a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)

\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)

\(\Leftrightarrow40x=46+9+25=80\)

\(\Leftrightarrow x=2\)

b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)

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

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

\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)

\(\Leftrightarrow2x=-2+6-5=-1\)

\(\Leftrightarrow x=\frac{1}{2}\)

( x² - 1 ).( x² + 4x + 3 ) = 192

\(\Leftrightarrow\) ( x - 1 ).( x + 1 ) .( x² + 3x + x + 3 ) = 192

\(\Leftrightarrow\) ( x - 1 ).( x + 1 ).[ x.( x + 3 )+ ( x + 3 ) ] = 192

\(\Leftrightarrow\) ( x - 1 .( x + 1 ).( x + 1 ).( x + 3 ) -192 = 0

\(\Leftrightarrow\) ( x + 1 )².( x - 1 ).( x +3 ) - 192 = 0

Đặt : x + 1 = a

 Khi đó phương trình trở thành :

\(\Rightarrow\) a².( a - 2 ).( a + 2 ) - 192 = 0

\(\Leftrightarrow\)a².( a² - 4 ) - 192 = 0

\(\Leftrightarrow\) a⁴ - 4a² - 192 = 0

\(\Leftrightarrow\) ( a⁴ - 4a² + 4 ) - 4 - 192  = 0

\(\Leftrightarrow\) ( a² - 2 )² - 196 = 0

\(\Leftrightarrow\)( a² - 2 )² - 14² = 0

\(\Leftrightarrow\)( a² - 2 - 14 ).( a² - 2 + 14 ) = 0

\(\Leftrightarrow\)( a² - 16 ).( a² + 12 ) = 0

\(\Leftrightarrow\) ( a - 4 ).( a + 4 ).( a² + 12 ) = 0

\(\Leftrightarrow\) \(\orbr{\begin{cases}\left(a-4\right).\left(a+4\right)=0\\a^2+12=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(a-4\right).\left(a+4\right)=0\\a^2=-12\left(vl\right)\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}a-4=0\\a+4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}a=4\\a=-4\end{cases}}\)

Với a = 4                                                           Với a = -4

\(\Rightarrow\) x + 1 = 4                                          \(\Rightarrow\) x + 1 = -4

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

Vậy phương trình có nghiệm là x = 3 , x = -5

(x² - 1)(x² + 4x + 3) = 192 
<=> (x - 1)(x + 1)(x + 1)(x + 3) = 192 
<=> (x - 1)(x + 3)(x + 1)² = 192 
<=> (x² + 2x - 3)(x² + 2x + 1) = 192 
Đặt t = x² + 2x + 1 => x² + 2x - 3 = t - 4 
ta có pt: (t - 4)t = 192 
<=> t² - 4t - 192 = 0 
<=> t = - 12 hoặc t = 16 
*t = x² + 2x + 1 = -12: vn 
*t = x² + 2x + 1 = 16 
<=> (x+1)² = 16 
<=> x = -5 hoặc x = 3 

Mãi mãi có một tương lai tươi sáng

a)(x+3)³-x(3x+1)²+(2x+1)(4x²-2x+1-3x²)=54

\(\Rightarrow\)x³+9x²+27x+27-x(9x²+6x+1)+(2x+1)(x²-2x+1)=54

\(\Rightarrow\)x³+9x²+27x+27-9x³-6x²-x+2x³-4x²+2x+x²-2x+1=54

\(\Rightarrow\)-6x³+26x+28=54

\(\Rightarrow\)-6x³+26x=54-28

\(\Rightarrow\)-6x³+26x=26

\(\Rightarrow\)-6x³+26x-26=0

\(\Rightarrow\)-2(3x³+13x+14)

=2 tk mình

1+1=2 nha

bằng 2 nhé bạn bạn tk cho mình rồi kết bạn nhé

a) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x-3\right)\left(x+3\right)=36\)

\(\Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\)

\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)

\(\Leftrightarrow8x+76=36\)

\(\Leftrightarrow8x=-40\)

\(\Leftrightarrow x=-5\)

\(192-\left(x^2-1\right)\left(x^2+4x+3\right)=0\)

\(\Leftrightarrow192-\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow192-\left[\left(x-1\right)\left(x+3\right)\right]\left[\left(x+1\right)\left(x+1\right)\right]=0\)

\(\Leftrightarrow192-\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)

Đặt \(x^2+2x-3=a\)

\(pt\Leftrightarrow192-a\left(a+4\right)=0\)

\(\Leftrightarrow192-a^2-4a=0\)

\(\Leftrightarrow-a^2-16a+12a+192=0\)

\(\Leftrightarrow-a\left(a+16\right)+12\left(a+16\right)=0\)

\(\Leftrightarrow\left(a+16\right)\left(-a+12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-16\\a=12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2x-3=-16\\x^2+2x-3=12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2x+13=0\\x^2+2x-15=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2x+1+12=0\\x^2+5x-3x-15=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=-12\\x\left(x+5\right)-3\left(x+5\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\in\varnothing\\\left(x+5\right)\left(x-3\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

Vậy.....

mong mọi người giải giúp em vs