2021) - x + 2021 = 0

<=> (2x - 1)(x - 2021) = 0

<=> 2x - 1 = 0 hoặc x - 2021 = 0

<=> x = 1/2 hoặc x = 2021

c) 4x^2 - 16x = 0

<=> 4x(x - 4) = 0

<=> x = 0 hoặc x = 4

d) (3x + 7)^2 - (x + 1)^2 = 0

<=> (3x + 7 + x + 1)(3x + 7 - x - 1) = 0

<=> (4x + 8)(2x + 6) = 0

<=> 4x + 8 = 0 hoặc 2x + 6 = 0

<=> x = -2 hoặc x = -3

mình giải tạm nha