a) đk: \(x\ge0\)

Ta có: \(P< -\frac{1}{3}\)

\(\Leftrightarrow\frac{-3}{\sqrt{x}+3}+\frac{1}{3}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-6}{\sqrt{x}+3}< 0\)

Nhận thấy \(\sqrt{x}-6< \sqrt{x}+3\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x}-6< 0\\\sqrt{x}+3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 6\\\sqrt{x}>-3\end{cases}}\Leftrightarrow0\le x< 36\)

b) Ta có: \(Q=5+P=5-\frac{3}{\sqrt{x}+3}\)

Mà \(\sqrt{x}+3\ge3\left(\forall x\ge0\right)\Leftrightarrow\frac{3}{\sqrt{x}+3}\le1\left(\forall x\ge0\right)\)

\(\Leftrightarrow5-\frac{3}{\sqrt{x}+3}\ge4\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(x=0\)

Vậy Min(Q) = 4 khi x = 0

ĐKXD: \(x>0\)

a/ \(C-5=\frac{x+3\sqrt{x}+1}{\sqrt{x}}-5=\frac{x-2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)

Do \(x>0\Rightarrow\sqrt{x}>0\) ; \(\left(\sqrt{x-1}\right)^2\ge0\)

\(\Rightarrow C-5\ge0\Rightarrow C\ge5\)

b/ Từ kết quả câu a \(\Rightarrow\frac{7}{C}\le\frac{7}{5}=1,4\)

Do \(x>0\Rightarrow C>0\Rightarrow\frac{7}{C}>0\)

\(\Rightarrow0< \frac{7}{C}\le1,4\) Nên Với mọi x thoả mãn ĐKXĐ thì \(\frac{7}{C}\) có đúng 1 giá trị nguyên là 1

Câu a đề hơi sai nha bạn, nên mình chỉ giải câu b thoi

Áp dụng AM-GM cho các bộ 3 số dương (x,y,z) và (1/x,1/y,1/z):

\(x+y+z\ge3\sqrt[3]{xyz}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{3}{\sqrt[3]{xyz}}\)

\(\Rightarrow P\ge6\sqrt[3]{xyz}+\frac{3}{\sqrt[3]{xyz}}\ge2\sqrt{6\sqrt[3]{xyz}.\frac{3}{\sqrt[3]{xyz}}}=6\sqrt{2}\)(BĐT Cô-si)

Dấu = xảy ra khi và chỉ khi \(x=y=z=\frac{1}{\sqrt{2}}\)( thỏa x,y,z thuộc (0;1))

Mình cần câu a ạ :<

+) Ta có: \(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\)    \(\left(ĐK:x\ge0\right)\)

        \(\Leftrightarrow4\sqrt{3x}+2\sqrt{3x}=3\sqrt{3x}+6\)

        \(\Leftrightarrow3\sqrt{3x}=6\)

        \(\Leftrightarrow\sqrt{3x}=2\)

        \(\Leftrightarrow3x=4\)

        \(\Leftrightarrow x=\frac{4}{3}\left(TM\right)\)

Vậy \(S=\left\{\frac{4}{3}\right\}\)

+) Ta có:\(\sqrt{x^2-1}-4\sqrt{x-1}=0\)    \(\left(ĐK:x\ge1\right)\)

        \(\Leftrightarrow\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)

        \(\Leftrightarrow\sqrt{x-1}.\left(\sqrt{x+1}-4\right)=0\)

        \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{x+1}=4\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+1=16\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=15\left(TM\right)\end{cases}}\)

 Vậy \(S=\left\{1,15\right\}\)

+) Ta có: \(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\)       \(\left(ĐK:x\ge0\right)\)

         \(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)

         \(\Leftrightarrow\frac{2.\left(\sqrt{x}-2\right)-\sqrt{x}}{4\sqrt{x}}< 0\)

         \(\Leftrightarrow\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)

         \(\Leftrightarrow\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)

   Để \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)mà \(4\sqrt{x}\ge0\forall x\)

    \(\Rightarrow\)\(\sqrt{x}-4< 0\)

   \(\Leftrightarrow\)\(\sqrt{x}< 4\)

   \(\Leftrightarrow\)\(x< 16\)

   Kết hợp ĐKXĐ \(\Rightarrow\)\(0\le x< 16\)

 Vậy \(S=\left\{\forall x\inℝ/0\le x< 16\right\}\)

\(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\)  (Đk: x \(\ge\)0)

<=> \(4\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=6\)

<=> \(3\sqrt{3x}=6\)

<=> \(\sqrt{3x}=2\)

<=> \(3x=4\)

<=> \(x=\frac{4}{3}\)

\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) (đk: x \(\ge\)1)

<=> \(\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)

<=> \(\sqrt{x-1}\left(\sqrt{x+1}-4\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\) 

<=> \(\orbr{\begin{cases}x-1=0\\x+1=16\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=15\end{cases}}\)(tm)

\(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) (Đk: x > 0)

<=> \(\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)

<=>\(\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)

<=>  \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)

Do \(4\sqrt{x}>0\) => \(\sqrt{x}-4< 0\)

<=> \(\sqrt{x}< 4\) <=> \(x< 16\)

Kết hợp với đk => S = {x|0 < x < 16}

+) Ta có: \(2\sqrt{75}-4\sqrt{27}+3\sqrt{12}\)

         \(=2\sqrt{25}.\sqrt{3}-4\sqrt{9}.\sqrt{3}+3\sqrt{4}.\sqrt{3}\)

         \(=10.\sqrt{3}-12.\sqrt{3}+6.\sqrt{3}\)

         \(=4\sqrt{3}\approx6,9282\)

+) Ta có:\(\sqrt{x+6\sqrt{x-9}}\)

        \(=\sqrt{x-9+6\sqrt{x-9}+9}\)

        \(=\sqrt{\left(\sqrt{x-9}-3\right)^2}\)

        \(=\left|\sqrt{x-9}-3\right|\)

\(\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{2-\sqrt{3}}=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

\(=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\frac{2+\sqrt{3}}{4-3}=\sqrt{5}-\sqrt{3}+2+\sqrt{3}=\sqrt{5}+2\)

x² + y² = 2x²y²

<=> 2x² + 2y² - 4x²y² = 0

<=> 2x²(1 - 2y²) - (1 - 2y²) = -1

<=> (2x² - 1)(2y² - 1) = 1 = 1.1

Lập bảng: 

Vậy ...

OK cảm ơn nha 

a) \(\sqrt{1-x^2}\) có nghĩa

\(\Leftrightarrow1-x^2\ge0\)

\(\Leftrightarrow\left(1-x\right)\left(x+1\right)\ge0\)

\(\Leftrightarrow-1\le x\le1\)

b) \(\sqrt{\frac{1}{\left(x-5\right)^2}}\)có nghĩa

\(\Leftrightarrow\frac{1}{\left(x-5\right)^2}>0\)

\(\Leftrightarrow x\ne5\)

Vậy .............

a) Để \(\sqrt{1-x^2}\)có nghĩa 

    \(\Rightarrow\)\(1-x^2\ge0\)

  \(\Leftrightarrow\)\(\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)\ge0\)

   Vì \(\sqrt{x}\ge0\forall x\)\(\Rightarrow\)\(\sqrt{x}+1\ge1>0\forall x\)

   mà \(\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)\ge0\)

   \(\Rightarrow\)\(1-\sqrt{x}\ge0\)

  \(\Leftrightarrow\)\(\sqrt{x}\le1\)

  \(\Leftrightarrow\)\(x\le1\)

Vậy để \(\sqrt{1-x^2}\)có nghĩa thì \(x\le1\)

b) Để \(\sqrt{\frac{1}{\left(x-5\right)^2}}\)có nghĩa

    \(\Rightarrow\)\(\sqrt{\frac{1}{\left(x-5\right)^2}}\ge0\)

   \(\Leftrightarrow\)\(\frac{1}{\left|x-5\right|}\ge0\)

   Vì \(1>0\)mà \(\frac{1}{\left|x-5\right|}\ge0\)

   \(\Rightarrow\)\(\left|x-5\right|>0\)( vì là mẫu số )

  \(\Leftrightarrow\)\(x-5>0\)

  \(\Leftrightarrow\)\(x>5\)

 Vậy để \(\sqrt{\frac{1}{\left(x-5\right)^2}}\)có nghĩa thì \(x>5\)

\(Q=\sqrt{x^2-4x+4}+\sqrt{x^2+4x+4}=\sqrt{\left(x+2\right)^2}+\sqrt{\left(2-x\right)^2}\)

\(\Leftrightarrow\left|x+2\right|+\left|2-x\right|\ge\left|x+2+2-x\right|=4\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)\left(2-x\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2\ge0\\2-x\ge0\end{cases}}\) hoặc \(\orbr{\begin{cases}x+2\le0\\2-x\le0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge-2\\x\le2\end{cases}}\) hoặc \(\orbr{\begin{cases}x\le-2\\x\ge2\end{cases}}\left(vo-ly\right)\)

Vậy minQ = 4 \(\Leftrightarrow-2\le x\le2\)

Bài 1 :

ĐKXĐ : \(x\ge2\)

\(2x+5=6\sqrt{2x-4}\)

\(\Leftrightarrow4x^2+20x+25=36\left(2x-4\right)\)

\(\Leftrightarrow4x^2+20x+25-72x+144=0\)

\(\Leftrightarrow4x^2-52x+159=0\)

Đến đây chịu :))

a) \(\sqrt{25x^2-10x+1}=x+2\)

<=> \(\sqrt{\left(5x-1\right)^2}=x+2\)

<=> \(\left|5x-1\right|=x+2\)

TH1: 5x - 1 \(\ge\)0 <=> x \(\ge\)1/5

Khi đó pt trở thành: 5x - 1 = x + 2

<=> 4x = 3 <=> x = 3/4 (tm)

TH2: 5x - 1 < 0 <=>  x < 1/5

Khi đó pt trở thành:  1 - 5x = x + 2

<=> -6x = 1 <=> x = -1/6 (tm)

Vậy S = {3/4; -1/6}

b) \(\sqrt{4x^2+12x+9}=7\)

<=> \(\sqrt{\left(2x+3\right)^2}=7\)

<=> \(\left|2x+3\right|=7\)

TH1: 2x + 3 \(\ge\)0 <=> x \(\ge\)-3/2

Khi đó pt trở thành: 2x + 3 = 7 <=> 2x = 4 <=> x = 2 (Tm)

TH2: 2x + 3 < 0 <=> x < -3/2

Khi đó pt trở thành: -2x - 3 = 7

<=> -2x = 10 <=> x = -5 (tm)

Vậy S = {-5; 2}