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\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
1.
= -(13 + 3 căn7 ) / 2 + -(7 + 3 căn7 ) / 2
= -7 + 3 căn7
f) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)
Lũy thừa 6 cả 2 vế lên PT tương đương:
\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)
Cái ngoặc to vô nghiệm vì nó tương đương:
\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))
Vậy x = 3.
PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra
@Akai Haruma, @Nguyễn Việt Lâm
giúp em vs ạ! Cần gấp ạ
em cảm ơn nhiều!
Bài 2:
Ta có: \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
\(=\frac{\sqrt{\sqrt{5}-1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}-\left(\sqrt{2}-1\right)\)
\(=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\frac{4}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\sqrt{2}-\sqrt{2}+1\)
=1
câu 3: C = \(\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)}{\left(\text{4+\sqrt{15}}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-\sqrt{15}}}\)
\(=\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}}{\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)
=\(\frac{\sqrt{9-\left(\sqrt{5}\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}}{\sqrt{16-\left(\sqrt{15}\right)^2}.\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{15}}}\)
\(=\frac{2\left(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\right)}{\sqrt{40+10\sqrt{15}}-\sqrt{24-6\sqrt{15}}}\)
\(=2.\frac{\left(\sqrt{5}+5\right)-\left(\sqrt{5}+1\right)}{\left(\sqrt{15}+5\right)-\left(\sqrt{15}+3\right)}\)
= 4