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Bài 1 :
a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)
\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)
\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)
\(A=\sqrt{7}-\sqrt{28}\)
\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)
Vậy \(A=-\sqrt{7}\)
b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)
\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)
\(B=\left(\sqrt{a}+\sqrt{b}\right).\frac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(B=a-b\)
Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)
_Minh ngụy_
Bài 2 :
a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)
Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))
Vậy \(x>1\)thì \(B>0\)
_Minh ngụy_
1.
a. ĐKXĐ : x lớn hơn hoặc bằng 1/2
b. A\(\sqrt{2}\)= \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
= \(\sqrt{2x-1+1+2\sqrt{2x-1}}-\sqrt{2x-1+1-2\sqrt{2x-1}}\)
=\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
= \(\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)
Nếu \(x\ge1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(\sqrt{2x-1}-1\right)=2\)
\(\Rightarrow A=2\)
Nếu 1/2 \(\le x< 1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(1-\sqrt{2x-1}\right)=2\sqrt{2x-1}\)
Do đó : A= \(\sqrt{4x-2}\)
Vậy ............
2.
a. \(x\ge2\)hoặc x<0
b. A= \(2\sqrt{x^2-2x}\)
c. A<2 \(\Leftrightarrow\)\(2\sqrt{x^2-2x}< 2\Leftrightarrow\sqrt{x^2-2x}< 1\Leftrightarrow x^2-2x< 1\Leftrightarrow\left(x-1\right)^2< 2\)
\(-\sqrt{2}< x-1< \sqrt{2}\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\)
Kết hợp vs đk câu a , ta đc : \(1-\sqrt{2}< x< 0và2\le x< 1+\sqrt{2}\)
Vậy...........
Sửa đề nha: \(\sqrt{x^3-1}\) thành \(\sqrt{x^3}-1\)
\(B=\left(\frac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(B=\left(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(B=\frac{\left(x+\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
b/ Để B= 3\(\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=16\)
Mấy bạn giải chi tiết dùm mình , mình cần gấp lắm ạk