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\(\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^1}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}+\frac{a}{\sqrt{a^2-b^2}}\right).\frac{a-\sqrt{a^2-b^2}}{b}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b.\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)
\(=\frac{a-b}{\sqrt{a^2-b^2}}\)
\(=\frac{a-b}{\sqrt{a-b}.\sqrt{a+b}}\)
\(=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
\(=\frac{\sqrt{a^2-b^2}}{a+b}\)
\(1,\sqrt{4\left(a-4\right)^2}\left(dkxd:a\ge4\right)\)
\(=\sqrt{4}.\sqrt{\left(a-4\right)^2}\)
\(=\sqrt{2^2}.\left|a-4\right|\)
\(=2\left(a-4\right)\)
\(=2a-8\)
\(2,\sqrt{9\left(b-5\right)^2}\left(dkxd:b< 5\right)\)
\(=\sqrt{9}.\sqrt{\left(b-5\right)^2}\)
\(=\sqrt{3^2}.\left|b-5\right|\)
\(=3\left(-b+5\right)\)
\(=-3b+15\)
ĐK: a,b>0 , a khác b
\(A=\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}.\frac{\sqrt{a}+\sqrt{b}}{\sqrt{b}}\right]:\left(\frac{a^2-b^2}{ab}\right)\)
\(=\frac{a-b}{b}:\frac{\left(a-b\right)\left(a+b\right)}{ab}=\frac{a-b}{b}.\frac{ab}{\left(a-b\right)\left(a+b\right)}=\frac{a}{a+b}\)
Với b=1, A=2 ta có:
\(\frac{a}{a+1}=2\Leftrightarrow a=2a+2\Leftrightarrow a=-2\) loại
vậy không tồn tại a để A=2 b=1
\(A=\left[\left(\sqrt{\frac{a}{b}}-1\right).\left(\sqrt{\frac{a}{b}}+1\right)\right]:\left(\frac{a}{b}-\frac{b}{a}\right)\)
\(A=\left[\left(\sqrt{\frac{a}{b}}\right)^2-1\right]:\left(\frac{a^2}{ab}-\frac{b^2}{ab}\right)\)
\(A=\left(\frac{a}{b}-1\right):\left[\frac{\left(a-b\right)\left(a+b\right)}{ab}\right]\)
\(A=\left(\frac{a-b}{b}\right).\left[\frac{ab}{\left(a-b\right)\left(a+b\right)}\right]\)
\(A=\frac{a}{a+b}\)
A = \(\frac{8}{\sqrt{5}-1}\) - (\(2\sqrt{5}-1\) ) ( chúng ta cần trục căn thức lên để khử mẫu )
= \(\frac{8\left(\sqrt{5}+1\right)}{5-1}\)- \(\left(2\sqrt{5}-1\right)\)
= \(2\sqrt{5}\)+ 2 - \(2\sqrt{5}\)+1
= 3
B = \(\frac{\left(1-\sqrt{x}\right)^2+4\sqrt{x}}{1+\sqrt{x}}\)( x \(\ge\)0 )
= \(\frac{1-2\sqrt{x}+x+4\sqrt{x}}{1+\sqrt{x}}\)
= \(\frac{1+2\sqrt{x}+x}{1+\sqrt{x}}\)
= \(\frac{\left(1+\sqrt{x}\right)^2}{1+\sqrt{x}}\)
= 1 +\(\sqrt{x}\)
#mã mã#
Dat \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Ta co: \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\ge8\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Ta d̃i CM:\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Ta co:\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\left(dpcm\right)\)
Dau '=' xay ra khi \(a=b=c\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\frac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(\frac{a\left(a-1\right)}{a-1}-\frac{a\left(-6a+5\right)}{a-1}=\frac{a^2-a+6a^2-5a}{a-1}\)
=\(\frac{7a^2-6a}{a-1}\)
quy đòng, xong phá ngoặc là xong, nhớ tìm ĐKXĐ nữa