2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)

 \(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)

\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )

Vậy \(x=\frac{17}{3}\)

b) \(ĐKXĐ:x\ge1\)

\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)

\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)

\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )

Vậy \(x=1\)hoặc \(x=2\)

3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)

Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)

\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)

\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)

\(=\sqrt{6}-1-\sqrt{6}+2=1\)

hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)

2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))

Bình phương hai vế

\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)

\(\Leftrightarrow3x-1=16\)

\(\Leftrightarrow3x=17\)

\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )

Vậy phương trình có nghiệm duy nhất là x = 17/3

b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))

Bình phương hai vế 

\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)

\(\Leftrightarrow x-1=x^2-2x+1\)

\(\Leftrightarrow x^2-2x+1-x+1=0\)

\(\Leftrightarrow x^2-3x+2=0\)

\(\Leftrightarrow x^2-x-2x+2=0\)

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)

Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2

3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)

\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)

\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)

\(=\sqrt{6}-1-\sqrt{6}+2\)

\(=1\)

đk: \(x\ge0\)

Ta có: \(3x^2-18x+32=4\sqrt{x}+1\)

\(\Leftrightarrow3x^2-18x+31=4\sqrt{x}\)

\(\Leftrightarrow\left(3x^2-18x+31\right)^2=\left(4\sqrt{x}\right)^2\)

\(\Leftrightarrow9x^4+324x^2+961-108x^3-1116x+186x^2=16x\)

\(\Leftrightarrow9x^4-108x^3+510x^2-1132x+961=0\)

Bấm nghiệm ta được: \(x\approx2.1978946\) ; \(x\approx4.18013426\)

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