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sao khó vậy,mình học lớp 9 mà tính mãi chẳng ra đáp án bài này từ lâu rùi
Bài 1 :
\(2+\sqrt{9}=2+3=5\)
Bài 2 :
Với \(x\ge0\)
\(B=\left(\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}+7}\right):\frac{5}{\sqrt{x}+7}\)
\(=\frac{\sqrt{x}+7-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}:\frac{5}{\sqrt{x}+7}\)
\(=\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+7\right)}.\frac{\sqrt{x}+7}{5}=\frac{1}{\sqrt{x}+2}\)
Bài 3 :
\(\hept{\begin{cases}x+2y=4\left(1\right)\\x-2y=0\left(2\right)\end{cases}}\)Lấy (1) - (2) ta được :
\(4y=4\Leftrightarrow y=1\)
Thay y = 1 vào (1) ta được : \(x+2=4\Leftrightarrow x=2\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
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 \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| \(\sqrt{x}-1\) | 1 | -1 | 2 | -2 |
| \(\sqrt{x}\) | 2 | 0 | 3 | -1 |
| x | 4 | 0 | 9 | vô lí |
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
a, Với \(x\ge0,x\ne4\)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
b, Ta có \(x=6+4\sqrt{2}=2^2+4\sqrt{2}+\left(\sqrt{2}\right)^2=\left(2+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2+\sqrt{2}\right|=2+\sqrt{2}\)do \(2+\sqrt{2}>0\)
\(\Rightarrow A=\frac{2+\sqrt{2}-4}{2+\sqrt{2}-2}=\frac{-2+\sqrt{2}}{\sqrt{2}}=\frac{-2\sqrt{2}+2}{2}=\frac{-2\left(\sqrt{2}-1\right)}{2}=1-\sqrt{2}\)
1, A = \(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
2 , A = \(1-\sqrt{2}\)
1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)
=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)
=\(\dfrac{-4}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}+1\right)^2}\)
=\(\dfrac{-4\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
=-4
2, A=\(\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{2}{x+\sqrt{x}-2}\)
=\(\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\left(\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
=\(\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\dfrac{-\sqrt{x}-2}{\sqrt{x}+1}\)
ĐKXĐ : \(y>-5\)
Đặt \(\left(x-2\right)^2=a>0\) và \(\frac{1}{\sqrt{y+5}=b}\)
Hệ phương trình đã cho trở thành : \(\hept{\begin{cases}2a+b=3\\a-2b=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}4a+2b=6\\a-2b=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}5a=5\\a-2b=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}a=1\\b=1\end{cases}}\)( Thỏa mãn )
\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^2=1\\\frac{1}{\sqrt{y+5}=1}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\\\sqrt{y+5}=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^2=1\\\frac{1}{\sqrt{y+5}=1}\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt{y+5}=1\\\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\end{cases}\Leftrightarrow}\hept{\begin{cases}y+5=1\\\orbr{\begin{cases}x=3\\x=1\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x=3\\y=-4\end{cases}}\\\hept{\begin{cases}x=1\\y=-4\end{cases}}\end{cases}}}\)
ĐKXĐ : y > -5
Đặt \(\hept{\begin{cases}\left(x-2\right)^2=a\\\frac{1}{\sqrt{y+5}}=b\end{cases}\left(a\ge0;b>0\right)}\)
Hpt đã cho trở thành \(\hept{\begin{cases}2a+b=3\\a-2b=-1\end{cases}}\)=> \(a=b=1\left(tm\right)\)
=> \(\hept{\begin{cases}\left(x-2\right)^2=1\\\frac{1}{\sqrt{y+5}}=1\end{cases}}\)<=> \(\hept{\begin{cases}x=3\\y=-4\end{cases}}or\hept{\begin{cases}x=1\\y=-4\end{cases}}\)(tm)
Vậy ...
a) \(P=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{4}{4-3}\)
\(=4\)
b) \(Q=\left(1+\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}vớix>0,x\ne4\)
\(=\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}\)
\(=\)\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2}{\sqrt{x}-2}\)
Ta có: \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\dfrac{8\sqrt{x}-8x+8x}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
ta có : \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
=\(\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{4-x}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-x\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
=\(\dfrac{8\sqrt{x}-4x+8x}{4-x}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{8\sqrt{x}+4x}{4-x}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x-2}\right)}\) =\(\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\) =\(\dfrac{4x\left(\sqrt{x}-2\right)}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)
=\(-\dfrac{4x\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\) =\(-\dfrac{4x}{3-\sqrt{x}}\) =\(\dfrac{4x}{\sqrt{x}-3}\)
này mới đúng !!
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.
\(P=\dfrac{x}{\sqrt{x}+2}\)
VT=√(√5−4)2−√5+√20=∣∣√5−4∣∣−√5+2√5=4−√5−√5+2√5=4=VPVT=(5−4)2−5+20=|5−4|−5+25=4−5−5+25=4=VP.
Vậy đẳng thức được chứng minh.
2.
ĐKXĐ: x>0x>0 và x≠4x≠4.
P=(1√x+2+1√x−2):
VT=(5−4)2−5+2=∣∣∣5−4∣∣∣−5+25=4−5−5+25=4=VP.
Vậy đẳng thức được chứng minh.
VT=√(√5−4)2−√5+√20=∣∣√5−4∣∣−√5+2√5=4−√5−√5+2√5=4=VPVT=(5−4)2−5+20=|5−4|−5+25=4−5−5+25=4=VP.
Vậy đẳng thức được chứng minh.
2.
ĐKXĐ: x>0x>0 và x≠4x≠4.
P=(1√x+2+1√x−2):
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\(P=\dfrac{x}{\sqrt{x}+2}\)
x >0P = \dfrac
1 \(\sqrt{\left(\sqrt{5}-4\right)^2}-\sqrt{5}+\sqrt{20}\) = \(4-\sqrt{5}-\sqrt{5}+2\sqrt{5}\)=4
2 \(\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{x-2\sqrt{x}}\)
=\(\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}\)
=\(\dfrac{x}{\sqrt{x}+2}\)
1.VT=√(√5−4)2−√5+√20=∣∣√5−4∣∣−√5+2√5=4−√5−√5+2√5=4=VPVT=(5−4)2−5+20=|5−4|−5+25=4−5−5+25=4=VP.
2.P=(1√x+2+1√x−2):2√x(√x