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A = (x4 + 2x3 + x2) + 4. ( x2 + x + 1) = (x2 + x)2 + 4. a = (a - 1)2 + 4a = a2 + 2a + 1 = (a + 1)2
\(A=\left(x^4+2x^3+x^2\right)+4.\left(x^2+x+1\right)\)
\(A=\left(x^2+x\right)^2+4.a\)
\(A=\left(a-1\right)^2+4a\)
\(A=a^2+2a+1\)
\(A=\left(a+1\right)^2\)
Ta có : \(A=x^4+2x^3+5x^2+4x+4\)
\(=\left(x^4+x^3+x^2\right)+\left(x^3+x^2+x\right)+\left(3x^2+3x+3\right)+1\)
\(=x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+3\left(x^2+x+1\right)+1\)
\(=\left(x^2+x+1\right)\left(x^2+x+3\right)+1\)\(=a\left(a+2\right)+1=a^2+2a+1=\left(a+1\right)^2\)
\(A=\left(7x^4-21x^3\right):\left(7x^2\right)+\left(10x+5x^2\right):\left(5x\right)\)
\(=x^2-3x+2x+x\)
\(=x^2\ge0\)
Vậy ...
Lời giải:
\(A=x^4+2x^3+5x^2+4x+4\)
\(=(x^4+2x^3+x^2)+4x^2+4x+4\)
\(=(x^2+x)^2+4(x^2+x)+4\)
\(=(x^2+x+2)^2=(x^2+x+1+1)^2=(a+1)^2\)
Bài 1:
a, x2-3xy-10y2
=x2+2xy-5xy-10y2
=(x2+2xy)-(5xy+10y2)
=x(x+2y)-5y(x+2y)
=(x+2y)(x-5y)
b, 2x2-5x-7
=2x2+2x-7x-7
=(2x2+2x)-(7x+7)
=2x(x+1)-7(x+1)
=(x+1)(2x-7)
Bài 2:
a, x(x-2)-x+2=0
<=>x(x-2)-(x-2)=0
<=>(x-2)(x-1)=0
<=>\(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
b, x2(x2+1)-x2-1=0
<=>x2(x2+1)-(x2+1)=0
<=>(x2+1)(x2-1)=0
<=>x2+1=0 hoặc x2-1=0
1, x2+1=0 2, x2-1=0
<=>x2= -1(loại) <=>x2=1
<=>x=1 hoặc x= -1
c, 5x(x-3)2-5(x-1)3+15(x+2)(x-2)=5
<=>5x(x-3)2-5(x-1)3+15(x2-4)=5
<=>5x(x2-6x+9)-5(x3-3x2+3x-1)+15x2-60=5
<=>5x3-30x2+45x-5x3+15x2-15x+5+15x2-60=5
<=>30x-55=5
<=>30x=55+5
<=>30x=60
<=>x=2
d, (x+2)(3-4x)=x2+4x+4
<=>(x+2)(3-4x)=(x+2)2
<=>(x+2)(3-4x)-(x+2)2=0
<=>(x+2)(3-4x-x-2)=0
<=>(x+2)(1-5x)=0
<=>\(\orbr{\begin{cases}x+2=0\\1-5x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\-5x=-1\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{-1}{-5}\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{1}{5}\end{cases}}\)
Bài 3:
a, Sắp xếp lại: x3+4x2-5x-20
Thực hiện phép chia ta được kết quả là x2-5 dư 0
b, Sau khi thực hiện phép chia ta được :
Để đa thức x3-3x2+5x+a chia hết cho đa thức x-3 thì a+15=0
=>a= -15
Câu 1:
\(3x\left(12x+4\right)+9x\left(4x+3\right)\)
\(\Leftrightarrow3x\left(12x+4\right)+3x\left[3.\left(4x+3\right)\right]\)
\(\Leftrightarrow3x\left(12x+4\right)+3x\left(12x+6\right)\)
\(\Leftrightarrow3x\left[12x+4+12x+6\right]\)
\(\Leftrightarrow3x.\left(24x+10\right)\)
\(\Leftrightarrow72x^2+30x\)
Câu 2:
\(x\left(5+2x\right)+2x^2\left(x-1\right)\)
\(\Leftrightarrow5x+2x^2+2x^3-2x^2\)
\(\Leftrightarrow2x^3+5x\)
Đề 1
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Input:
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Inequality plot:
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Alternate forms:
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Expanded form:
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Solution:
- Approximate form
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Integer solution:
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x4 +2x3+5x2+4x+4 = (x2 + x + 2)2 = (a + 1)2
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