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a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)
\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)
\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)
\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
\(a.K=\left(\dfrac{\sqrt{x}+2}{3\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}-3\right):\dfrac{2-4\sqrt{x}}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)+6\sqrt{x}-9\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{2\left(1-2\sqrt{x}\right)}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}=\dfrac{2\left(1-2\sqrt{x}\right)\left(1+2\sqrt{x}\right)}{3\sqrt{x}}.\dfrac{1}{2\left(1-2\sqrt{x}\right)}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}=\dfrac{x-\sqrt{x}}{3\sqrt{x}}=\dfrac{\sqrt{x}-1}{3}\) \(b.x=\dfrac{1}{4}\left(KTMĐKXĐ\right)\) nên tại \(x=\dfrac{1}{4}\) giá trị của K không xác định .
\(c.K< 1\) ⇔ \(\dfrac{\sqrt{x}-1}{3}< 1\)
⇔ \(\sqrt{x}-1< 3\text{⇔}x< 16\)
Kết hợp với ĐKXĐ : \(0< x< 16\) ( x # \(\dfrac{1}{4}\) )
\(d.Để:\) K ∈ Z ⇔ \(\sqrt{x}-1\text{∈}\left\{1;-1;3;-3\right\}\)
+) \(\sqrt{x}-1=1\text{⇔ }x=4\left(TM\right)\)
+) \(\sqrt{x}-1=-1\text{⇔ }x=0\left(KTM\right)\)
+) \(\sqrt{x}-1=3\text{⇔ }x=16\left(TM\right)\)
+) \(\sqrt{x}-1=-3\text{⇔ }vô-nghiem\)
KL...............
Bài 1. Ta có : \(xy+\dfrac{1}{xy}=16xy-15xy+\dfrac{1}{xy}\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(x+y\) ≥ \(2\sqrt{xy}\)
⇔ \(\left(x+y\right)^2\) ≥ \(4xy\)
⇔ \(\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\) ≥ xy
⇔ - 15xy ≥ \(\dfrac{1}{4}.\left(-15\right)=\dfrac{-15}{4}\)
CMTT , \(16xy+\dfrac{1}{xy}\) ≥ \(2\sqrt{16xy.\dfrac{1}{xy}}=2.\sqrt{16}=8\)
⇒ \(16xy+\dfrac{1}{xy}\) - 15xy ≥ \(8-\dfrac{15}{4}=\dfrac{17}{4}\)
a/\(x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{\left(x-3\right)^2}=x+3+\left|x-3\right|=x+3+3-x=6\)
b/ \(\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{\left(x+2\right)^2}-\left|x\right|=\left|x+2\right|-\left|x\right|=-x-2-\left(-x\right)=-x-2+x=-2\)
c/ \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\cdot\left(x-1\right)=\sqrt{x^2-2x+1}=\sqrt{\left(x-1\right)^2}=\left|x-1\right|\)
d/ \(\left|x-2\right|+\dfrac{\sqrt{x^2-4x+4}}{x-2}=2-x+\dfrac{\sqrt{\left(x-2\right)^2}}{x-2}=2-x+\dfrac{\left|x-2\right|}{x-2}=2-x+\dfrac{-\left(x-2\right)}{x-2}=2-x-1=1-x\)