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b) Đặt a+b=s và ab=p. Ta có: \(a^2+b^2=4-\left(\frac{ab+2}{a+b}\right)^2\Leftrightarrow\left(a+b\right)^2-2ab+\frac{\left(ab+2\right)^2}{\left(a+b\right)^2}=4\)
\(\Leftrightarrow s^2-2p+\frac{\left(p+2\right)^2}{s^2}=4\Leftrightarrow s^4-2ps^2+\left(p+2\right)^2=4s^2\)
\(\Leftrightarrow s^4-2s^2\left(p+2\right)+\left(p+2\right)^2=0\Leftrightarrow\left(s^2-p-2\right)^2=0\)
\(\Leftrightarrow s^2-p-2=0\Leftrightarrow p+2=s^2\Leftrightarrow\sqrt{p+2}=\left|s\right|\Leftrightarrow\sqrt{ab+2}=\left|a+b\right|\)
Vì a, b là số hữu tỉ nên |a+b| là số hữu tỉ. Vậy \(\sqrt{ab+2}\)là số hữu tỉ
5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
a) ĐK: \(x\inℝ\).
Đặt \(\sqrt{x^2-3x+4}=a>0\)
\(x^2-5x+4-\left(2x-1\right)a=0\)
\(\Leftrightarrow a^2-\left(2x-1\right)a-2x=0\)
\(\Leftrightarrow-\left(a+1\right)\left(2x-a\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-1\left(L\right)\\2x=a\left(C\right)\end{cases}}\)
Xét \(2x=a\Leftrightarrow\hept{\begin{cases}x>0\\a^2=4x^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\-3x^2-3x+4=0\end{cases}}\Leftrightarrow x=\frac{-3+\sqrt{57}}{6}\) ( đã loại 1 nghiệm vì ko t/m x> 0)
P/s: em ko chắc:v
Ta có: \(x\left(x+1\right)=\frac{\sqrt{5}-1}{2}.\frac{\sqrt{5}+1}{2}=1\)
Ta có: x5 + x4 - x3 + 1 = (x5 + x4) - x3 + 1 = x3 - x3 + 1 = 1
x2 + x - 3 = x(x + 1) - 3 = - 2
x5 + x4 - x3 - 22016 = - 22016
Từ đó ta có
\(=1^{2017}+\frac{\left(-2\right)^{2016}}{-2^{2016}}=1-1=0\)
Ta có: \(x^2\text{+}x-1=...=0 \)
\(=>x^3\left(x^2\text{+}x-1\right)=0\)
=> \(x^5\text{+}x^4-x^3=0\)
=> A=\(\left(\left(x^5\text{+}x^4-x^3\right)\text{+}1\right)^{2017}\text{+}\frac{\left(\left(x^2\text{+}x-1\right)-2\right)^{2016}}{\left(x^5\text{+}x^4-x^3\right)-2^{2016}}\)
=\(1^{2017}\text{+}\frac{2^{2016}}{-2^{2016}}=1-1=0\)