\(\frac{\sqrt{12+8x-x^2-x^3}}{\sqrt{3-x}}\)
help me
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AD
23 tháng 7 2023
\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)
\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)
\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\) (vì \(x>4\) )
@seven
16 tháng 8 2023
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
AN
0
17 tháng 8 2019
a, ĐKXĐ: x>0 (1)
b,T= (\(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)+6\sqrt{x}-9\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\)).(\(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\))+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{x+3\sqrt{x}+2+6\sqrt{x}-9x-9\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{2-8x}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{2\left(1-2\sqrt{x}\right)\left(1+2\sqrt{x}\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2\left(1-2\sqrt{x}\right)}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\frac{1+2\sqrt{x}}{3\sqrt{x}}\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\) = \(\frac{x-\sqrt{x}}{3\sqrt{x}}\)=\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{3\sqrt{x}}\)=\(\frac{\sqrt{x}-1}{3}\)
c, Để T<0 \(\Leftrightarrow\)\(\frac{\sqrt{x}-1}{3}\) <0 \(\Leftrightarrow\) \(\sqrt{x}\)-1<0 \(\Leftrightarrow\) \(\sqrt{x}\)<1\(\Leftrightarrow\) x<1 mà do ĐK (1)
=> Để T<0 \(\Leftrightarrow\) 0<x<1
17 tháng 8 2019
Cho mk hỏi là bước t2 từ dưới lên phần b thì \(\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)\) sao lại khai triển đc như vậy
\(\sqrt{12+8x-x^2-x^3}\)
\(=\sqrt{3\left(x^2+4x+4\right)-x\left(x^2+4x+4\right)}\)
\(=\sqrt{\left(3-x\right)\left(x+2\right)^2}\)
\(=\sqrt{3-x}.\left(x+2\right)\)
Chia tử cho mẫu còn :
\(x+2\)
thank you