Giải phương trình : \(\left(x^2+x\right)^2+4.\left(x^2+x\right)=12\)
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\(\left(x^2+7x+12\right).\left(4x-16\right)-\left(x+3\right)\left(x^2-5x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^2+3x+4x+12\right).4.\left(x-4\right)-\left(x+3\right)\left(x^2-x-4x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow4\left(x+4\right)\left(x+3\right)\left(x-4\right)-\left(x+3\right)\left(x-4\right)\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(4-x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(8-x\right)=0\)
\(\Leftrightarrow\frac{\orbr{\begin{cases}x+4=0\\x-4=0\end{cases}}}{\orbr{\begin{cases}x+3=0\\8-x=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-4\\x=4\end{cases}}}{\orbr{\begin{cases}x=-3\\x=8\end{cases}}}\)
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x^8 + 2x^6 + 2x^4 + x^2 + 1 - 4x^6 = 12( x^4 - 2x^2 - 1 ) - 4
x^8 + 2x^4 + x^2 + 1 - 2x^6 = 12x^4 - 24x^2 - 12 - 4
x^8 - 2x^6 = 12x^4 - 2x^4 - 24x^2 - x^2 - 16 - 1
x^8 - 2x^6 = 10x^4 - 25x^2 - 17
( x^2 )^4 - 2( x^2 )^3 = 10(x^2)^2 - 25x^2 - 17
0 = 10(x^2)^2 - ( x^2)^4 - 25x^2 + 2(x^2)^3 - 17
17 = (x^2)[ 10x^2 - (x^2)^3 - 25 + 2(x^2)^2 ]
17 = ( x^2 )[ 10x^2 - x^6 - 25 + 2x^4 ]
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ta có: \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
<=> \(\left(x^2+x\right)+2.2.\left(x^2+x\right)+4-16=0\)
<=> \(\left[\left(x^2+x\right)^2+2.2\left(x^2+x\right)+4\right]=16\)
<=> \(\left(x^2+x+2\right)^2=16\)
<=> \(x^2+x+2=4\)hoặc \(x^2+x+2=-4\)
TH1: \(x^2+x+2=4\)=> x=1 ;-2
TH2 : \(x^2+x+2=-4\)=> vô nghiệm
Vậy S ={ -2;1}
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a, \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow x^4+2x^3+x^2+4x^2+4x+12=0\)
\(\Leftrightarrow x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3+3x^2+8x+12\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+x^2+2x+6x+12\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)=0\)
có : \(x^2+x+6>0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
b, \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)-297=0\)
\(\Leftrightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]-297=0\)
\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+7x-21\right)-297=0\)
đặt \(x^2+4x-13=t\)
\(\Leftrightarrow\left(t+8\right)\left(t-8\right)-297=0\)
\(\Leftrightarrow t^2-64-297=0\)
\(\Leftrightarrow t^2=361\)
\(\Leftrightarrow t=\pm19\)
có t rồi tìm x thôi
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Ta có : \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x\right)+2.2.\left(x^2+x\right)+4-16=0\)
\(\Leftrightarrow\left[\left(x^2+x\right)^2+2.2\left(x^2+x\right)+4\right]=16\)
\(\Leftrightarrow\left(x^2+x+2\right)^2=16\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+x+2=4\\x^2+x+2=-4\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1;-2\\vônghiệm\end{array}\right.\)
Vậy \(S=\left\{-2;1\right\}\)
Ta có : \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
Đặt \(t=x^2+x\) , pt trở thành \(t^2+4t-12=0\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}t=2\\t=-6\end{array}\right.\)
Nếu t = 2 ta có pt : \(x^2+x=2\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)
Nếu t = -6 , ta có pt : \(x^2+x=-6\Leftrightarrow x^2+x+6=0\Leftrightarrow\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)
mà \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}>0\) . Dấu đẳng thức không xảy ra nên pt này vô nghiệm.
Vậy tập nghiệm của pt : S={-2;1}
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b) Đặt \(x-7=a\) ta có:
\(\left(a+1\right)^4+\left(a-1\right)^4=16\)
\(\Leftrightarrow\)\(a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=16\)
\(\Leftrightarrow\)\(2a^4+12a^2+2-16=0\)
\(\Leftrightarrow\)\(2\left(a^4+6a^2-7\right)=0\)
\(\Leftrightarrow\)\(a^4+6a^2-7=0\)
\(\Leftrightarrow\)\(\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)
Vì \(a^2+7>0\) nên \(\orbr{\begin{cases}a-1=0\\a+1=0\end{cases}}\)
Thay trở lại ta có: \(\orbr{\begin{cases}x-8=0\\x-6=0\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=6\end{cases}}\)
Vậy...
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Đặt \(t=x^2+x\) ta có pt sau:
\(t^2+4t=12\Rightarrow t^2+4t-12=0\)
\(\Rightarrow t^2-2t+6t-12=0\)
\(\Rightarrow t\left(t-2\right)+6\left(t-2\right)=0\)
\(\Rightarrow\left(t-2\right)\left(t+6\right)=0\)\(\Rightarrow\orbr{\begin{cases}t=2\\t=-6\end{cases}}\)
*)Xét \(x^2+x=2\Rightarrow x^2+x-2=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)=0\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
*)Xét \(x^2+x=-6\Rightarrow x^2+x+6=0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\) (vô nghiệm)
Ta đặt \(x^2+x=a\)
Khi đó pt trở thành :
\(a^2+4a=12\)
\(\Leftrightarrow a^2+4a-12=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=2\\a=-6\end{cases}}\)
Với \(a=2\Leftrightarrow x^2+x=2\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
Với \(a=-6\Leftrightarrow x^2+x=-6\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=-\frac{23}{4}\) ( vô lí )
Vậy pt đã cho có tập nghiêm \(S=\left\{1,-2\right\}\)
Ta có: \(\Delta=4^2+4.12=64,\sqrt{\Delta}=8\)
\(\Rightarrow\orbr{\begin{cases}x^2+x=\frac{-4+8}{2}=2\\x^2+x=\frac{-4-8}{2}=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2+x-2=0\\x^2+x+6=0\end{cases}}\)
+) \(x^2+x-2=0\)
Ta có: \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\x=\frac{-1-3}{2}=-2\end{cases}}\)
+) \(x^2+x+6=0\)
Ta có: \(\Delta=1^2-4.6=-25< 0\)
Vậy pt có 2 nghiệm\(\left\{1;-2\right\}\)