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\(\Rightarrow\left(\sqrt{a+\sqrt{b}}\mp\sqrt{a-\sqrt{b}}\right)^2=\left(\sqrt{2\left(a\mp\sqrt{a^2-b}\right)}\right)^2\Leftrightarrow a+\sqrt{b}+a-\sqrt{b}\mp2\sqrt{\left(a+\sqrt{b}\right)\cdot\left(a-\sqrt{b}\right)}=2a\mp2\sqrt{a^2-b}\Leftrightarrow2a\mp2\sqrt{a^2-b}=2a\mp2\sqrt{a^2-b}\) (luôn đúng) \(\Rightarrowđpcm\)
a, \(VT=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b=VP\) đpcm
b,\(VT=1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}-\dfrac{a^2-a}{a-1}=1-\sqrt{a}+\sqrt{a}-a=1-a=VP\) đpcm
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=2
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
=\(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)(đpcm)
Lời giải:
Đặt \((\sqrt{a}, \sqrt{b}, \sqrt{c})=(x,y,z)\). Bài toán trở thành
Cho $x,y,z$ dương thỏa mãn \(y^2\neq z^2; x+y\neq z; x^2+y^2=(x+y-z)^2\)
CMR: \(\frac{x^2+(x-z)^2}{y^2+(y-z)^2}=\frac{x-z}{y-z}\)
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Ta có:
\(x^2+y^2=(x+y-z)^2=[y+(x-z)]^2\)
\(\Leftrightarrow x^2+y^2=y^2+(x-z)^2+2y(x-z)\)
\(\Leftrightarrow x^2=(x-z)^2+2y(x-z)\)
\(\Leftrightarrow x^2+(x-z)^2=2(x-z)^2+2y(x-z)=2(x-z)(x-z+y)\)
Tương tự:
\(y^2+(y-z)^2=2(y-z)^2+2x(y-z)=2(y-z)(y-z+x)\)
Do đó: \(\frac{x^2+(x-z)^2}{y^2+(y-z)^2}=\frac{2(x-z)(x-z+y)}{2(y-z)(y-z+x)}=\frac{x-z}{y-z}\)
Ta có đpcm.