Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
10. a)
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Leftrightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(x^4+y^4\right)=ab\left(x^2+y^2\right)^2\Leftrightarrow\left(bx^2-ay^2\right)^2=0\Leftrightarrow bx^2=ay^2\)
b) Từ \(ay^2=bx^2\Rightarrow\frac{y^2}{b}=\frac{x^2}{a}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2008}}{a^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\); \(\frac{y^{2008}}{b^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\)
\(\Rightarrow\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)
25. Ta có \(\left(ax+by+cz\right)^2=0\Leftrightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(abxy+bcyz+acxz\right)\)
Xét mẫu số của P : \(bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ac\left(x^2-2xz+z^2\right)+ab\left(x^2-2xy+y^2\right)\)
\(=y^2bc-2bcyz+bcz^2+acx^2-2xzac+acz^2+abx^2-2abxy+aby^2\)
\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(abxy+xzac+bcyz\right)\)
\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\)
\(=c\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+a\left(ax^2+by^2+cz^2\right)=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)
\(\Rightarrow P=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{2007}\)
8. \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)^3-3.\frac{xy}{ab}\left(\frac{x}{a}+\frac{y}{b}\right)=1^3-3.\left(-2\right).1=7\)
\(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(B=\frac{x-\sqrt{x}+3\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)
b/ \(C=\left(\frac{\sqrt{x}-1}{\sqrt{x}-5}.\frac{\sqrt{x}+6}{\sqrt{x}-1}\right).\frac{\sqrt{x}-5}{\sqrt{x}}\)
\(C=\frac{\sqrt{x}+6}{\sqrt{x}-5}.\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}+6}{\sqrt{x}}=1+\frac{6}{\sqrt{x}}\)
Cai này thì so sánh \(\frac{6}{\sqrt{x}}\) vs 2
Nếu0< x<9\(\Rightarrow\frac{6}{\sqrt{x}}< 2\)
Nếu x=9\(\Rightarrow\frac{6}{\sqrt{x}}=2\)
Nếu x>9\(\Rightarrow\frac{6}{\sqrt{x}}>2\)
bài tập nâng cao thì 3=1+2
Mà vế kia cx có 1 thì so sánh 2 cái còn lại chứ!
\(A=\frac{3}{x^4-x^3+x-1}-\frac{1}{x^4+x^3-x-1}-\frac{4}{x^5-x^4+x^3-x^2+x-1}\)
\(=\frac{3}{\left(x-1\right)\left(x^3+1\right)}-\frac{1}{\left(x+1\right)\left(x^3-1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{3}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\left[\frac{3}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\right]-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\left[\frac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{x^2-x+1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\right]-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)\(=\frac{3x^2+3x+3-x^2+x-1}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2+4x+2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2+4x+2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}-\frac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2+4x+2-4x-4}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2x^2-2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2}{x^4+x^2+1}\)
\(\Rightarrow A=\frac{2}{x^4+x^2+1}\left(x\ne\pm1\right)\)
Ta có: \(x^4+x^2+1=\left(x^2\right)^2+2.x^2.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
Vậy A > 0 với mọi \(x\ne\pm1\)