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a/ đk: a\(\ne b\), b\(\ne0,a\ne-b\)
= \(\frac{a\left(a-b\right)-a^2-b^2}{a-b}.\frac{a+b+2b}{b\left(a+b\right)}\)
= \(\frac{a^2-ab-a^2-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)
= \(\frac{-ab-b^2}{a-b}.\frac{a+3b}{b\left(a+b\right)}\)
= \(\frac{-b\left(a+b\right)\left(a+3b\right)}{b\left(a+b\right)\left(a-b\right)}\)
= \(\frac{-a-3b}{a-b}\)
b/ đk: a\(\ne0,a\ne\pm3\)
= \(\left[\frac{3a+1}{a\left(a-3\right)}+\frac{3a-1}{a\left(a+3\right)}\right].\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{\left(3a+1\right)\left(a+3\right)+\left(3a-1\right)\left(a-3\right)}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{6a^2+6}{a\left(a-3\right)\left(a+3\right)}.\frac{\left(a-3\right)\left(a+3\right)}{a^2+1}\)
= \(\frac{6\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}{a\left(a^2+1\right)\left(a-3\right)\left(a+3\right)}\)
= \(\frac{6}{a}\)
a) Biến đổi vế phải, ta có :\(\frac{-3x\left(x-y\right)}{y^2-x^2}=\frac{3x\left(x-y\right)}{x^2-y^2}=\frac{3x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{3x}{x+y}\) = vế trái \(\Rightarrowđpcm\)
c)Biến đổi vế phải ta có: \(\frac{3a\left(x+y\right)^2}{9a^2\left(x+y\right)}=\frac{x+y}{3a}=vt\Rightarrowđpcm\)
\(=\frac{\left(a+2\right)\left(a-1\right)}{a^n\left(a-3\right)}.\left[\frac{\left(a+2-a\right)\left(a+2+a\right)}{4\left(a-1\right)\left(a+1\right)}-\frac{3}{a.\left(a-1\right)}\right]\) (Đk : x khác 0 ; 3 ; - 1 ; 1
\(=\frac{\left(a+2\right)\left(a-1\right)}{a^n\left(a-3\right)}.\left[\frac{4\left(a+1\right)}{4\left(a-1\right)\left(a+1\right)}-\frac{3}{a\left(a-1\right)}\right]\)
\(=\frac{\left(a+2\right)\left(a-1\right)}{a^n\left(a-3\right)}.\left[\frac{1}{a-1}-\frac{3}{a\left(a-1\right)}\right]\)
\(=\frac{\left(a+2\right)\left(a-1\right)}{a^n\left(a-3\right)}.\frac{a-3}{a\left(a-1\right)}=\frac{a+2}{a^{n+1}}\)
9) bài này nhiều cách thay lắm. chả biết cách nào nhanh hơn.
ĐK : ...
\(N=\frac{a+x+1}{a+x}:\frac{a^2+ax-a}{a+x}.\left[\frac{2ax-1+\left(a^2+x^2\right)}{2ax}\right]\)
\(N=\frac{a+x+1}{a+x}.\frac{a+x}{a\left(a+x-1\right)}.\frac{\left(a+x\right)^2-1}{2ax}\)
\(N=\frac{a+x+1}{a\left(a+x-1\right)}.\frac{\left(a+x-1\right)\left(a+x+1\right)}{2ax}\)
\(N=\frac{\left(a+x+1\right)^2}{2a^2x}=\frac{\left(a+1+\frac{1}{a-1}\right)^2}{\frac{2a^2}{a-1}}\)
\(N=\frac{\left(\frac{\left(a+1\right)\left(a-1\right)+1}{a-1}\right)^2}{\frac{2a^2}{a-1}}=\frac{\left(\frac{a^2}{a-1}\right)^2}{\frac{2a^2}{a-1}}=\frac{\frac{a^4}{\left(a-1\right)^2}}{\frac{2a^2}{a-1}}=\frac{a^2}{2\left(a-1\right)}\)
10) \(3a^2+3b^2=10ab\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(3a^2-9ab\right)-\left(ab-3b^2\right)=0\)
\(\Leftrightarrow3a\left(a-3b\right)-b\left(a-3b\right)=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-3b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3a=b\\a=3b\left(loai-vi-b>a>0\right)\end{cases}}\)
Thay 3a = b vào biểu thức, ta có :
\(P=\frac{a-b}{a+b}=\frac{a-3a}{a+3a}=\frac{-2a}{4a}=\frac{-1}{2}\)
a) đặt \(t=x^2\) (t\(\ge0\))
=>\(t^2-t-2=0\)=>\(\orbr{\begin{cases}t=2\\t=-1\left(loại\right)\end{cases}}\)
=>\(x^2=2\)=>\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
a) \(\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\end{cases}}\)
b)\(\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
c)\(x=\frac{47}{6}\)