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Đề viết mệt quá nên thay \(\sqrt{a}=a;\sqrt{b}=b;\sqrt{c}=c\) viết lại đề tiện thể sửa đề luôn.
\(a^2+b^2=\left(a+b-c\right)^2\)
Chứng minh:
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
Ta có: \(a^2+b^2=\left(a+b-c\right)^2\)
\(\Leftrightarrow c^2-2ac-2bc+2ab=0\)
\(\Leftrightarrow a=\frac{c^2-2bc}{2c-2b}\)
Thế vô bài toán ta được
\(VT=\frac{\left(\frac{c^2-2bc}{2c-2b}\right)^2+\left(\frac{c^2-2bc}{2c-2b}-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\frac{\left(\frac{c^2-2bc}{2c-2b}\right)^2+\left(\frac{c^2-2bc}{2c-2b}-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\frac{\left(\frac{c^2-2bc}{2c-2b}\right)^2+\left(c^2\right)^2}{b^2+\left(b-c\right)^2}=\frac{2c^2\left(2b^2+c^2-2bc\right)}{\left(2b^2+c^2-2bc\right)4\left(c-b\right)^2}=\frac{c^2}{2\left(c-b\right)^2}\)
Ta lại có:
\(VP=\frac{\frac{c^2-2bc}{2c-2b}-c}{b-c}=\frac{-c^2}{-2\left(c-b\right)^2}=\frac{c^2}{2\left(c-b\right)^2}\)
\(\Rightarrow\)ĐOCM
Ta có : \(\frac{\frac{\left(a-b\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}\)
\(=\frac{\frac{\left(\sqrt{a}-\sqrt{b}\right)^3\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}+2a\sqrt{a}-b\sqrt{b}}{\sqrt{a}^3-\sqrt{b}^3}+\frac{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{-\left(a-b\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^3+2a\sqrt{a}-b\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{-\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}+b\sqrt{b}+2a\sqrt{a}-b\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{3a\sqrt{b}+3\sqrt{a}b+3a\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{3\sqrt{a}\left(\sqrt{ab}+b+a\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=-\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}=0\)
Vậy ...
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.
Lời giải:
\(\frac{\frac{(a-b)^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}=\frac{\frac{[(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})]^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}\)
\(=\frac{(\sqrt{a}+\sqrt{b})^3-b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}\)
\(=\frac{a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}+b\sqrt{b}-b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}=\frac{3\sqrt{a}(a+\sqrt{ab}+b)}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}=\frac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)
\(\frac{3a+3\sqrt{ab}}{b-a}=\frac{3\sqrt{a}(\sqrt{a}+\sqrt{b})}{(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})}=\frac{3\sqrt{a}}{\sqrt{b}-\sqrt{a}}\)
Do đó:
\(\frac{\frac{(a-b)^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}=\frac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}+\frac{3\sqrt{a}}{\sqrt{b}-\sqrt{a}}=0\)
Ta có đpcm.
