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NV
2 tháng 3 2020

a. \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)\left(x+1\right)\left(2x-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x+5=0\\x+1=0\\2x-9=0\end{matrix}\right.\) \(\Rightarrow x=\)

b. \(\Leftrightarrow x^3+x+3x^2+3=0\)

\(\Leftrightarrow x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)

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

c. \(\Leftrightarrow2x\left(3x-1\right)^2-\left(9x^2-1\right)=0\)

\(\Leftrightarrow\left(6x^2-2x\right)\left(3x-1\right)-\left(3x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(6x^2-5x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-1=0\\6x+1=0\end{matrix}\right.\)

NV
2 tháng 3 2020

d.

\(\Leftrightarrow x^3-3x^2+2x-3x^2+9x-6=0\)

\(\Leftrightarrow x\left(x^2-3x+2\right)-3\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x-2\right)=0\)

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

e.

\(\Leftrightarrow x^3+2x^2+x+3x^2+6x+3=0\)

\(\Leftrightarrow x\left(x^2+2x+1\right)+3\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\)

a, \(x^2-49x-50=0\Leftrightarrow\left(x-1\right)\left(x+50\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-50\end{cases}}\)

b, \(3x^2-7x-10=0\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\Leftrightarrow\left(3x-10\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-10=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=10\\x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{10}{3}\\x=-1\end{cases}}}\)

c, \(x^2-4x-5=0\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

d, \(x^2+2x-3=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

e, \(x^2+2020x-2021=0\)

=> vô nghiệm 

f, \(x^2+9x-10=0\Leftrightarrow\left(x-1\right)\left(x+10\right)\Leftrightarrow\orbr{\begin{cases}x=1\\x=-10\end{cases}}\)

g, \(-5x^2+4x+1=0\Leftrightarrow5x^2+x-5x-1=0\Leftrightarrow x\left(5x+1\right)-1\left(5x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{5}\end{cases}}\)

h, \(4x^2+3x-7=0\Leftrightarrow x\left(4x+7\right)-1\left(4x+7\right)=0\Leftrightarrow\left(x-1\right)\left(4x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{7}{4}\end{cases}}\)

21 tháng 4 2020

a) (x-50)(x+1)=0

<=>x=50 hoặc x=1

b) (x+1)(x-10/3)=0

<=>x=-1 hoặc x=10/3

c)  (x-5)(x+1)=0

<=>x=5 hoặc x=-1

d)  (x+3)(x-1)=0

<=>x=-3 hoặc x=1

e) (x-1)(x+2021)=0

<=>x=1 hoặc x=-2021

f) (x-1)(x+10)=0

<=> x=1 hoặc x=-10

g) (x+1/5)(x-1)=0

<=>x=1 hoặc x=-1/5

h) (x-1)(x+7/4)=0

<=> x=1 hoặc x=-7/4

Học tốt. tk vs ạ

26 tháng 7 2020

a, \(x^2=\frac{1}{9}\)

=> \(x=\pm\frac{1}{3}\)

b, \(x^2=\frac{1}{3}\)

=> \(x=\pm\frac{1}{\sqrt{3}}\)

c, \(x^2=\frac{2}{7}\)

=> \(x=\pm\sqrt{\frac{2}{7}}\)

d, Vô nghiệm vì \(x^2+2019\ge2019>0\forall x\)

e, \(x=\pm\sqrt{3}\)

g, Vô nghiệm vì -2 < 0

h, \(x=0\)

4 tháng 4 2017

a) 2x2 – 7x + 3 = 0 có a = 2, b = -7, c = 3

∆ = (-7)2 – 4 . 2 . 3 = 49 – 24 = 25, \(\sqrt{\text{∆}}\) = 5

x1 = \(\dfrac{-\left(-7\right)-5}{2.2}\) = \(\dfrac{2}{4}\) = \(\dfrac{1}{2}\), x2 =\(\dfrac{-\left(-7\right)+5}{2.2}=\dfrac{12}{4}=3\)

b) 6x2 + x + 5 = 0 có a = 6, b = 1, c = 5

∆ = 12 - 4 . 6 . 5 = -119: Phương trình vô nghiệm

c) 6x2 + x – 5 = 0 có a = 6, b = 5, c = -5

∆ = 12 - 4 . 6 . (-5) = 121, \(\sqrt{\text{∆}}\) = 11

x1 = \(\dfrac{-5-1}{2.3}\) = -1; x2 = \(\dfrac{-1+11}{2.6}\) =

d) 3x2 + 5x + 2 = 0 có a = 3, b = 5, c = 2

∆ = 52 – 4 . 3 . 2 = 25 - 24 = 1, \(\sqrt{\text{∆}}\) = 1

X1 = \(\dfrac{-5-1}{2.3}\) = -1, x2 = \(\dfrac{-5+1}{2.3}\) = \(\dfrac{-2}{3}\)

e) y2 – 8y + 16 = 0 có a = 1, b = -8, c = 16

∆ = (-8)2 – 4 . 1. 16 = 0

y1 = y2 = \(-\dfrac{-8}{2.1}\) = 4

f) 16z2 + 24z + 9 = 0 có a = 16, b = 24, c = 9

∆ = 242 – 4 . 16 . 9 = 0

z1 = z2 = \(\dfrac{-24}{2.16}\) = \(\dfrac{3}{4}\)