Bài 3: 1) $2x^2+5x+3=0\\ \Leftrightarrow\left(3+2x\right)\cdot\left(1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3+2x=0\\1+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=-1\end{matrix}\right.$ 2) $x^2+4x+3=0\\ \Leftrightarrow\left(3+x\right)\cdot\left(1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3+x=0\\1+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.$ 3) $x^2-x-12=0\\ \Leftrightarrow\left(-3-x\right)\cdot\left(4-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-x=0\\4-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=4\end{matrix}\right.$ 4) $x^2-3x+2=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.$ 5) $-x^2+5x-6=0\\ \Leftrightarrow\left(-3+x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3+x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.$ 6) $4x^2-12x+5=0\\ \Leftrightarrow\left(1-2x\right)\cdot\left(5-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-2x=0\\5-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.$ 7) $4x^2+4x-3=0\\ \Leftrightarrow\left(-3-2x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-2x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.$Câu trả lời: Bài 3: 1) $2x^2+5x+3=0\\ \Leftrightarrow\left(3+2x\right)\cdot\left(1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3+2x=0\\1+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=-1\end{matrix}\right.$ 2) $x^2+4x+3=0\\ \Leftrightarrow\left(3+x\right)\cdot\left(1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3+x=0\\1+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.$ 3) $x^2-x-12=0\\ \Leftrightarrow\left(-3-x\right)\cdot\left(4-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-x=0\\4-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=4\end{matrix}\right.$ 4) $x^2-3x+2=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.$ 5) $-x^2+5x-6=0\\ \Leftrightarrow\left(-3+x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3+x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.$ 6) $4x^2-12x+5=0\\ \Leftrightarrow\left(1-2x\right)\cdot\left(5-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-2x=0\\5-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.$ 7) $4x^2+4x-3=0\\ \Leftrightarrow\left(-3-2x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-2x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.$

8) $x^2-3x+2=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.$ 9) $3x^2-22x-16=0\\ \Leftrightarrow\left(-2-3x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2-3x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{2}{3}\\x=8\end{matrix}\right.$ 10) $2x^2+7x-15=0\\ \Leftrightarrow\left(-5-x\right)\cdot\left(3-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-5-x=0\\3-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.$ 11) $\left(x-5\right)^2-16=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=1\end{matrix}\right.$ 12) $\left(x-4\right)^2-25=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=-1\end{matrix}\right.$ 13) $25-\left(3-x\right)^2=0\\ \Leftrightarrow\left(2+x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2+x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=8\end{matrix}\right.$ 14) $\left(x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow-4\cdot\left(2x-2\right)=0\\ \Rightarrow2x-2=0\\ \Rightarrow x=1$Câu trả lời: 8) $x^2-3x+2=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.$ 9) $3x^2-22x-16=0\\ \Leftrightarrow\left(-2-3x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2-3x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{2}{3}\\x=8\end{matrix}\right.$ 10) $2x^2+7x-15=0\\ \Leftrightarrow\left(-5-x\right)\cdot\left(3-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-5-x=0\\3-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.$ 11) $\left(x-5\right)^2-16=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=1\end{matrix}\right.$ 12) $\left(x-4\right)^2-25=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=-1\end{matrix}\right.$ 13) $25-\left(3-x\right)^2=0\\ \Leftrightarrow\left(2+x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2+x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=8\end{matrix}\right.$ 14) $\left(x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow-4\cdot\left(2x-2\right)=0\\ \Rightarrow2x-2=0\\ \Rightarrow x=1$

Bài 3 1, 2x2 + 5x + 3 = 0 $\Leftrightarrow$ 2x2 + 2x + 3x + 3 = 0 $\Leftrightarrow$ 2x(x + 1) + 3(x + 1) = 0 $\Leftrightarrow$ (x + 1)(2x + 3) = 0 $\Leftrightarrow$ x + 1 = 0 hoặc 2x + 3 = 0 $\Leftrightarrow$ x = -1 và x = $\frac{-3}{2}$ Vậy S = {-1; $\frac{-3}{2}$} 2, x2 + 4x + 3 = 0 $\Leftrightarrow$ x2 + x + 3x + 3 = 0 $\Leftrightarrow$ x(x + 1) + 3(x + 1) = 0 $\Leftrightarrow$ (x + 1)(x + 3) = 0 $\Leftrightarrow$ x + 1 = 0 hoặc x + 3 = 0 $\Leftrightarrow$ x = -1 hoặc x = -3 Vậy S = {-1; -3}Câu trả lời: Bài 3 1, 2x2 + 5x + 3 = 0 $\Leftrightarrow$ 2x2 + 2x + 3x + 3 = 0 $\Leftrightarrow$ 2x(x + 1) + 3(x + 1) = 0 $\Leftrightarrow$ (x + 1)(2x + 3) = 0 $\Leftrightarrow$ x + 1 = 0 hoặc 2x + 3 = 0 $\Leftrightarrow$ x = -1 và x = $\frac{-3}{2}$ Vậy S = {-1; $\frac{-3}{2}$} 2, x2 + 4x + 3 = 0 $\Leftrightarrow$ x2 + x + 3x + 3 = 0 $\Leftrightarrow$ x(x + 1) + 3(x + 1) = 0 $\Leftrightarrow$ (x + 1)(x + 3) = 0 $\Leftrightarrow$ x + 1 = 0 hoặc x + 3 = 0 $\Leftrightarrow$ x = -1 hoặc x = -3 Vậy S = {-1; -3}

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Hoàng Ngọc Anh

29 tháng 3 2020 lúc 20:16

Bài 3:

1)

$2x^2+5x+3=0\\ \Leftrightarrow\left(3+2x\right)\cdot\left(1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3+2x=0\\1+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=-1\end{matrix}\right.$

2)

$x^2+4x+3=0\\ \Leftrightarrow\left(3+x\right)\cdot\left(1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3+x=0\\1+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.$

3)

$x^2-x-12=0\\ \Leftrightarrow\left(-3-x\right)\cdot\left(4-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-x=0\\4-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=4\end{matrix}\right.$

4)

$x^2-3x+2=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.$

5)

$-x^2+5x-6=0\\ \Leftrightarrow\left(-3+x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3+x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.$

6)

$4x^2-12x+5=0\\ \Leftrightarrow\left(1-2x\right)\cdot\left(5-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-2x=0\\5-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.$

7)

$4x^2+4x-3=0\\ \Leftrightarrow\left(-3-2x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-2x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.$

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Hoàng Ngọc Anh

29 tháng 3 2020 lúc 20:38

8)

$x^2-3x+2=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.$

9)

$3x^2-22x-16=0\\ \Leftrightarrow\left(-2-3x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2-3x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{2}{3}\\x=8\end{matrix}\right.$

10)

$2x^2+7x-15=0\\ \Leftrightarrow\left(-5-x\right)\cdot\left(3-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-5-x=0\\3-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.$

11)

$\left(x-5\right)^2-16=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=1\end{matrix}\right.$

12)

$\left(x-4\right)^2-25=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=-1\end{matrix}\right.$

13)

$25-\left(3-x\right)^2=0\\ \Leftrightarrow\left(2+x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2+x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=8\end{matrix}\right.$

14)

$\left(x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow-4\cdot\left(2x-2\right)=0\\ \Rightarrow2x-2=0\\ \Rightarrow x=1$

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Trương Huy Hoàng

29 tháng 3 2020 lúc 20:42

Bài 3

1, 2x² + 5x + 3 = 0

$\Leftrightarrow$ 2x² + 2x + 3x + 3 = 0

$\Leftrightarrow$ 2x(x + 1) + 3(x + 1) = 0

$\Leftrightarrow$ (x + 1)(2x + 3) = 0

$\Leftrightarrow$ x + 1 = 0 hoặc 2x + 3 = 0

$\Leftrightarrow$ x = -1 và x = $\frac{-3}{2}$

Vậy S = {-1; $\frac{-3}{2}$}

2, x² + 4x + 3 = 0

$\Leftrightarrow$ x² + x + 3x + 3 = 0

$\Leftrightarrow$ x(x + 1) + 3(x + 1) = 0

$\Leftrightarrow$ (x + 1)(x + 3) = 0

$\Leftrightarrow$ x + 1 = 0 hoặc x + 3 = 0

$\Leftrightarrow$ x = -1 hoặc x = -3

Vậy S = {-1; -3}

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Trương Huy Hoàng

29 tháng 3 2020 lúc 20:45

tiếp bài 3

14, (x - 3)² - (x + 1)² = 0

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

$\Leftrightarrow$ (-4)(2x - 2) = 0

$\Leftrightarrow$ (-8)(x - 1) = 0

$\Leftrightarrow$ x - 1 = 0

$\Leftrightarrow$ x = 1

Vậy S = {1}

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Hoàng Ngọc Anh

29 tháng 3 2020 lúc 20:57

15)

$\left(3x-7\right)^2-4\cdot\left(x+1\right)^2=0\\ \Leftrightarrow5x^2-50x+45\\ \Leftrightarrow5\cdot\left(1-x\right)\cdot\left(9-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\9-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=9\end{matrix}\right.$

16)

$\left(7x-4\right)^2\cdot\left(2x+1\right)^2=0\\ \Leftrightarrow\left(5x-5\right)\cdot\left(9x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x-5=0\\9x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.$

17)

$\left(x^2-4\right)\cdot\left(2x+3\right)=\left(x^2-4\right)\cdot\left(x-1\right)\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(2x+3\right)-\left(x^2-4\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+2\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.$

18)

$x^3-25x=0\\ \Leftrightarrow x\cdot\left(x-5\right)\cdot\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.$

19)

$x^3-16x=0\\ \Leftrightarrow x\cdot\left(x-4\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.$

20)

$x^3-5x^2+6x=0\\ \Leftrightarrow x\cdot\left(2-x\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2-x=0\\3-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.$

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Hoàng Ngọc Anh

30 tháng 3 2020 lúc 9:15

Bài 3 (thêm bước):

3)

$x^2-x-12=0\\ \Leftrightarrow-12+3x-4x+x^2=0\\ \Leftrightarrow-3\cdot\left(4-x\right)-x\cdot\left(4-x\right)=0\\ \Leftrightarrow\left(4-x\right)\cdot\left(-3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}4-x=0\\-3-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-3\end{matrix}\right.$

4)

$x^2-3x+2=0\\ \Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow2\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.$

5)

$-x^2+5x-6=0\\ \Leftrightarrow-x^2+2x+3x-6=0\\ \Leftrightarrow x\cdot\left(2-x\right)-3\cdot\left(2-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.$

6)

$4x^2-12x+5=0\\ \Leftrightarrow5-2x-10x+4x^2=0\\ \Leftrightarrow5\cdot\left(1-2x\right)-2x\cdot\left(1-2x\right)=0\\ \Leftrightarrow\left(5-2x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}5-2x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.$

7)

$4x^2+4x-3=0\\ \Leftrightarrow-3+6x-2x+4x^2=0\\ \Leftrightarrow-3\cdot\left(1-2x\right)-2x\cdot\left(1-2x\right)=0\\ \Leftrightarrow\left(-3-2x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-2x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.$

8)

$x^2-3x+2=0\\ \Leftrightarrow2-3x+x^2=0\\ \Leftrightarrow2-x-2x+x^2=0\\ \Leftrightarrow2\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.$

9)

$3x^2-22x-16=0\\ \Leftrightarrow-16-22x+3x^2=0\\ \Leftrightarrow-16+2x-24x+3x^2=0\\ \Leftrightarrow-2\cdot\left(8-x\right)-3x\cdot\left(8-x\right)=0\\ \Leftrightarrow\left(-2-3x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2-3x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{2}{3}\\x=8\end{matrix}\right.$

10)

$2x^2+7x-15=0\\ \Leftrightarrow-15+7x+2x^2=0\\ \Leftrightarrow-15+10x-30+2x^2=0\\ \Leftrightarrow-2\cdot\left(3-2x\right)-x\cdot\left(3-2x\right)=0\\ \Leftrightarrow\left(-5-x\right)\cdot\left(3-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-5-x=0\\3-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.$

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Hoàng Ngọc Anh

30 tháng 3 2020 lúc 19:36

11)

$\left(x-5\right)^2-16=0\\ \Leftrightarrow\left(x-5\right)^2-4^2=0\\ \Leftrightarrow\left(x-5-4\right)\cdot\left(x-5+4\right)=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=1\end{matrix}\right.$

12)

$\left(x-4\right)^2-25=0\\ \Leftrightarrow\left(x-4\right)^2-5^2=0\\ \Leftrightarrow\left(x-4-5\right)\cdot\left(x-4+5\right)=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=-1\end{matrix}\right.$

13)

$25-\left(3-x\right)^2=0\\ \Leftrightarrow5^2-\left(3-x\right)^2=0\\ \Leftrightarrow\left(5-3+x\right)\cdot\left(5+3-x\right)=0\\ \Leftrightarrow\left(2+x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2+x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=8\end{matrix}\right.$

14)

$\left(x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(x-3-x-1\right)\cdot\left(x-3+x+1\right)=0\\ \Leftrightarrow-4\cdot\left(2x-2\right)=0\\ \Rightarrow2x-2=0\\ \Rightarrow x=1$

15)

$\left(3x-7\right)^2-4\cdot\left(x+1\right)^2=0\\ \Leftrightarrow9x^2-42x+49-4x^2-8x-4=0\\ \Leftrightarrow5x^2-50x+45=0\\ \Leftrightarrow5\cdot\left(x^2-10x+9\right)=0\\ \Leftrightarrow5\cdot\left(9-9x-x+x^2\right)=0\\ \Leftrightarrow5\cdot\left[9\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\right]\\ \Leftrightarrow5\cdot\left(9-x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}9-x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=1\end{matrix}\right.$

16)

$\left(7x-4\right)^2-\left(2x+1\right)^2=0\\ \Leftrightarrow\left(7x-4-2x-1\right)\cdot\left(7x-4+2x+1\right)=0\\ \Leftrightarrow\left(5x-5\right)\cdot\left(9x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x-5=0\\9x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.$

17)

$\left(x^2-4\right)\cdot\left(2x+3\right)=\left(x^2-4\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(2x+3\right)-\left(x^2-4\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(2x+3-x+1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(x+4\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+2\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.$

18)

$x^3-25x=0\\ \Leftrightarrow x\cdot\left(x^2-25\right)=0\\ \Leftrightarrow x\cdot\left(x-5\right)\cdot\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.$

19)

$x^3-16x=0\\ \Leftrightarrow x\cdot\left(x^2-16\right)=0\\ \Leftrightarrow x\cdot\left(x-4\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.$

20)

$x^3-5x^2+6x=0\\ \Leftrightarrow x\cdot\left(x^2-5x+6\right)=0\\ \Leftrightarrow x\cdot\left(6-2x-3x+x^2=0\right)\\ \Leftrightarrow x\cdot\left[2\cdot\left(3-x\right)-x\cdot\left(3-x\right)=0\right]\\ \Leftrightarrow x\cdot\left(2-x\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2-x=0\\3-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.$

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