Lời giải:

Áp dụng BĐT Bunhiacopxky:

$\frac{47}{15}(3x^2+5y^2)=[(\sqrt{3}x)^2+(-\sqrt{5}y)^2][(\frac{2}{\sqrt{3}})^2+(\frac{3}{\sqrt{5}})^2]\geq (2x-3y)^2$

$\Leftrightarrow \frac{47}{15}(3x^2+5y^2)\geq 49$

$\Rightarrow 3x^2+5y^2\geq \frac{735}{47}$

Ta có đpcm.

Ta có: 

Vì \(\frac{2}{3}< x< \frac{13}{2}\Rightarrow\hept{\begin{cases}3x-2>0\\10-x>0\\13-2x>0\end{cases}}\)

Khi đó: \(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\)

\(=\frac{1}{3x-2}+\frac{1}{10-x}+\frac{1}{13-2x}\) \(\left(1\right)\)

Áp dụng BĐT Cauchy Schwarz ta được:

\(\left(1\right)\ge\frac{\left(1+1+1\right)^2}{3x-2+10-x+13-2x}\)

\(=\frac{3^2}{21}=\frac{3}{7}\)

Vậy với \(\frac{2}{3}< x< \frac{13}{2}\) thì \(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\ge\frac{3}{7}\)

a) \(x\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(x\left(2x+7\right)>0\)

\(TH1:\left\{{}\begin{matrix}x>0\\2x+7>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x>-\dfrac{7}{2}\end{matrix}\right.\) \(\Leftrightarrow x>0\)

\(TH2:\left\{{}\begin{matrix}x< 0\\2x+7< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x< -\dfrac{7}{2}\end{matrix}\right.\) \(\Leftrightarrow x< -\dfrac{7}{2}\)

Vậy \(x>0\) hay \(x< -\dfrac{7}{2}\)

c) \(x\left(2x+7\right)< 0\)

\(TH1:\left\{{}\begin{matrix}x>0\\2x+7< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< -\dfrac{7}{2}\end{matrix}\right.\) (Vô lý nên loại)

\(TH2:\left\{{}\begin{matrix}x< 0\\2x+7>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x>-\dfrac{7}{2}\end{matrix}\right.\) \(\Leftrightarrow-\dfrac{7}{2}< x< 0\)

Vậy \(-\dfrac{7}{2}< x< 0\)

a, x² + 5x + 10 = x² + 5x + 11

\(\Leftrightarrow\) x² - x² + 5x - 5x = 11 - 10

\(\Leftrightarrow\) 0 = 1 (KTM)

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

b, 2x² - 6x + 7 = 0

\(\Leftrightarrow\) x² - 6x + 9 + x² - 2 = 0

\(\Leftrightarrow\) (x - 3)² + (x - \(\sqrt{2}\))(x + \(\sqrt{2}\)) = 0

\(\Leftrightarrow\) (x - 3)² = 0 và (x - \(\sqrt{2}\))(x + \(\sqrt{2}\))

Mà (x - 3)² \(\ne\) (x - \(\sqrt{2}\))(x + \(\sqrt{2}\))

nên không có x nào TM để (x - 3)² = (x - \(\sqrt{2}\))(x + \(\sqrt{2}\))

Hay (x - 3)² + (x - \(\sqrt{2}\))(x + \(\sqrt{2}\)) = 0 hay 2x² - 6x + 7 = 0

Vậy S = \(\varnothing\)

Bạn ơi câu c hình như đề sai rồi thì phải, VD nếu x = 3 thì nó vẫn lớn hơn 0 mà bạn

c, \(|x^2+3x+20|+|x-3|\ge0\)

\(|3^2+3\cdot3+20|+|3-3|\)

= \(38+0=38>0\)

Câu c vẫn có nghiệm mà, đâu có vô nghiệm đâu!

Chúc bn học tốt!!

1: \(x^2+x+1\)

\(=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

2: \(2x^2+2x+1\)

\(=2\left(x^2+x+\dfrac{1}{2}\right)\)

\(=2\left(x^2+x+\dfrac{1}{4}+\dfrac{1}{4}\right)\)

\(=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\forall x\)

3: 

\(x^2+y^2=\left(x-y\right)^2+2xy=7^2+2\cdot60=169\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2\cdot\left(xy\right)^2\)

\(=169^2-2\cdot60^2=21361\)

\(a,0,1^{2-x}>0,1^{4+2x}\\ \Leftrightarrow2-x>2x+4\\ \Leftrightarrow3x< -2\\ \Leftrightarrow x< -\dfrac{2}{3}\)

\(b,2\cdot5^{2x+1}\le3\\ \Leftrightarrow5^{2x+1}\le\dfrac{3}{2}\\ \Leftrightarrow2x+1\le log_5\left(\dfrac{3}{2}\right)\\ \Leftrightarrow2x\le log_5\left(\dfrac{3}{2}\right)-1\\ \Leftrightarrow x\le\dfrac{1}{2}log_5\left(\dfrac{3}{2}\right)-\dfrac{1}{2}\\ \Leftrightarrow x\le log_5\left(\dfrac{\sqrt{30}}{10}\right)\)

c, ĐK: \(x>-7\)

\(log_3\left(x+7\right)\ge-1\\ \Leftrightarrow x+7\ge\dfrac{1}{3}\\ \Leftrightarrow x\ge-\dfrac{20}{3}\)

Kết hợp với ĐKXĐ, ta có:\(x\ge-\dfrac{20}{3}\)

d, ĐK: \(x>\dfrac{1}{2}\)

\(log_{0,5}\left(x+7\right)\ge log_{0,5}\left(2x-1\right)\\ \Leftrightarrow x+7\le2x-1\\ \Leftrightarrow x\ge8\)

Kết hợp với ĐKXĐ, ta được: \(x\ge8\)

\(VT=\left(x^2-3x+\dfrac{9}{4}\right)+\left(y^2+\dfrac{z^2}{4}+4-yz-4y+2z\right)+\dfrac{3}{4}\left(z^2-\dfrac{8z}{3}+\dfrac{16}{9}\right)-\dfrac{91}{12}\)

\(VT=\left(x-\dfrac{3}{2}\right)^2+\left(y-\dfrac{z}{2}-2\right)^2+\dfrac{3}{4}\left(z-\dfrac{4}{3}\right)^2-\dfrac{91}{12}\ge-\dfrac{91}{12}>-7\)

\(x\in R\)

\(\left|x^2+2x+7\right|\ge-10\)

\(\Rightarrow x\in Q\)