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VT là căn bậc 2 hay 3?

VT không có căn VP căn bậc 2

b: ĐKXĐ: 8/3<=x<=7

Ta có: \(\sqrt{x+3} - \sqrt{7-x} > \sqrt{3x-8}\)

=>\(\sqrt{x+3}>\sqrt{3x-8}+\sqrt{7-x}\)

=>\(x + 3 > (3x - 8) + (7 - x) + 2\sqrt{(3x-8)(7-x)}\)

=>\(x + 3 > 2x - 1 + 2\sqrt{(3x-8)(7-x)}\)

=>\(4 - x > 2\sqrt{-3x^2 + 29x - 56}\) (1)

Để (1) có nghiệm thì 4-x>0

=>x<4

=>8/3<=x<4

(1) sẽ tương đương: \((4 - x)^2 > 4(-3x^2 + 29x - 56)\)

=>\(16 - 8x + x^2 > -12x^2 + 116x - 224\)

=>\(13x^2 - 124x + 240 > 0\)

=>\(x^2-\frac{124}{13}x+\frac{240}{13}>0\)

=>\(x^2-2\cdot x\cdot\frac{62}{13}+\frac{3844}{169}-\frac{3844}{169}+\frac{240}{13}>0\)

=>\(\left(x-\frac{62}{13}\right)^2>\frac{3844}{169}-\frac{240}{13}=\frac{724}{169}\)

=>\(\left[\begin{array}{l}x-\frac{62}{13}>\frac{\sqrt{724}}{13}\\ x-\frac{62}{13}<-\frac{\sqrt{724}}{13}\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\frac{62+\sqrt{724}}{13}\\ x<\frac{62-\sqrt{724}}{13}\end{array}\right.\)

mà 8/3<=x<4

nên \(\frac{8}{3} \le x < \frac{62 - 2\sqrt{181}}{13}\)

ĐKXĐ: \(x^2-4x+1\ge0\)

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

\(\Leftrightarrow2x+2-5\sqrt{x}+2\sqrt{x^2-4x+1}-\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{4x^2-17x+4}{2x+2+5\sqrt{x}}+\dfrac{4x^2-17x+4}{2\sqrt{x^2-4x+1}+\sqrt{x}}=0\)

\(\Leftrightarrow\left(4x^2-17x+4\right)\left(\dfrac{1}{2x+2+5\sqrt{x}}+\dfrac{1}{2\sqrt{x^2-4x+1}+\sqrt{x}}\right)=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

\(\Leftrightarrow...\)

ĐKXĐ:  \(x\ge1\)

\(\Rightarrow\left(\sqrt{x-1}+\sqrt{2x+1}\right)^2=1\Leftrightarrow x-1+2x+1+2\sqrt{\left(x-1\right)\left(2x+1\right)}=1\Leftrightarrow3x+2\sqrt{2x^2-x-1}=1\) \(\Leftrightarrow2\sqrt{2x^2-x-1}=1-3x\Rightarrow\left(2\sqrt{2x^2-x-1}\right)^2=\left(1-3x\right)^2\Leftrightarrow8x^2-4x-4=9x^2-6x+1\) \(\Leftrightarrow x^2-2x+5=0\Leftrightarrow\left(x-1\right)^2+4=0\Leftrightarrow\left(x-1\right)^2=-4\) vô lí vì VT\(\ge0\) mà VP<0 \(\Rightarrow\) ko có x Vậy...

Thanks Broo