Tìm giá trị nguyên x để đa thức f(x)=x^3-3x^2-3x-1 chia hết cho g(x)=x^2+x+1
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\(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^5+1}-\frac{16}{x^{16}+1}\)
\(=\frac{x+1-x+1}{\left(x+1\right)\left(x-1\right)}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2}{x^2-1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2\left(x^2+1\right)-2.\left(x^2-1\right)}{x^2-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{2x^2+2-2x^2+2}{\left(x^2+1\right)\left(x^2-1\right)}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{4}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{4\left(x^4+1\right)-4\left(x^4-1\right)}{\left(x^4-1\right)\left(x^4+1\right)}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{8}{x^8-1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
\(=\frac{8.\left(x^8+1\right)-8\left(x^8-1\right)}{\left(x^8-1\right)\left(x^8+1\right)}-\frac{16}{x^{16}+1}\)
\(=\frac{16}{x^{16}-1}-\frac{16}{x^{16}+1}\)
\(=\frac{16.\left(x^{16}+1\right)-16.\left(x^{16}-1\right)}{\left(x^{16}-1\right)\left(x^{16}+1\right)}\)
\(=\frac{32}{x^{32}-1}\)
a)\(M=\frac{y^4+1}{x^2y^4+2y^4+x^2+2}\)
\(M=\frac{y^4+1}{y^4\left(x^2+2\right)+\left(x^2+2\right)}\)
\(M=\frac{y^4+1}{\left(y^4+1\right)\left(x^2+2\right)}\)
\(M=\frac{1}{\left(x^2+2\right)}\left(y^4+1\ne0\right)\)
b) M<1 thì phải~
Ta có: \(x^2\ge0\forall x\Rightarrow x^2+2\ge2\forall x\)
\(\Rightarrow\frac{1}{x^2+2}\le\frac{1}{2}< 1\)
\(\Rightarrow M< 1\)
đpcm
Ta có: \(M\le\frac{1}{2}\)( ý b)
\(M=\frac{1}{2}\Leftrightarrow x^2+2=2\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Vậy \(M_{max}=\frac{1}{2}\Leftrightarrow x=0\)
Tham khảo nhé~
Với mọi x; y thì phân thức M đều xác định ( vì mẫu lớn hơn 0 )
a) \(M=\frac{y^4+1}{x^2y^4+2y^4+x^2+2}\)
\(M=\frac{y^4+1}{y^4\left(x^2+2\right)+\left(x^2+2\right)}\)
\(M=\frac{y^4+1}{\left(x^2+2\right)\left(y^4+1\right)}\)
\(M=\frac{1}{x^2+2}\)
b) *đề phải là c/m M luôn bé hơn 1*
Dễ thấy \(x^2+2>1\forall x\)
\(\Rightarrow M< 1\forall x;y\) ( vì tử số bé hơn mẫu số )
c) \(M=\frac{1}{x^2+2}\)
Vì \(x^2\ge0\forall x\)
\(\Rightarrow M\le\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Vậy Mmax = 1/2 khi và chỉ khi x = 0
\(B=\frac{x+2}{x+3}\cdot\frac{x+3}{x+4}:\frac{x+4}{x+5}\cdot\frac{\left(x+4\right)^2}{x+5}\)
\(B=\frac{x+2}{x+3}\cdot\frac{x+3}{x+4}\cdot\frac{x+5}{x+4}\cdot\frac{\left(x+4\right)^2}{x+5}\)
\(B=\frac{\left(x+2\right)\left(x+3\right)\left(x+5\right)\left(x+4\right)^2}{\left(x+3\right)\left(x+4\right)\left(x+4\right)\left(x+5\right)}\)
\(B=\frac{x+2}{1}\)
\(B=x+2\)
\(B=\frac{x+2}{x+3}.\frac{x+3}{x+4}.\frac{x+4}{x+5}.\frac{\left(x+4\right)^2}{x+5}\)
\(B=\frac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+4\right)^2}{\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+5\right)}\)
\(B=\frac{\left(x+2\right)\left(x+4\right)^2}{\left(x+5\right)^2}\)
Đặt biểu thức cần tìm =A
Theo bài ra ta có:
A=2.1+3.(-1)+...+199(-1)+200.1
=2-3+4-5+...-199+200
=(-1)+(-1)+...+(-1)+200(có [(200-2):2=99 số -1]
=-99+200
=101
\(4x^3-3x^2+5x-21=4x^3-7x^2+4x^2-7x+12x-21\)
\(=x^2\left(4x-7\right)+x\left(4x-7\right)+3\left(4x-7\right)=\left(4x-7\right)\left(x^2+x+3\right)\)
\(\frac{x^2+y^2+z^2-2xy-2yz+2zx}{x^2-2xy+y^2-z^2}=\frac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}=\frac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}=\frac{x-y+z}{x-y-z}\)
\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\frac{3x^2\left(x-1\right)-4x\left(x-1\right)+\left(x-1\right)}{2x^2\left(x-1\right)+x\left(x-1\right)-3\left(x-1\right)}\)
\(=\frac{\left(x-1\right)\left(3x^2-4x+1\right)}{\left(x-1\right)\left(2x^2+x-3\right)}=\frac{3x^2-4x+1}{2x^2+x-3}\)
\(=\frac{3x\left(x-1\right)-\left(x-1\right)}{2x\left(x-1\right)+3\left(x-1\right)}=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(2x+3\right)}=\frac{3x-1}{2x+3}\)
Áp dụng bất dẳng thức Cauchy - Schwartz dạng engel, ta có:
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{a+b+b+c+c+a}=\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi: \(\frac{a}{a+b}=\frac{b}{b+c}=\frac{c}{c+a}\)
\(\frac{\left(2a-1\right)^3+\left(a-1\right)^3}{a^3-\left(a-1\right)^3}=\frac{8a^3-12a^2+6a-1+a^3-3a^2+3a-1}{a^3-\left(a^3-3a^2+3a-1\right)}\)
\(=\frac{9a^3-15a^2+9a-2}{3a^2-3a+1}=\frac{3a\left(3a^2-3a+1\right)-2\left(3a^2-3a+1\right)}{3a^2-3a+1}=3a-2\)
f(x) = x3 - 3x2 - 3x - 1 ⋮ x2 + x + 1
f(x) = x3 + x2 - 4x2 + x - 4x - 4 + 3 ⋮ x2 + x + 1
f(x) = ( x3 + x2 + x ) - ( 4x2 + 4x + 4 ) + 3 ⋮ x2 + x + 1
f(x) = x ( x2 + x + 1 ) - 4 ( x2 + x + 1 ) + 3 ⋮ x2 + x + 1
f(x) = ( x2 + x + 1 ) ( x - 4 ) + 3 ⋮ x2 + x + 1
Mà ( x2 + x + 1 ) ( x - 4 ) ⋮ x2 + x + 1
=> 3 ⋮ x2 + x + 1
=> x2 + x + 1 thuộc Ư(3) = { 1; 3; -1; -3 }
Tự thay vào rồi tìm x thôi bạn
VD :
x2 + x + 1 = 1
<=> x2 + x = 0
<=> x ( x + 1 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
Xét tiếp 3 t/h còn lại nha bạn