a, \(\sqrt{\left(\sqrt{5}-4\right)^2}-\sqrt{5}+\sqrt{20}=4\)
\(VT=\sqrt{\left(4-\sqrt{5}\right)^2}-\sqrt{5}+\sqrt{20}=\left|4-\sqrt{5}\right|-\sqrt{5}+\sqrt{20}\)
\(=4-\sqrt{5}-\sqrt{5}+2\sqrt{5}=4\) hay \(VT=VP\)
Vậy ta có đpcm
b, Với \(x>0,x\ne4\)
\(P=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right):\frac{2}{x-2\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\frac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}=\frac{x}{\sqrt{x}+2}\)
1.
Giả sử điều trên là đúng ta có:
\( \left | \sqrt{5}-4 \right |-\sqrt{5}+\sqrt{20}=4\)
Ta có: \(4>\sqrt{5}\)
\(\Rightarrow 4-\sqrt{5}- \sqrt{5}+\sqrt{20}=4\)
\(\Leftrightarrow 4-\sqrt{20}+\sqrt{20}=4\)
\(\Rightarrow đpcm\)
2.
\(P=\dfrac{x}{\sqrt{x}+2}\)
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Vậy đẳng thức được chứng minh.
2.
ĐKXĐ: và .
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Vậy đẳng thức được chứng minh.
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Vậy đẳng thức được chứng minh.
2.
ĐKXĐ: và .
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1.
=4-√5-√5+2√5
=4-2√5+2√5=4
2.=√x-2+√x+2/(√x-2) (√x+2) ×√x(√x+2) /2
=2√x×√x/2(√x+2)
=x/√x+2
Ta có :√(√5-4)2 - √5 + √20 = 4 - √5 - √5 + √20
=4 - 2√5 + 2√5
=4
Vậy √(√5-4)2 - √5 + √20 = 4 (dpcm)
\(\sqrt{\left(\sqrt{5}-4\right)^2}-\sqrt{5}+\sqrt{20}=\left|\sqrt{5}-4\right|-\sqrt{5}+2\sqrt{5}=4-\sqrt{5}-\sqrt{5}+2\sqrt{5}=4\) Vậy \(\sqrt{\left(\sqrt{5}-4\right)^2}-\sqrt{5}+\sqrt{20}=4\) 2.
\(P=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{x-2\sqrt{x}}\left(ĐK:x>0,x\ne4\right)\)
\(=[\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}]:\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}\)
\(=\dfrac{x}{\sqrt{x}+2}\)
Vậy P=\(\dfrac{x}{\sqrt{x}+2}\)
Ta có :
VT=\(\sqrt{\left[\sqrt{5}-4\right]^2}-\sqrt{5}+\sqrt{20}\)
=|\(\sqrt{5}-4\left|-\sqrt{5}\right|+2\sqrt{5}\)
=\(4-\sqrt{5}-\sqrt{5}+2\sqrt{5}\)
4
2.
Ta có P=(\(\dfrac{\sqrt{x}-2}{\left[\sqrt{x}+2\right]\left[\sqrt{x}-2\right]}+\dfrac{\sqrt{x}+2}{\left[\sqrt{x}-2\right]\left[\sqrt{x}+2\right]}\left(\right).\dfrac{x-2\sqrt{x}}{2}\)
=\(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}\)
\(\dfrac{x}{\sqrt{x}+2}\)
\(1.\sqrt{(\sqrt{5}-4)}^2-\sqrt{5}+\sqrt{20}=4\) \(\Leftrightarrow|\sqrt{5}-4|-\sqrt{5}+2\sqrt{5}=4\)\(\Leftrightarrow4-\sqrt{5}-\sqrt{5}+2\sqrt{5}=4(đúng)\)
Vậy \(\sqrt{(\sqrt{5}-4)}^2-\sqrt{5}+\sqrt{20}=4(đpcm)\)
2.\(P=[\dfrac{1(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}+\dfrac{1(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}]\div\dfrac{2}{\sqrt{x}(\sqrt{x}-2)}\)
\(P=[\dfrac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}]\times\dfrac{\sqrt{x}(\sqrt{x}-2)}{2}\)\(=\dfrac{x}{\sqrt{x}+2}\)
Vậy\(P=\dfrac{x}{\sqrt{x}+2}\)
Câu 1
1.
=\(\left|\sqrt{5}-4\right|-\sqrt{5}+2\sqrt{5}\)
=\(4-\sqrt{5}-\sqrt{5}+2\sqrt{5}\)
=4
\(\Rightarrow\)đẳng thức trên đúng
2
P=\(\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{x-2\sqrt{x}}\)(ĐK : x>0,x\(\ne\)4)
=\(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{2}\)
=\(\dfrac{x}{\sqrt{x}+2}\)
1.
\(\sqrt{(\sqrt{5}-4)^2}\) \(-\)\(\sqrt{5}\)\(+\)\(\sqrt{20}\)
\(=\) \(\left|\sqrt{5}-4\right|\)\(-\)\(\sqrt{5}\)\(+\)\(2\sqrt{5}\)
\(=\) \(4-\sqrt{5}-\sqrt{5}+2\sqrt{5}\)
\(=4\)
2.
\(P=(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2})\)\(:\dfrac{2}{x-2\sqrt{x}}\) \((x>0,\) \(x\ne4)\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}\begin{matrix}. &\dfrac{\sqrt{x}(\sqrt{x}-2)}{2}&\end{matrix}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}. \dfrac{\sqrt{x}}{2}\)
\(=\dfrac{x}{\sqrt{x}+2}\)
1. Ta có \(\sqrt{\left(\sqrt{5}-4\right)^2}\) -\(\sqrt{5}+\sqrt{20}=\) 4 - \(\sqrt{5}\) - \(\sqrt{5}+2\sqrt{5}\) = 4
2. Với x>0 , x≠ 4 ta có
P= \(\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\): \(\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\times\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}\)
= \(\dfrac{x}{\sqrt{x}+2}\)
ý 1: căn (căn 5 -4)2 -căn 5 + căn 5 +căn 20 =4
(=) 4-căn 5-căn 5 +2 căn 5=4=Vp
ý 2 x trên căn x +2
Bài 1:
\(VT=\sqrt{\left(\sqrt{5}-4\right)^2}-\sqrt{5}+\sqrt{20}=|\sqrt{5}-4|-\sqrt{5}+2\sqrt{5}=4-\sqrt{5}-\sqrt{5}+2\sqrt{5}=4=VP\)
Bài 2:
\(ĐKXĐ:x>0,x\ne4\)
\(P=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}=\dfrac{x}{\sqrt{x}+2}\)
Vậy\(P=\dfrac{x}{\sqrt{x}+2}\)
1 \(\sqrt{\left(\sqrt{5}-4\right)^2}-\sqrt{5}+\sqrt{20}\) = \(4-\sqrt{5}-\sqrt{5}+2\sqrt{5}\)=4
2 \(\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{x-2\sqrt{x}}\)
=\(\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}\)
=\(\dfrac{x}{\sqrt{x}+2}\)
