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c) Ta có: \(C=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
d)
Sửa đề: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
Ta có: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)
\(=\dfrac{x+4}{2x-8}\)
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
a) \(\sqrt{\left(2x-3\right)^2}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)
\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)
\(\Leftrightarrow5\sqrt{x+2}=20\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
a. \(\sqrt{\left(2x-3\right)^2}=7\)
<=> \(\left|2x-3\right|=7\)
<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\) ĐK: \(x\ge-2\)
<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)
<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)
<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)
<=> \(5\sqrt{x+2}=20\)
<=> \(\sqrt{x+2}=4\)
<=> \(\left(\sqrt{x+2}\right)^2=4^2\)
<=> \(\left|x+2\right|=16\)
<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)
c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\) ĐK: \(x\ge3\)
<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)
<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)
Bài 1 :
a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)
= \(2\left|a-3\right|+2\left|a+2\right|\)
\(=2.\left(-a+3\right)+2\left(-a-2\right)\)
b) có sai đề ko ?
c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)
a) Đề có lẽ là:
đk: \(x\ge0\)
\(\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}+2\right)x=x\sqrt{x}-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}+1+x\sqrt{x}+2x-x\sqrt{x}+\sqrt{x}-3=0\)
\(\Leftrightarrow3x+3\sqrt{x}-2=0\)
\(\Leftrightarrow3\left(x+\sqrt{x}+\frac{1}{4}\right)-\frac{11}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{11}{12}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{3+\sqrt{33}}{6}\right)\left(\sqrt{x}+\frac{3-\sqrt{33}}{6}\right)=0\)
Vì \(\sqrt{x}\ge0\left(\forall x\right)\)
=> \(\sqrt{x}=\frac{3-\sqrt{33}}{6}\Rightarrow x=\frac{7-\sqrt{33}}{6}\)
b) đk: \(x\ge1\)
Ta có: \(\sqrt{4\left(x^2-1\right)}-2\sqrt{15}=0\)
\(\Leftrightarrow\sqrt{x^2-1}=\sqrt{15}\)
\(\Leftrightarrow x^2-1=15\)
\(\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
Cô hướng dẫn nhé :)
a. ĐK: \(x>0;x\ne1\)
Ta có \(E=\frac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)+4\sqrt{x}\left(x-1\right)}{x-1}:\frac{x-1}{\sqrt{x}}\)
\(\Leftrightarrow E=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}=\frac{4x^2}{\left(x-1\right)^2}\)
b. Để \(E=2\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Leftrightarrow2x^2+4x-2=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}-1\\x=-\sqrt{2}-1\left(L\right)\end{cases}}\)
c. \(x=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=2\)
Vậy E = 16.
a)Rút gọn E ta đc:
\(\frac{4x^2+\sqrt{x}\left(2x+2\right)-4x}{x^2-2x+1}\)
b)Với E=2\(\Leftrightarrow\)\(\frac{4x^2+\sqrt{x}\left(2x+2\right)-4x}{x^2-2x+1}=2\)
\(\Leftrightarrow\frac{4x^2}{x^2-2x+1}+\frac{2\sqrt{x^3}}{x^2-2x+1}-\frac{4x}{x^2-2x+1}+\frac{2\sqrt{x}}{x^2-2x+1}-2=0\)
\(\Leftrightarrow\frac{2\left(x^2\sqrt{x^3}+\sqrt{x}-1\right)}{x^2-2x+1}=0\)
\(\Leftrightarrow x^2+\sqrt{x^3}+\sqrt{x}-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{-\sqrt{x^3}-\sqrt{x}+1}=0\left(tm\right)\\\sqrt{-\sqrt{x^3}-\sqrt{x}+1}+x=0\left(loai\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\sqrt{5}-3=0\left(loai\right)\\2x+\sqrt{5}-3=0\left(tm\right)\end{cases}}\)
\(\Leftrightarrow x=-\frac{\sqrt{5}-3}{2}\left(tm\right)\)
\(\sqrt{4\left(x+1\right)}=\sqrt{8}\)
⇒4(x+1)=8
⇒x+1=2
⇒x=1
a. \(\sqrt{4\left(x+1\right)}=\sqrt{8}\) ĐKXĐ: \(x\ge-1\)
<=> \(\left(\sqrt{4\left(x+1\right)}\right)^2=\left(\sqrt{8}\right)^2\)
<=> 4(x + 1) = 8
<=> 4x + 4 = 8
<=> 4x = -4
<=> x = -1 (TM)
Vậy nghiệm của PT là S = \(\left\{-1\right\}\)