K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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 \(x=1;x=3\)

29 tháng 5 2018

tai sao cho xcan bac hai lai bang 2

29 tháng 7 2017

gõ lại đề 

28 tháng 1 2020

\(ĐKXĐ:x\ne-1;x\ne2\)

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5x+5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow x-2-5x-5=15\)

\(\Leftrightarrow-4x=22\Leftrightarrow x=\frac{-11}{2}\)

Vậy \(S=\left\{\frac{-11}{2}\right\}\)

28 tháng 1 2020

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\left(ĐKXĐ:x\ne-1;x\ne2\right)\)

\(\Leftrightarrow\frac{1\left(x-2\right)-5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{-4x-7}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow-4x-7=15\)

\(\Leftrightarrow-4x=22\)

\(\Leftrightarrow x=22:\left(-4\right)\)

\(\Leftrightarrow x=\frac{-22}{4}=\frac{-11}{2}\)

Vậy tập nghiệm \(S=\left\{\frac{-11}{2}\right\}\)

9 tháng 8 2015

\(-2=\frac{2}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{2}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{2}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{2}{\left(x^2+2\right)\left(x^2+1\right)}\)

<=>\(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)

<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+2}+\frac{1}{x^2+2}-\frac{1}{x^2+3}+...+\frac{1}{x^2+4}-\frac{1}{x^2+5}=-1\)

<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+5}=-1\)

<=>(x2+5)-(x2+1)=-(x2+1)(x2+5)

<=>4=-x4-6x2-5

<=>x4+6x2+9=0

<=>(x2+3)2=0

<=>x2+3=0

Do x2>0

=>x2+3>0 nên PT vô nghiệm

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

21 tháng 7 2019

\(=>\frac{3}{4}\left[\left(x^2+1\right)^2+4\left(x^2+1\right)+4\right]-12=0\)

\(=>\frac{3}{4}\left(x^2+1+2\right)^2-12=0\)

\(=>\left(x^2+3\right)^2=16\)

Đến đây tự tìm nha 

 Hok tốt 

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 =))

21 tháng 7 2019

\(\frac{\left(x-2\right)^2}{3}-\frac{2x-1}{4}=4-\frac{\left(2x-3\right)^2}{6}.\)

\(\Rightarrow\frac{4\left(x-2\right)^2}{12}-\frac{3\left(2x-1\right)^2}{12}=\frac{48}{12}-\frac{2\left(2x-3\right)^2}{12}\)

\(\Rightarrow4\left(x^2-4x+4\right)-3\left(4x^2-4x+1\right)=48-2\left(4x^2-12x+9\right)\)

\(\Rightarrow4x^2-16x+16-12x^2+12x-3=48-8x^2+24x-18\)

\(\Rightarrow-16x+12x+16-3=24x+48-18\)

\(\Rightarrow28x=-17\Leftrightarrow x=-\frac{17}{28}\)

21 tháng 7 2019

nhung sao lai binh phuong len vay

-------------------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

12 tháng 6 2016

điều kiền x # 0

đặt \(t=x+\frac{1}{x};đk:t\ge2\)=>\(x^2+\frac{1}{x^2}=t^2-2\)

Ta được phương trình mới ẩn t :  \(t^2-2t-5=0\)

tự giải phương trình nhé. lấy nghiệm t>= 2