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(\(\sqrt{1+x}+\sqrt{1-x}\))\(\left(2+2\sqrt{1-x^2}\right)=8\)(1)(đk: \(-1\le x\le1\))
đặt \(\sqrt{1+x}+\sqrt{1-x}\) =a (\(a\ge0\)
=> \(a^2=2+2\sqrt{1-x^2}\)
khi đó
(1)\(\Leftrightarrow a^3=8\Leftrightarrow a=\sqrt{8}=2\) (tm)
=>\(\sqrt{1+x}+\sqrt{1-x}\) =2
\(\Leftrightarrow2+2\sqrt{1-x^2}=4\)
\(\Leftrightarrow2\sqrt{1-x^2}=2\)
\(\Leftrightarrow\sqrt{1-x^2}=1\Leftrightarrow1-x^2=1\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)(tm)
vậy x=0 là nghiệm của phương trình
\(A=\left(\frac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\frac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right).\sqrt{3}\)
\(=\left(\frac{1+\sqrt{3}-1+\sqrt{3}}{-2}\right).\sqrt{3}=-3\)
\(B=\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để \(A=\frac{B}{6}\Leftrightarrow B=6A\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}=-18\)
\(\Rightarrow\sqrt{x}-1=-18\sqrt{x}\Rightarrow\sqrt{x}=\frac{1}{19}\Rightarrow x=\frac{1}{361}\)
c/ \(C'=\frac{1}{\frac{1}{3-2\sqrt{x}}}.\frac{1}{\frac{1}{\sqrt{3-2\sqrt{x}}}+1}=\frac{\sqrt{\left(3-2\sqrt{x}\right)^3}}{1+\sqrt{\left(3-2\sqrt{x}\right)}}\)
Đặt \(\sqrt{\left(3-2\sqrt{x}\right)}=a\)
\(\Rightarrow C'=\frac{a^3}{a+1}=a^2-a+1-\frac{1}{a+1}\)
Đế C' nguyên thì a + 1 là ước của 1
\(\Rightarrow a=0\)
\(\Rightarrow\sqrt{\left(3-2\sqrt{x}\right)}=0\)
\(\Rightarrow x=\frac{9}{4}\left(l\right)\)
Vậy không có x.
Không biết có nhầm chỗ nào không nữa. Lam biếng kiểm tra lại quá. You kiểm tra lại hộ nhé. Thanks
a/ \(C=\left(\frac{2\sqrt{x}}{2x-5\sqrt{x}+3}-\frac{5}{2\sqrt{x}-3}\right):\left(3+\frac{2}{1-\sqrt{x}}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\frac{5}{2\sqrt{x}-3}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-1}\right)\)
\(=\left(\frac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-1}\right)\)
\(=\frac{5-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}.\frac{\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\frac{1}{3-2\sqrt{x}}\)
Câu b, c tự làm nhé
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\frac{-x+x\sqrt{x}+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x-\sqrt{x}-x+x\sqrt{x}+6-x-\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x\sqrt{x}-x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{\left(x-4\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\sqrt{x}-2\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(Q=\frac{\left(x+27\right)P}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\frac{\left(x+27\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\frac{x+27}{\sqrt{x}+3}\)
\(Q=\frac{x+27}{\sqrt{x}+3}\ge6\\ \Leftrightarrow\frac{x+27}{\sqrt{x}+3}-6\ge0\\ \Leftrightarrow\frac{x+27-6\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\ge0\\ \Leftrightarrow\frac{x-6\sqrt{x}+45}{\sqrt{x}+3}\ge0\)
Dễ thấy \(x-6\sqrt{x}+45=\left(\sqrt{x}-3\right)^2+36\ge36>0\forall x\ge0\)
\(\sqrt{x}+3\ge3>0\forall x\ge0\)
=> Ko có giá trị nào của x thỏa mãn yêu cầu
P/s: Nếu đề là \(x\sqrt{x}+27\)thì sẽ khác một chút :v
Bạn ơi chỗ kia phải là \(\frac{x-6\sqrt{x}+9}{\sqrt{x}+3}\)
A=\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right)\):\(\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)Đk x>0 x#0 x#1
=\(\frac{x-1}{\sqrt{x}\left(\sqrt{x-1}\right)}\):\(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{\sqrt{x}+1}{\left(\sqrt{x-1}\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}.\sqrt{x}-1\)
=\(\frac{x-1}{\sqrt{x}}\)
Ta có 3+\(2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)(thay và A ta dc
=>\(\frac{3+2\sqrt{2}-1}{\sqrt{2}+1}\)
= \(\frac{2\sqrt{2}+2}{\sqrt{2}+1}\)
=2
mk nhầm....\(\frac{x-1}{\sqrt{x}}>0\)=> \(x-1>0\Rightarrow x>1\)
mk làm r nhé
Ta có :
\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\frac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right).\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}+2\right)\)
\(=\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+2\right)\)
\(=\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=15+12+4\sqrt{45}+12\sqrt{5}=27+24\sqrt{5}\)
\(B=\left(2\sqrt{3}+6\sqrt{3}\right).\frac{\sqrt{3}}{2}-5\sqrt{6}=\frac{8\sqrt{3}.\sqrt{3}}{2}-5\sqrt{6}=12-5\sqrt{6}\)
\(C=4\sqrt{3}+\frac{4}{\sqrt{3}}+10\sqrt{5}-\frac{10}{\sqrt{5}}=\frac{16}{\sqrt{3}}+8\sqrt{5}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ge0\\6-x\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge1\\x\le6\end{matrix}\right.\)
=> \(1\le x\le6\)
Vậy phương trình trên có ĐKXĐ là \(1\le x\le6\) .
- Ta có :\(\sqrt{x-1}+\sqrt{6-x}=3\)
<=> \(\left(\sqrt{x-1}+\sqrt{6-x}\right)^2=3^2\)
<=> \(\left(x-1\right)+2\sqrt{x-1}\sqrt{6-x}+\left(6-x\right)=9\)
<=> \(x-1+2\sqrt{x-1}\sqrt{6-x}+6-x=9\)
<=> \(x-1+6-x-9=-2\sqrt{x-1}\sqrt{6-x}\)
<=> \(-4=-2\sqrt{x-1}\sqrt{6-x}\)
<=> \(4=2\sqrt{x-1}\sqrt{6-x}\)
<=> \(4^2=\left(2\sqrt{x-1}\sqrt{6-x}\right)^2\)
<=> \(16=4\left(x-1\right)\left(6-x\right)\)
<=> \(4=\left(x-1\right)\left(6-x\right)\)
<=> \(4=6x-6-x^2+x\)
<=> \(4+6=7x-x^2=10\)
<=> \(-x^2+7x-10=0\)
<=> \(x^2-7x+10=0\)
<=> \(\left(x^2-2.3,5.x+3,5^2\right)-2,25=0\)
<=> \(\left(x-3,5\right)^2-1,5^2=0\)
<=> \(\left(x-3,5-1,5\right)\left(x-3,5+1,5\right)=0\)
<=> \(\left(x-5\right)\left(x-2\right)=0\)
<=> \(\left\{{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\) ( TM )
Vậy phương trình trên có nghiệm là \(x=\left\{2;5\right\}\)