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\(ĐKXĐ:a\ne1;a\ge0;\)
\(B=\left(1+\dfrac{\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)-\left(a+1\right)}\right]\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\left[\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\dfrac{\sqrt{a}-1}{a+1}\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}.\dfrac{a+1}{\sqrt{a}-1}\)
\(B=\dfrac{a+1+\sqrt{a}}{\sqrt{a}-1}\)
b.
\(B=\dfrac{a+1+\sqrt{a}}{\sqrt{a}-1}>1\)
\(\Rightarrow B=\dfrac{a+1+\sqrt{a}}{\sqrt{a}-1}-1>0\)
\(\Rightarrow B=\dfrac{a+1+\sqrt{a}-\sqrt{a}+1}{\sqrt{a}-1}>0\)
\(\Rightarrow B=\dfrac{a+2}{\sqrt{a}-1}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+2>0\\\sqrt{a}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}a+2< 0\\\sqrt{a}-1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a>1\\a< -2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow a>1\)
c.
\(a=2007-2\sqrt{2006}=2006-2\sqrt{2006}+1=\left(\sqrt{2006}-1\right)^2\)Thay \(a\) vào ta được:
\(B=\dfrac{2007-2\sqrt{2006}+1+\sqrt{\left(\sqrt{2006}-1\right)^2}}{\sqrt{\left(\sqrt{2006}-1\right)^2}-1}\)
\(B=\dfrac{2007-2\sqrt{2006}+1+\sqrt{2006}-1}{\sqrt{2006}-1-1}\)
\(B=\dfrac{2007-\sqrt{2006}}{\sqrt{2006}-2}\)
\(B=\dfrac{\left(2007-\sqrt{2006}\right)\left(\sqrt{2006}+2\right)}{\left(\sqrt{2006}-2\right)\left(\sqrt{2006}+2\right)}\)
\(B=\dfrac{2007\sqrt{2006}+4014-2006-2\sqrt{2006}}{2006-4}\)
\(B=\dfrac{2005\sqrt{2006}+2008}{2002}\)
Cho A=[(x+1)/(x-1)-(x-1)/(x+1)+(x^2-4x-1)/(x^2-1)](x+2006)/x
a) tìm ĐKXĐ,rút gọn
b) tìm x để A thuộc Z
a)\(P=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}-1}{a-2\sqrt{a}+1}\)
\(P=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}-1}{a-2\sqrt{a}+1}\)
\(P=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)^2}\)
\(P=\left(\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{1}{\left(\sqrt{a}-1\right)}\)
\(P=\frac{\sqrt{a}+1}{\sqrt{a}}\)
b) Để \(P=\frac{1}{4}\Leftrightarrow\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{1}{4}\)
\(\Rightarrow4\left(\sqrt{a}+1\right)=\sqrt{a}\)
\(\Leftrightarrow3\sqrt{a}+1=0\)
<=> a ko có giá trị
P/s tha m khảo nha
\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(A=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(A=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(A=a+\sqrt{a}-2\sqrt{a}-1+1\)
\(A=a-\sqrt{a}\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)