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\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
Đặt \(a=x^2+x\)
\(\Leftrightarrow a^2+4a=12\)
\(\Leftrightarrow a^2+4a-12=0\)
\(\Leftrightarrow a^2+6a-2a-12=0\)
\(\Leftrightarrow a\left(a+6\right)-2\left(a+6\right)=0\)
\(\Leftrightarrow\left(a+6\right)\left(a-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-6\\a=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+x=-6\\x^2+x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{23}{4}=0\\x^2+2x-x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\\\left(x+2\right)\left(x-1\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
Vậy....
Ta có : \(3x\left(1-x\right)+\left(x+3\right)\left(x-2\right)=-2\left(x-4\right)^2\)
=> \(3x\left(1-x\right)+\left(x+3\right)\left(x-2\right)=-2\left(x^2-8x+16\right)\)
=> \(3x-3x^2+x^2+3x-2x-6=-2x^2+16x-32\)
=> \(3x-3x^2+x^2+3x-2x-6+2x^2-16x+32=0\)
=> \(-12x+26=0\)
=> \(x=\frac{26}{12}=\frac{13}{6}\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{\frac{13}{6}\right\}\)
\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
<=>\(\left(x^2-4\right)\left(x^2-10\right)=72\) (1)
Đặt \(x^2-7=t\)
=> pt (1) <=> \(\left(t+3\right)\left(t-3\right)=72\)
<=> \(t^2-9=72\)
<=> \(t^2-81=0\)
<=> \(\left(t-9\right)\left(t+9\right)=0\)
Tự làm nốt
\(8x^2-\left(4x+3\right)^3+\left(2x+3\right)^3=0\)
\(\Leftrightarrow8x^2+\left(2x+3-4x-3\right)\left[\left(4x+3\right)^2+\left(2x+3\right)\left(4x+3\right)+\left(2x+3\right)^2\right]=0\)
\(\Leftrightarrow8x^2-2x\left(16x^2+24x+9+8x^2+18x+9+4x^2+12x+9\right)=0\)
\(\Leftrightarrow2x\left(4x-28x^2-54x-27\right)=0\)
\(\Leftrightarrow2x\left(28x^2+50x+27\right)=0\)
Tự làm nốt
( x-2) ( x+2 ) ( x2 - 10 ) = 72
<=> ( x2-4 ) ( x2 - 10 ) = 72
<=> ( x2-7+3) ( x2-7-3)=72
<=> ( x^2-7)^2 -9 = 72
<=> ( x^2 -7)^2 = 81
<=> x^2-7 = -9 hoặc 9
mà x^2-7 luôn lớn hơn hoặc bằng -7
<=> x^2-7 = 9
<=> x^2 = 16
<=> x = 4 hoặc -4
Đúng thì nhấn đáng hộ nhé
a) \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+5x^2-10x+2x-4=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x^2\left(x-2\right)+5x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+5x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+x^2+3x^2+3x+2x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)+2\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2+3x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2+2x+x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left[x\left(x+2\right)+\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)^2\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{2;-1;-2\right\}\)
Vậy....
c, \(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow2\left(x^3+1\right)+7x\left(x+1\right)=0\Leftrightarrow2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[2\left(x^2-x+1\right)+7x\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(2x+1\right)=0\)
Tập nghiệm của pt: \(S=\left\{-1;-2;-\frac{1}{2}\right\}\)
b, \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\) (1)
Đặt: \(x^2-7=t\left(t\ge-7\right)\)
Khi đó (1) trở thành: \(\left(t+3\right)\left(t-3\right)=72\Leftrightarrow t^2-9=72\Leftrightarrow\orbr{\begin{cases}t=9\\t=-9\left(loai\right)\end{cases}}\)
\(t=9\Rightarrow x^2-7=9\Leftrightarrow x=\pm4\)
Tập nghiệm của pt là \(S=\left\{\pm4\right\}\)
a, \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow x^3\left(x+1\right)+x^2\left(x+1\right)-4x\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm2\end{cases}}\)
Đặt x-7=a ta có \(a\left(a+2\right)\left(a+3\right)\left(a+5\right)=72\)\(\Rightarrow\left(a^2+5a\right)\left(a^2+5a+6\right)=72\) Đặt \(a^2+5a=b\)ta có \(b\left(b+6\right)=72\)từ đó tìm ra b, suy ra a và tìm x nha bn!
Lời giải:
$3x(1-x)+(x+3)(x-2)=-2(x-4)^2$
$\Leftrightarrow (3x-3x^2)+(x^2-2x+3x-6)=-2(x^2-8x+16)$
$\Leftrightarrow -2x^2+4x-6=-2x^2+16x-32$
$\Leftrightarrow 12x=26\Rightarrow x=\frac{13}{6}$
Vậy........
Bài làm:
Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)
\(\Leftrightarrow x^4-14x^2+40-72=0\)
\(\Leftrightarrow x^4-14x^2-32=0\)
\(\Leftrightarrow\left(x^4-16x^2\right)+\left(2x^2-32\right)=0\)
\(\Leftrightarrow x^2\left(x^2-16\right)+2\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-16\right)=0\)
Mà \(x^2+2\ge2>0\left(\forall x\right)\)
\(\Rightarrow x^2-16=0\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow x=\pm4\)
( x + 2 )( x - 2 )( x2 - 10 ) = 72
<=> ( x2 - 4 )( x2 - 10 ) = 72
<=> x4 - 14x2 + 40 - 72 = 0
<=> x4 - 14x2 - 32 = 0
Đặt t = x2 ( \(t\ge0\))
Pt <=> t2 - 14t - 32 = 0
<=> t2 + 2t - 16t - 32 = 0
<=> t( t + 2 ) - 16( t + 2 ) = 0
<=> ( t - 16 )( t + 2 ) = 0
<=> \(\orbr{\begin{cases}t-16=0\\t+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}t=16\\t=-2\end{cases}}\)
\(t\ge0\Rightarrow t=16\)
=> x2 = 16
=> \(x=\pm4\)