BÀI 1:

\(\dfrac{a}{k}=\dfrac{x}{a}\Rightarrow a^2=kx\)

\(\dfrac{b}{k}=\dfrac{y}{b}\Rightarrow b^2\)=ky

Vay \(\dfrac{a^2}{b^2}=\dfrac{kx}{ky}=\dfrac{x}{y}\)

Bài 2:

Vì a=b+c nên ad=(b+c)d=bd+cd (1)

Vi c=\(\dfrac{bd}{b-d}\)nen \(bd=\)c.(b-d)=bc-cd hay bc=bd+cd (2)

Từ (1),(2) =>ad=bc=>\(\dfrac{a}{b}=\dfrac{c}{d}\)

1 , ĐKXĐ : \(x\ge0,x\ne1\)

Với điều kiện xác định trên phương trình đã cho thánh :

\(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}+\dfrac{x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-1-2\left(\sqrt{x}+1\right)+x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

Ta có:

+) \(\dfrac{a}{k}=\dfrac{b}{k}\Rightarrow a=b\)

+) \(\dfrac{x}{a}=\dfrac{y}{b}\)mà a=b \(\Rightarrow x=y\)

Ta lại có:

+)a=b \(\Rightarrow\) \(\dfrac{a^2}{b^2}=\left(\dfrac{a}{b}\right)^2=1^2=1\)(1)

+)x=y \(\Rightarrow\dfrac{x}{y}=1\)(2)

* Từ (1) và (2) \(\Rightarrow\)\(\dfrac{a^2}{b^2}=\dfrac{x}{y}\)

Vậy \(\dfrac{a^2}{b^2}=\dfrac{x}{y}\)

CHÚC BẠN HỌC TỐT!

a ,rút gọn P (dkxd x\(\ge0,x\ne0\)

P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

=\(\dfrac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}\)+\(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

=\(\dfrac{15\sqrt{x}-11}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x-1}\right)}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

=\(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

=\(\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{7\sqrt{x}-5x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

..............=\(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

a: ĐKXĐ: x>=0; x<>1

b: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

c: Để K=1/2 thì \(\dfrac{-5x+5\sqrt{x}-8}{x+2\sqrt{x}-3}=\dfrac{1}{2}\)

=>\(-10x+10\sqrt{x}-16-x-2\sqrt{x}+3=0\)

=>\(-11x+8\sqrt{x}-13=0\)

hay \(x\in\varnothing\)

a, ĐKXĐ:\(\left\{{}\begin{matrix}X-1\ne0\\X+1\ne0\\X^2-1\ne0\\X\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}X\ne0\\X\ne\pm1\end{matrix}\right.\)

b,Ta có: \(K=\left(\dfrac{\left(X+1\right)^2-\left(X-1^2\right)+X^2-4X-1}{X^2-1}\right).\dfrac{X+2003}{X}\)

\(=\dfrac{X^2+2X+1-X^2+2X-1+X^2-4X-1}{X^2-1}.\dfrac{X+2003}{X}\)

\(=1.\dfrac{X+2003}{X}=\dfrac{X+2003}{X}\)

a) K có nghĩa khi \(\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\) b) Rút gọn: \(K=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}=\left[\dfrac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right].\dfrac{x+2003}{x}=\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}.\dfrac{x+2003}{x}=\dfrac{x^2-1}{x^2-1}.\dfrac{x+2003}{x}=\dfrac{x+2003}{x}\)

c) ta có: \(K=\dfrac{x+2003}{x}=1+\dfrac{2003}{x}\), để K có giá trị nguyên thì \(\dfrac{2003}{x}\) có giá trị nguyên, hay \(x\inƯ\left(2003\right)=\left\{2003;-2003;1;-1\right\}\),vì x\(\ne\)1 nên x nhận các giá trị 2003;-1;-2003.

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=k\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=k^2\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}+\dfrac{2}{bc}+\dfrac{1}{c^2}+\dfrac{2}{ac}=k^2\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1\left(a+b+c\right)}{abc}=k^2\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=k^2-k\)

áp dụng tính chất dãy tỉ số băng nhau ta có

\(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{a}\)=\(\dfrac{a+b+c}{b+c+a}\)=1

=>k=1