GIAI PHUONG TRINH:
\(2\left(\sqrt{\frac{x-1}{4}-3}\right)=2\sqrt{\frac{4x-4}{9}}-\frac{1}{3}\)
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16 tháng 6 2019
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\) (ĐK: \(x\ge\frac{-1}{2}\) )
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2x\left(x^2+1\right)+\left(x^2+1\right)\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow2x+1=\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+1\right)-\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2+1-1\right)=0\)
\(\Leftrightarrow x^2\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\end{cases}}\) (nhận)
Vậy .....
16 tháng 6 2019
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+2x+1\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)(1)
Vì VT > 0 nên VP >0
\(\Leftrightarrow\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\ge0\)
\(\Leftrightarrow x\ge-\frac{1}{2}\)
Khi đó \(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-x-\frac{3}{4}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x^2-x-\frac{3}{4}=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x-3=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\end{cases}}\)
Cần cù bù thông minh , phá tung pt dưới ra được cái phương trình bậc 5, sau đó dùng Wolfram|Alpha: Computational Intelligence để tính nghiệm rồi phân tích nhân tử =))
LP
3
28 tháng 5 2018
\(DKXD:x>0\)
\(PT\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)
\(\Leftrightarrow\frac{x+\frac{3}{x}-4}{\sqrt{x+\frac{3}{x}}+2}=\frac{x^2-4x-4+7}{2\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{x^2-4x+3}{2\left(x+1\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(\frac{1}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{1}{2\left(x+1\right)}\right)=0\)
\(\Rightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x\sqrt{x+\frac{3}{x}}=2\text{ }\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\left(\text{ }x-1\right)\left(x^2+x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy PT có 2 nghiệm \(x=1;x=3\)
27 tháng 7 2017
\(\frac{15}{2}\left(30x^2-4x\right)=2004\left(\sqrt{30060x+1}+1\right)\)
\(\Leftrightarrow5\left(15x^2-2x\right)=668\left(\sqrt{30060x+1}+1\right)\)
\(\Leftrightarrow75x^2-10x-1340008=668\left(\sqrt{30060x+1}-2005\right)\)
\(\Leftrightarrow\left(5x+668\right)\left(15x-2006\right)=\frac{1338672\left(15x-2006\right)}{\left(\sqrt{30060x+1}+2005\right)}\)
\(\Leftrightarrow\left(15x-2006\right)\left(5x+668-\frac{1338672}{\left(\sqrt{30060x+1}+2005\right)}\right)=0\)
Tới đây tự làm tiếp nhá.
27 tháng 7 2017
câu này chỉ cần đưa ề đối xúng là được thôi
\(\Leftrightarrow\left(15x\right)^2-30x=2004\sqrt{30060x+1}+2004\)
\(\Leftrightarrow\left(15x-1\right)^2=2004\sqrt{30060x+1}+2005\)
đặt \(\sqrt{30060x+1}=15y-1\)
\(\Leftrightarrow\hept{\begin{cases}\left(15x-1\right)^2=2004\left(15y-1\right)+2005\\\left(15y-1\right)^2=30060x+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(15x-1\right)^2=30060y+1\\\left(15y-1\right)^2=30060x+1\end{cases}}\)
đến đây thì lấy cái đầu trừ cái thứ 2 là ra
KS
2
21 tháng 7 2019
\(\frac{\left(x-2\right)^2}{3}-\frac{2x-1}{4}=4-\frac{\left(2x-3\right)^2}{6}.\)
\(\Rightarrow\frac{4\left(x-2\right)^2}{12}-\frac{3\left(2x-1\right)^2}{12}=\frac{48}{12}-\frac{2\left(2x-3\right)^2}{12}\)
\(\Rightarrow4\left(x^2-4x+4\right)-3\left(4x^2-4x+1\right)=48-2\left(4x^2-12x+9\right)\)
\(\Rightarrow4x^2-16x+16-12x^2+12x-3=48-8x^2+24x-18\)
\(\Rightarrow-16x+12x+16-3=24x+48-18\)
\(\Rightarrow28x=-17\Leftrightarrow x=-\frac{17}{28}\)
KS
1
21 tháng 7 2019
-------------------ko chép đề nha---------
\(\Leftrightarrow\frac{4\left(x^2-4x+4\right)-3\left(2x+1\right)}{12}=\frac{12-2\left(4x^2-12x+9\right)}{12}\)
\(\Rightarrow4x^2+16x+16-6x-3=12-8x^2+24x-18\)
\(\Leftrightarrow4x^2+10x+13=-8x^2+24x-6\)
\(\Leftrightarrow4x^2+8x^2+10x-24x+13+6=0\)
\(\Leftrightarrow12x-14x+19=0\)
Ta có :\(\Delta'=7^2-12.19=-179< 0\)
\(\Rightarrow\)phương trình vô nghiệm
KS
2
21 tháng 7 2019
\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)
<=> \(3\left(x^2+1\right)^2.4+3\left(x^2+x\right).4-9.4=0.4\)
<=> \(3\left(x^2+1\right)^2+12\left(x^2+x\right)-36=0\)
<=> \(3x^4+18x^2+12x-33=0\)
<=> \(3\left(x-1\right)\left(x^3+x^2+7x+11\right)=0\)
<=> \(x-1=0\)
<=> \(x=1\)
Mà vì: \(x^3+x^2+7x+11\ne0\)
=> x = 1
23 tháng 3 2019
\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)
\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)
\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{6\left(x-1\right)}{2\left(x^2+4x+3\right)}\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{3\left(x-1\right)}{x^2+4x+3}\)
\(\Leftrightarrow-x^2-3=3x-3\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\)
Vậy x = 0
23 tháng 3 2019
\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)
\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow-6x+3=8\)
\(\Leftrightarrow x=-\frac{5}{6}\)
Vậy ...
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