Trả lời :

Vì \(\frac{x}{a}+\frac{y}{b}=\frac{z}{c}=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}=1^2\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}=1\left(dpcm\right)\)

Study ưell

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cj mai>>>>

Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)

\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)

\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)

Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)

\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)

\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)

Xét \(\left(1\sqrt{a-1}+1\sqrt{b-1}\right)^2\le\left(a-1+1\right)\left(b-1+1\right)=ab\)

=> \(\sqrt{a-1}+\sqrt{b-1}\le\sqrt{ab}\)

Xét \(\left(1\sqrt{ab}+1.\sqrt{c-1}\right)^2\le\left(ab+1\right)\left(c-1+1\right)=\left(ab+1\right)c\)

=> \(\sqrt{ab}+\sqrt{c-1}\le\sqrt{c\left(ab+1\right)}\)

=> \(\sqrt{a-1}+\sqrt{b-1}\sqrt{c-1}\le\sqrt{c\left(ab+1\right)}\)

\(c^2=\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)

\(\Rightarrow x^2+y^2\ge\frac{c^2}{a^2+b^2}\) (đpcm)

Nếu làm như kia thì fải là nhỏ hơn hoặc bằng chứ
Nhân chia đổi chiều mà

â) \(A=\left(\frac{x}{x+4}+\frac{4}{x-4}\right):\frac{x^2+16}{x+2}\) 

\(=\left(\frac{x\left(x-4\right)+4\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}\right)=\left(\frac{x^2+16}{x^2-16}\right):\frac{x^2+16}{x+2}\)  

\(=\frac{x+2}{x^2-16}\left(đpcm\right)\)

a) \(A=\left(\frac{x}{x+4}+\frac{4}{x-4}\right):\frac{x^2+16}{x+2}\)

\(A=\frac{x\left(x-4\right)+4\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}.\frac{x+2}{x^2+16}\)

\(A=\frac{x^2-4x+4x+16}{x^2-16}.\frac{x+2}{x^2+16}\)

\(A=\frac{x^2+16}{x^2-16}.\frac{x+2}{x^2+16}\)

\(A=\frac{x+2}{x^2-16}\left(đpcm\right)\)

a) \(B=\frac{1}{x+3}+\frac{x}{x-1}-\frac{4x}{x^2+2x-3}=\frac{x-1}{x^2+2x-3}+\frac{x^2+3x}{x^2+2x-3}-\frac{4x}{x^2+2x-3}\)

\(\Leftrightarrow B=\frac{x-1+x^2+3x-4x}{x^2+2x-3}=\frac{x^2-1}{x^2+2x+1-4}=\frac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2-2^2}\)

\(\Leftrightarrow B=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{x+1}{x+3}\)

b) \(\frac{A-1}{B}=\frac{\frac{x-1}{x+3}-1}{\frac{x+1}{x+3}}=\frac{\frac{-4}{x+3}}{\frac{x+1}{x+3}}=\frac{-4}{x+1}\le\frac{1}{2}\Leftrightarrow-8\le x+1\Leftrightarrow x\ge-9\)