Câu 5:
\(S=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\\ =\left(x^2+5x+4\right)\left(x^2+5x+6\right)=\left[\left(x^2+5x+5\right)+1\right]\left[\left(x^2+5x+5\right)-1\right]\\ =\left(x^2+5x+5\right)^2-1\ge-1\)
Dấu = xảy ra \(\Leftrightarrow x=\dfrac{-5\pm\sqrt{5}}{2}\)
Bài 1:
b) Ta có: \(\sqrt{52+30\sqrt{3}}-\sqrt{50-30\sqrt{3}}\)
\(=\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}\)
\(=\sqrt{\left(3\sqrt{3}+5\right)^2}-\sqrt{\left(3\sqrt{3}-5\right)^2}\)
\(=3\sqrt{3}+5-3\sqrt{3}+5=10\)(đpcm)
Bài 2:
a) ĐKXĐ : x khác 2,6
Ta có:
\(\dfrac{1}{x-2}+\dfrac{3}{6-x}=2\\ \Rightarrow\left(6-x\right)+3\left(x-2\right)=2\left(x-2\right)\left(6-x\right)\Leftrightarrow2x=2\left(-x^2+8x-12\right)\\ \Leftrightarrow x^2-7x+12=0\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow x\in\left\{3;4\right\}\left(t.m\right)\)
Câu 3
b) ĐKXĐ \(x\ge-\dfrac{5}{2}\)
\(\sqrt{2x+5}=2x-1\\ 2x+5=\left(2x-1\right)^2\left(x\ge\dfrac{1}{2}\right)\Rightarrow2x+5=4x^2-4x+1\\ \Leftrightarrow4x^2-6x-4=0\Leftrightarrow2x^2-3x-2=0\\ \Leftrightarrow\left(2x+1\right)\left(x-2\right)=0\Leftrightarrow x=2\left(do.x\ge\dfrac{1}{2}\right)\)
