(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

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...

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

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

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

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

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

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

không đọc được đề lun đó 

\(a,2^{x+1}=32\\ 2^{x+1}=2^5\\ x+1=5\\ x=4\\ b,2^{2x}+2^{2x+1}=48\\ 2^{2x}+2\cdot2^{2x}=48\\ 3\cdot2^{2x}=48\\ 2^{2x}=16\\ 2^{2x}=2^4\\ 2x=4\\ x=2\)

\(c,3^x+5\cdot3^{x+1}=144\\ 3^x+15\cdot3^x=144\\ 16\cdot3^x=144\\ 3^x=9\\ 3^x=3^2\\ x=2\\ d,3^{x+5}=9^{x+1}\\ 3^{x+5}=3^{2x+2}\\ x+5=2x+2\\ x=3\)

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

\(\Leftrightarrow4x^2-4x+1-2x+1=0\)

\(\Leftrightarrow4x^2-6x+2=0\)

\(\Delta=\left(-6\right)^2-4.4.2=4\)

\(x_1=\dfrac{1}{2};x_2=1\)

b) \(\left(4x-3\right)^2-12x+9=0\)

\(\Leftrightarrow16x^2-24x+9-12x+9=0\)

\(\Leftrightarrow16x^2-36x+18=0\)

\(\Delta=\left(-36\right)^2-4.16.18=144\)

\(x_1=\dfrac{3}{4};x_2=\dfrac{3}{2}\)

c) \(3x^2-48=0\)

\(\Leftrightarrow3x^2=48\)

\(\Leftrightarrow x^2=16\)

\(x_1=4;x_2=-4\)

Tick cho mk nha