\(pt\Leftrightarrow\left(x+2\right)\left(3x+4\right)\left(3x^2+15x+8\right)\)

1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

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

hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)

2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)

hay \(x\in\left\{1;5\right\}\)

3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

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

hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)

hay \(x\in\left\{-4;3;-3\right\}\)

5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)

\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)

\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)

hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)

1.

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)

\(\Leftrightarrow x+3=5x-2\)

\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)

2.

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)

\(\Leftrightarrow x^2+x+1=x^2-2x+16\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

3.

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

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

1: ĐKXĐ: x<>0

\(\sqrt{x^2+x+2} + \frac{1}{x} = \frac{13-7x}{2}\)

\(\Leftrightarrow \sqrt{x^2+x+2} = \frac{13}{2} - \frac{7x}{2} - \frac{1}{x} = \frac{13x - 7x^2 - 2}{2x}\)

=>\(\left(\sqrt{x^2+x+2} - 2\right) + \left(\frac{1}{x} - 1\right) + \frac{7x - 7}{2} = 0\)

=>\(\frac{x^2+x-2}{\sqrt{x^2+x+2} + 2}+\frac{1-x}{x}+\frac{7(x-1)}{2}=0\)

=>\((x-1)\left[\frac{x+2}{\sqrt{x^2+x+2} + 2}-\frac{1}{x}+\frac{7}{2}\right]=0\)

=>x-1=0

=>x=1(nhận)

7: ĐKXĐ: \(16-x^2>0\)

=>\(x^2<16\)

=>-4<x<4

\(\frac{x^3}{\sqrt{16-x^2}} + x^2 - 16 = 0\)

\(\Leftrightarrow \frac{x^3}{\sqrt{16-x^2}} - \left(\sqrt{16-x^2}\right)^2 = 0\)

\(\Leftrightarrow x^3 - \left(\sqrt{16-x^2}\right)^3 = 0\)

\(\Leftrightarrow x^3 = \left(\sqrt{16-x^2}\right)^3\)

\(\Leftrightarrow x = \sqrt{16-x^2} \quad (x > 0)\)

\(\Leftrightarrow x^2 = 16 - x^2\) và x>0

=>\(2x^2=16\) và x>0

=>\(x^2=8\) và 0<x<4

=>\(x=2\sqrt2\)

1, \(\Delta=\left(-11\right)^2-4.1.38=121-152=-31< 0\)

\(\Rightarrow\) pt vô nghiệm

2, \(\Delta=71^2-4.6.175=5041-4200=841\)

\(x_1=\dfrac{-71+\sqrt{841}}{2.6}=\dfrac{-71+29}{12}=\dfrac{-42}{12}=-\dfrac{7}{2}\)

\(x_2=\dfrac{-71-\sqrt{841}}{2.6}=\dfrac{-71-29}{12}=\dfrac{-10}{12}=-\dfrac{25}{3}\)

3, \(\Delta=\left(-3\right)^2-5.27=9-135=-126< 0\)

⇒ pt vô nghiệm

4, \(\Delta=15^2-\left(-30\right)\left(-7,5\right)=225-225=0\)

\(\Rightarrow x_1=x_2=\dfrac{-30}{2.\left(-30\right)}=\dfrac{1}{2}\)

5, \(\Delta'=\left(-8\right)^2-4.17=64-68=-4\)

⇒ pt vô nghiệm

6, \(\Delta=4^2-4.1.\left(-12\right)=16+48=64\)

\(x_1=\dfrac{-4+\sqrt{64}}{2.1}=\dfrac{-4+8}{2}=\dfrac{4}{2}=2\)

​\(x_2=\dfrac{-4-\sqrt{64}}{2.1}=\dfrac{-4-8}{2}=\dfrac{-12}{2}=-6\)​

Em cần làm gì với các đa thức này vậy em

a) Phương trình  4 x 2 + 2 x − 5 = 0

Có a = 4; b = 2; c = -5, a.c < 0

⇒ Phương trình có hai nghiệm  x 1 ;   x 2

Theo hệ thức Vi-et ta có: 

b) Phương trình . 9 x 2 − 12 x + 4 = 0

Có a = 9; b' = -6; c = 4  ⇒ Δ 2 = ( - 6 ) 2 - 4 . 9 = 0

⇒ Phương trình có nghiệm kép  x 1   =   x 2 .

Theo hệ thức Vi-et ta có: 

c) Phương trình  5 x 2 + x + 2 = 0

Có a = 5; b = 1; c = 2  ⇒ Δ = 1 2 − 4.2.5 = − 39 < 0

⇒ Phương trình vô nghiệm.

d) Phương trình  159 x 2 − 2 x − 1 = 0

Có a = 159; b = -2; c = -1; a.c < 0

⇒ Phương trình có hai nghiệm phân biệt  x 1 ;   x 2 .

Theo hệ thức Vi-et ta có: 

a. (3x - 1)² - (x + 3)² = 0

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

\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

a: =>(2x-5x-1)(2x+5x+1)=0

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

=>x=-1/3 hoặc x=-1/7

b: =>(5x-5)^2-(x+2)^2=0

=>(5x-5-x-2)(5x-5+x+2)=0

=>(4x-7)(6x-3)=0

=>x=1/2 hoặc x=7/4

c: =>(x^2+4x-1-x^2+3x-2)(x^2+4x-1+x^2-3x+2)=0

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

=>7x-3=0

=>x=3/7

c: \(x^2+4x+4=\left(x+2\right)^2\)

d: \(9x^2+6x+1=\left(3x+1\right)^2\)