Question

ba số dương a,b,c thỏa mãn \(b\ne c,\sqrt{a}+\sqrt{b}\ne\sqrt{c}\) và\(a+b=\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)^2\).chứng minh đẳng thức

\(\dfrac{a+\left(\sqrt{a}-\sqrt{c}\right)^2}{b+\left(\sqrt{b}-\sqrt{c}\right)^2}=\dfrac{\sqrt{a}-\sqrt{c}}{\sqrt{b}-\sqrt{c}}\)

Answer 1

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: \(\frac{x^2+(x-z)^2}{y^2+(y-z)^2}=\frac{x-z}{y-z}\)

--------------------------------------------------

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.