(2x² + 1)(x-3)=0

\(\Rightarrow\orbr{\begin{cases}2x^2+1=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x^2=-1\Rightarrow x^2=-\frac{1}{2}\left(vl\right)\\x=3\end{cases}}\)

Vậy x=3

48-(15-x)⁵=48

     (15-x)⁵=48-48

     (15-x)⁵=0

=>  15-x   =0

           x    =15-0

           x    =15

Vậy x=15

Ta có: 2x³ + 3x² - 32x =48

<=>    2x³  + 3x² - 32x - 48 =0

<=>   x²(2x+3) - 16(2x+3)   =0

<=>   (x²-16)(2x+3)             =0

<=>   (x-4)(x+4)(2x+3)        =0

<=>  x-4=0 hoặc x+4=0 hoặc 2x+3=0

<=>  x=4 hoặc x=-4 hoặc x= \(\dfrac{-3}{2}\)

Vậy phương trình trên có tập nghiệm là S={4;-4;\(\dfrac{-3}{2}\)}

2x³+3x²-32x=48

⇔2x³+3x²-32x-48=0

⇔x²(2x+3)-16(2x+3)=0

⇔(2x+3)(x²-16)=0

⇔(2x+3)(x-4)(x+4)=0

⇔2x+3=0 hoặc x-4=0 hoặc x+4=0

1.2x+3=0⇔2x=-3⇔x=-3/2

2.x-4=0⇔x=4

3.x+4=0⇔x=-4

phương trình có 3 nghiệm:x=-3/2 và x=4 và x=-4

Ta có: \(2x^3+3x^2-32x=48\)

\(\Leftrightarrow2x^3+3x^2-32x-48=0\)

\(\Leftrightarrow x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x^2-16\right)=0\)

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

vậy: \(S=\left\{-\dfrac{3}{2};4;-4\right\}\)

3 x (2x-1)²=48

=> (2x-1)²=48:3

=>(2x-1)²=16

Mà: 4²=16 ; (-4)²=16

=> (2x-1)² =4² hoặc (2x-1)²= (-4)²

=> 2x-1=4 hoặc 2x-1= -4

=> x= 2,5 hoặc x= -2,5

3 x ( 2x - 1 )² = 48

=> ( 2x - 1 )² = 48 : 3

=> ( 2x - 1 )² = 16

=> ( 2x - 1 )² = 4²

=> 2x - 1 = 4 hoặc 2x - 1 = -4

+) 2x - 1 = 4

=> 2x = 5

=> x = 5/2

+) 2x - 1 = -4

=> 2x = -3

=> x = -3/2

Vậy x = 5/2 hoặc x = -3/2

3 x ( 2x - 1 )2 = 48

=> ( 2x - 1 )2 = 48 : 3

=> ( 2x - 1 )2 = 16

Ta thấy: 16 = 42

=> ( 2x - 1 )2 = 42

=> 2x - 1 = 4

=> 2x = 4 + 1

=> 2x = 5

=> x = 5 : 2

=> x = 2,5

\(5\left(7+48:x\right)=45\)

 \(\Leftrightarrow\left(7+48x\right)=9\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=2\)

________________

\(5^2x-3.2.5^2=5^2.3\)

\(\Leftrightarrow5x^2-3-2.5^2=75\)

\(\Leftrightarrow5x^2-3-50=75\)

\(\Leftrightarrow5x^2-3=125\)

\(\Leftrightarrow5x^2=128\)

\(\Leftrightarrow x^2=\frac{128}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{128}{5}\\x=-\frac{128}{5}\end{cases}}\)

___________________

\(2+x:5=0\)

\(\Leftrightarrow x:5=-2\)

\(\Leftrightarrow x=-10\)

\(10+2x=16\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

sai câu 3 rồi 

giai ho cai cac ban oi 

giup cai

x sẽ bằng 1

giải thích giùm mk

Ta có: (x+1)(2x+2)(3x+3) = 48

      => (x+1).2(x+1).3(x+1) = 48

      => (x+1)³ . 6 = 48

      => (x+1)³ = 8 = 2³

      =>  x + 1 = 2

      =>  x = 1

a)        \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)

Vậy...

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

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

3) \(x^2+2x-48=\left(x-6\right)\left(x+8\right)\)

4) \(x^2-2x-48=\left(x-8\right)\left(x+6\right)\)

5) \(x^2+x-42=\left(x-6\right)\left(x+7\right)\)

6) \(x^2-x-42=\left(x-7\right)\left(x+6\right).\)

bài này tìm x hay tìm cực trị vậy

1) \(x^2+x-6\)

\(=x^2-2x+3x-6\)

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

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

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

2) \(x^2-x-6\)

\(=x^2+2x-3x-6\)

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

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

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

3) \(x^2+2x-48\)

\(=x^2+8x-6x-48\)

\(=\left(x^2+8x\right)-\left(6x+48\right)\)

\(=x\left(x+8\right)-6\left(x+8\right)\)

\(=\left(x+8\right)\left(x-6\right)\)

4) \(x^2-2x-48\)

\(=x^2-8x+6x-48\)

\(=\left(x^2-8x\right)+\left(6x-48\right)\)

​\(=x\left(x-8\right)+6\left(x-8\right)\)​

\(=\left(x-8\right)\left(x+6\right)\)

5) \(x^2+x-42\)

\(=x^2+7x-6x-42\)

\(=\left(x^2+7x\right)-\left(6x+42\right)\)

\(=x\left(x+7\right)-6\left(x+7\right)\)

\(=\left(x+7\right)\left(x-6\right)\)

6) \(x^2-x-42\)

\(=x^2-7x+6x-42\)

\(=\left(x^2-7x\right)+\left(6x-42\right)\)

\(=x\left(x-7\right)+6\left(x-7\right)\)

\(=\left(x-7\right)\left(x+6\right)\)