(=)2 căn x + 2 căn (3x -2)= 2x²+2

(=) 2(x-1)²+( căn x -1)²+(căn (3 -2 ) -1)²=0

vì bình phương luôn lớn hơn bằng 0 nên x-1=0,căn x -1=0,căn (3-2)-1=0

suy ra x=1

ĐKXĐ: x \ge \dfrac23x≥32​.

\sqrt x + \sqrt{3x-2} = x^2 + 1x​+3x−2​=x2+1

\Leftrightarrow 2\sqrt x + 2\sqrt{3x-2} = 2x^2 + 2⇔2x​+23x−2​=2x2+2

\Leftrightarrow 2(x-1)^2 + (\sqrt x - 1)^2 + (\sqrt{3x-2}-1)^2 = 0.⇔2(x−1)2+(x​−1)2+(3x−2​−1)2=0.

Phương trình có nghiệm duy nhất $x=1.$

a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)

\(\Rightarrow A=\frac{2+1}{2+2}=\frac{3}{4}\)

Vậy với x = 4 thì A = 3/4 

b, \(B=\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}+5}{x-1}=\frac{3\left(\sqrt{x}+1\right)-\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

a,Ta có  \(x=4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)

\(\Rightarrow A=\frac{1}{\sqrt{3}-1-1}=\frac{1}{\sqrt{3}-2}\)

b, Với \(x\ge0;x\ne1\)

 \(B=\left(\frac{-3\sqrt{x}}{x\sqrt{x}-1}-\frac{1}{1-\sqrt{x}}\right):\left(1-\frac{x+2}{1+\sqrt{x}+x}\right)\)

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

\(=\left(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\right)\)

\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}.\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=1\)

Vậy biểu thức ko phụ thuộc biến x 

c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên 

thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Áp dụng bđt bunhiacopxki, ta có:

\(\left(x^2+\frac{1}{x^2}\right)\left(1+16\right)\ge\left(x+\frac{4}{x}\right)^2\) => \(x^2+\frac{1}{x^2}\ge\frac{\left(x+\frac{4}{x}\right)^2}{17}\)

=> \(\sqrt{x^2+\frac{1}{x^2}}\ge\frac{x+\frac{4}{x}}{\sqrt{17}}=\frac{x}{\sqrt{17}}+\frac{4}{x\sqrt{17}}\)

CMTT: \(\sqrt{y^2+\frac{1}{y^2}}\ge\frac{y}{\sqrt{17}}+\frac{4}{\sqrt{17}y}\)

\(\sqrt{z^2+\frac{1}{z^2}}\ge\frac{z}{\sqrt{17}}+\frac{4}{\sqrt{17}z}\)

=> A \(\ge\frac{x+y+z}{\sqrt{17}}+\frac{4}{\sqrt{17}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{x+y+z}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}\)(bđt: 1/a + 1/b + 1/c > = 9/(a+b+c)

=> A \(\ge\frac{16\left(x+y+z\right)}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}-\frac{15\left(x+y+z\right)}{\sqrt{17}}\)

A \(\ge2\sqrt{\frac{16\left(x+y+z\right)}{\sqrt{17}}\cdot\frac{36}{\sqrt{17}\left(x+y+z\right)}}-\frac{15\cdot\frac{3}{2}}{\sqrt{17}}\)(Bđt cosi + bđt: x + y + z < = 3/2)

A \(\ge\frac{48}{\sqrt{17}}-\frac{45}{2\sqrt{17}}=\frac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra <=> x = y= z = 1/2

Vậy MinA = \(\frac{3\sqrt{17}}{2}\) <=> x = y = z = 1/2

Ta có: \(\left(x+2\right)^2-3x-9=\left(1-x\right)\left(1+x\right)\)

\(\Leftrightarrow x^2+4x+4-3x-9=1-x^2\)

\(\Leftrightarrow2x^2+x-6=0\)

\(\Leftrightarrow\left(2x^2+4x\right)-\left(3x+6\right)=0\)

\(\Leftrightarrow2x\left(x+2\right)-3\left(x+2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}\)

pt <=> x² + 4x + 4 - 3x - 9 = 1 - x²

<=> 2x² + x - 6 = 0

Δ = b² - 4ac = 1 + 48 = 49

Δ > 0, áp dụng công thức nghiệm thu được x₁ = 3/2 ; x₂ = -2

Vậy ...