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\(\hept{\begin{cases}x+y=a+b\\x^2+y^2=a^2+b^2\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=a+b\\x^2-a^2=b^2-y^2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x-a=b-y\\\left(x-a\right)\left(x+a\right)=\left(y-b\right)\left(y+b\right)\end{cases}}\) (1)
Nếu \(x=a;y=b\Rightarrow x^n+y^n=a^n+b^n\)
Nếu \(x\ne a;x\ne b\) Từ \(\left(1\right)\Rightarrow x+a=-y-b\Rightarrow x+y=-a-b\)
Mà \(x+y=a+b\Rightarrow-a-b=a+b\Leftrightarrow2\left(a+b\right)=0\Rightarrow\hept{\begin{cases}a=-b\\x=-y\end{cases}}\)
\(\Rightarrow x^n+y^n=a^n+b^n=0\)
\(P=\frac{\sqrt{a^2}\left(\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}\right)}{\sqrt{a^2-2a+1}}=\frac{a\left(\sqrt{a-1}+1+\sqrt{a-1}-1\right)}{a-1}\)
\(=\frac{2a\sqrt{a-1}}{a-1}=\frac{2a}{\sqrt{a-1}}\)
Từ a \(\ge\)2 \(\Leftrightarrow2a\ge4\left(1\right)\)
\(a\ge2\Leftrightarrow a-1\ge1\Leftrightarrow\sqrt{a-1}\ge1\Leftrightarrow\frac{1}{\sqrt{a-1}}\le1\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\frac{2a}{\sqrt{a-1}}\ge4\)hay \(P\ge4\)
ĐKXĐ: x \(\ge\)0; x \(\ne\)1
a) P = \(\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
P = \(\left(\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{5}{x+2\sqrt{x}-\sqrt{x}-2}\right):\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\frac{2\sqrt{x}+4-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\)
P = \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b) P = \(\frac{1}{\sqrt{x}}\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\)
=> \(\sqrt{x}\left(2\sqrt{x}+1\right)-\sqrt{x}-1=0\)
<=> \(2x+\sqrt{x}-\sqrt{x}-1=0\)
<=> \(x=\frac{1}{2}\)(tm)
c)Với đk: x \(\ge\)0 và x \(\ne\)1
\(x-2\sqrt{x-1}=0\) (đk: \(x\ge1\))
<=> \(x-1-2\sqrt{x-1}+1=0\)
<=> \(\left(\sqrt{x-1}-1\right)^2=0\)
<=> \(\sqrt{x-1}-1=0\)
<=> \(\sqrt{x-1}=1\)
<=> \(\left(\sqrt{x-1}\right)^2=1\)
<=> \(\left|x-1\right|=1\)
<=> \(\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Với x = 2 => P = \(\frac{2\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left(2\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{4-2\sqrt{2}+\sqrt{2}-1}{2-1}=3-\sqrt{2}\)
a) P = \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)(sửa lại)
b) \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\) => \(2x-\sqrt{x}-\sqrt{x}-1=0\)
<=> \(2x-2\sqrt{x}-1=0\)<=> \(2\left(x-\sqrt{x}+\frac{1}{4}\right)-\frac{3}{4}=0\)
<=> \(2\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{4}\) <=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{8}\)....(tiếp tự lm)
Chứng minh:
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
Ta có: \(a+b\in Z\)
và \(a^2+b^2=\left(a+b\right)^2-2ab\in Z\Rightarrow2ab\in Z\)
\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\in Z\Rightarrow2a^2b^2\in Z\)
Đặt 2ab=k , k thuộc Z => \(4a^2b^2=k^2\Rightarrow2a^2b^2=\frac{k^2}{2}\in Z\Rightarrow\frac{k}{2}\in Z\)=> ab thuộc Z
=> \(a^3+b^3\in Z\)
Em chưa hiểu chỗ này: \(\frac{k^2}{2}\inℤ\Rightarrow\frac{k}{2}\inℤ\)