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1. ( x2 - x + 2 )4 - 3x2 ( x2 - x + 2 )2 + 2x4
Đặt t = x2 - x + 2 , ta có :
t4 - 3x2t2 + 2x4
= t4 - 2x2t2 - x2t2 + 2x4
= t2 ( t2 - 2x2 ) - x2 ( t2 - 2x2 )
= ( t2 - x2 ) ( t2 - 2x2 )
= ( t - x ) ( t + x ) ( t2 - 2x2 )
= ( x2 - x + 2 - x ) ( x2 - x + 2 + x ) [ ( x2 - x + 2 )2 - 2x2 ]
= ( x2 - 2x + 2 ) ( x2 + 2x ) ( x2 - 3x + 2 ) ( x2 + x + 2 )
2. 3 ( - x2 + 2x + 3 )4 - 26x2 ( - x2 + 2x + 3 )2 - 9x4
Đặt y = - x2 + 2x + 3 , ta có :
3y4 - 26x2y2 - 9x4
= x2y2 + 3y4 - 9x4 - 27x2y2
= y2 ( x2 + 3y2 ) - 9x2 ( x2 + 3y2 )
= ( y2 - 9x2 ) ( x2 + 3y2 )
= ( y - 3x ) ( y + 3x ) ( x2 + 3y2 )
= ( - x2 + 2x + 3 - 3x ) ( - x2 + 2x + 3 + 3x ) [ x2 + 3 ( - x2 + 2x + 3 )2 ]
= ( - x2 - x + 3 ) ( - x2 + 5x + 3 ) ( 3x4 - 12x3 - 5x2 + 36x + 27 )
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
1, (x-1)(x+2)(x+3)(x-6)+32x^2
= (x^2 - 7x + 6)(x^2 + 5x + 6) + 32x^2
đặt x^2 - x + 6 = a ta có
(a - 6x)(a + 6x) + 32x^2
= a^2 - 36x^2 + 32x^2
= a^2 - 4x^2
= (a - 2x)(a + 2x)
= (x^2 - x + 6 - 2x)(x^2 - x + 6 + 2x)
= (x^2 - 3x + 6)(x^2 + x + 6)
2, (x+1)(x-4)(x+2)(x-8)+4x^2
= (x^2 + 7x - 8)(x^2 - 2x - 8) + 4x^2
đặt x^2 + 2,5x - 8 = a ta có
(a + 4,5x)(a - 4,5x) + 4x^2
= a^2 - 81/4x^2 + 4x^2
= a^2 - 65/4x^2
\(=\left(a-\sqrt{\frac{65}{4}}x\right)\left(a+\sqrt{\frac{65}{4}}x\right)=\left(x^2+\frac{5}{2}x-8+\sqrt{\frac{65}{4}}x\right)\left(x^2+\frac{5}{2}x-8-\sqrt{\frac{65}{4}x}\right)\)
a)\(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x^3+2x+3x^2+3=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\x^2+1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=-3\\x^2+1>0\left(loai\right)\end{array}\right.\)
\(\Leftrightarrow x=-\frac{3}{2}\)
b)\(x\left(2x-1\right)\left(1-2x\right)=0\)
\(\Leftrightarrow-x\left(2x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow-x\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\\left(2x-1\right)^2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\2x=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)
\(2x^3+3x^2+2x+3=0\)
\(2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\left(2x+3\right)\left(x^2+1\right)=0\)
\(2x+3=0\left(x^2+1\ge1>0\right)\)
\(2x=-3\)
\(x=-\frac{3}{2}\)
\(x\left(2x-1\right)\left(1-2x\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\2x-1=0\\1-2x=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\2x=1\\2x=1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)
\(\left(x^2+x\right)^2-2x^2-2x-15\)
\(=\left(x^2+x\right)^2-\left(2x^2+2x+15\right)\)
\(=\left(x^2+x\right)^2-\left[\left(2x^2+2x\right)+15\right]\)
\(=\left(x^2+x\right)^2-\left[2.\left(x^2+x\right)+15\right]\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\) \(\left(1\right)\)
đặt \(x^2+x=t\)
\(\left(1\right)\)\(=\) \(t^2-2t-15\)
\(=\left(t-1\right)^2-16\)
\(=\left(t-1-4\right)\left(t-1+4\right)\)
\(=\left(t-5\right)\left(t+3\right)\)
thay \(t=x^2+x\) ta có
\(\left(1\right)=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
các câu còn lại tương tự nha
học tốt
câu này mih biết làm nhưng pp nhẩm nghiệm là sao bạn
bạn có thể cho mih vd đi\ược ko