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Điều kiện: x khác (-3,-2,1,4)
PT <=>
\(1+\frac{2}{x-1}+1-\frac{4}{x+2}+1-\frac{6}{x+3}+1+\frac{8}{x-4}=4\)
<=> \(\frac{1}{x-1}-\frac{2}{x+2}-\frac{3}{x+3}+\frac{4}{x-4}=0\)
<=> (x+2)(x+3)(x-4)-2(x-1)(x+3)(x-4)-3(x-1)(x+2)(x-4)+4(x-1)(x+2)(x+3)=0
<=> (x3+x2-14x-24)-2(x3 - 2x2-11x+12) - 3(x3 - 3x2- 6x+8) + 4(x3+4x2 + x-6) = 0
<=> x3+x2-14x-24-2x3 + 4x2+22x-24 - 3x3 + 9x2+ 18x-24 + 4x3+16x2 + 4x-24 = 0
<=> 30x2 + 30x -96=0
<=> 5x2 + 5x -16 = 0
Giải ra được: \(\orbr{\begin{cases}x_1=\frac{-5-\sqrt{345}}{10}\\x_2=\frac{-5+\sqrt{345}}{10}\end{cases}}\)
\(\frac{x+1}{97}\) + \(\frac{x+1}{98}\) - \(\frac{x+1}{99}\) - \(\frac{x+1}{100}\) \(\Leftrightarrow\) (x+1).(1/97 + 1/98 - 1/99 - 1/100) . Vì (1/97 = 1/ 98 - 1/99 - 1/100) \(\ne\) 0 \(\Rightarrow\) x+ 1= 0 \(\Leftrightarrow\) x= -1
Theo đề bài ta có: \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}-\frac{x-4}{5}-\frac{x-5}{6}>0\)
=> \(\frac{x-1}{2}+1+\frac{x-2}{3}+1+\frac{x-3}{4}+1-\left(\frac{x-4}{5}+1\right)-\left(\frac{x-5}{6}+1\right)>1\)
<=> \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}>1\)
<=>\(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>1\)
<=> \(\left(x+1\right)\cdot\frac{43}{60}>1\)
<=>\(x+1>\frac{60}{43}\)
<=> x>\(\frac{17}{43}\)
Vậy x>17/43
b) đk: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
pt (1) \(\Leftrightarrow\left(x^2-2x\right)\left(x^2-2x+4\right)=0\Leftrightarrow x\left(x-2\right)\left(x^2-2x+4\right)=0\Leftrightarrow x=0\left(L\right),x=2\left(T\right)\)\(,x^2-2x+4=0\left(3\right)\)
pt(3) VÔ NGHIỆM vì \(\Delta'=1-4=-3< 0\)
Thay x=2 vào pt (2) ta được: \(\frac{1}{2}+\frac{1}{y-1}=\frac{3}{2}\Leftrightarrow\frac{1}{y-1}=1\Leftrightarrow y-1=1\Leftrightarrow x=2\left(tm\right)\)
Vậy nghiệm của hệ pt là(x;y)=(2;2)
\(S=\frac{-1+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{99}+\sqrt{100}}{100-99}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{99}+\sqrt{100}\)
\(=-1+\sqrt{100}\)
\(\hept{\begin{cases}a=\left(x^2-x+1\right)^2\\b=x^2\end{cases}}\)
\(a^2-\left(b+1\right)a+b=0\Leftrightarrow\left(a-1\right)\left(a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=b\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x^2-x+1\right)^2=1\\\left(x^2-x+1\right)^2=x^2\end{cases}}\)(easy)
b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)
\(\Leftrightarrow\left(\frac{4}{x}-x\right)+\frac{x-\frac{1}{x}-\frac{3}{2}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}=\frac{2x-\frac{5}{x}-\frac{3}{2}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\)
\(\Leftrightarrow\frac{-\left(x-2\right)\left(x+2\right)}{x}+\frac{\frac{\left(x-2\right)\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(x-2\right)\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{-\left(x+2\right)}{x}+\frac{\frac{\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\right)=0\)
Pt trong ngoặc VN suy ra x=2
a)\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)
\(\Leftrightarrow x^2+3\sqrt{x^2-1}-1=\sqrt{x^4-x^2+1}-1\)
\(\Leftrightarrow\frac{x^2\left(3\sqrt{x^2-1}+1\right)}{3\sqrt{x^2-1}+1}+\frac{9\left(x^2-1\right)-1}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2+1-1}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)
pt trong căn vô nghiệm
suy ra x=1; x=-1
\(\frac{1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}=\frac{1}{4}\)
\(\Leftrightarrow\frac{\left(x+1\right)2}{4\left(x+1\right)\left(x-1\right)}+\frac{3\cdot4}{4\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)\left(x-1\right)}{4\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow2\left(x+1\right)+12=x^2-1\)
\(\Leftrightarrow2x+2+12-x^2+1=0\)
\(2x-x^2+15=0\Leftrightarrow16-\left(x-1\right)^2=0\Leftrightarrow\left(4-x+1\right)\left(4+x-1\right)=0\Leftrightarrow\left(5-x\right)\left(3+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=0\\3+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
\(\frac{X+1}{99}+1+\frac{X+2}{98}+1+\frac{x+3}{97}+1+\frac{X+4}{96}+1=0\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{X+100}{98}+\frac{X+100}{97}+\frac{X+100}{96}=0\Leftrightarrow\left(X+100\right)\times\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0 \)\(\Leftrightarrow X+100=0\Leftrightarrow x=-100\)