x+1/98+x+2/97+x+90/9+x+84/15>-4
giải bất phương trình
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Lời giải:
ĐKXĐ: $x\geq 1$
Đặt $\sqrt{x+1}=a; \sqrt{x-1}=b$ (ĐK: $a,b\geq 0$)
PT đã cho trở thành:
$\frac{a^2+b^2}{2}+ab=a+b+4$
$\Leftrightarrow a^2+b^2+2ab=2(a+b)+8$
$\Leftrightarrow (a+b)^2-2(a+b)-8=0$
$\Leftrightarrow (a+b-4)(a+b+2)=0$
Với $a\geq 0; b\geq 0$ thì $a+b+2\geq 2>0$
$\Rightarrow a+b-4=0$
$\Leftrightarrow a+b=4$
$\Leftrightarrow \sqrt{x+1}+\sqrt{x-1}=4$
$\Leftrightarrow \sqrt{x+1}=4-\sqrt{x-1}$
$\Rightarrow x+1=15+x-8\sqrt{x-1}$ (bp 2 vế)
$\Leftrightarrow 14=8\sqrt{x-1}$
$\Leftrightarrow x-1=(\frac{7}{4})^2=\frac{49}{16}$
$\Leftrightarrow x=\frac{65}{16}$ (tm)
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10) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/86 + 1/85 + 1/84 + 1/83 + 1/4 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy....
sửa đề đến đây thôi bạn nhé, do nếu thêm vào thì mình cũng ko biết có quy luật gì nữa :<
\(\dfrac{x-1}{99}-1+\dfrac{x-3}{97}-1+\dfrac{x-5}{95}-1=\dfrac{x-2}{98}-1+\dfrac{x-4}{96}-1\)
\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{97}+\dfrac{x-100}{95}=\dfrac{x-100}{98}+\dfrac{x-100}{96}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}\ne0\right)=0\Leftrightarrow x=100\)
\(\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)
\(\Leftrightarrow\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
=> x + 100 = 0
=> x = -100
Vậy x = -100
1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow-7x+3=-4x-15\)
\(\Leftrightarrow-7x+4x=-15-3\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=6\)( tmđk )
Vậy x = 6 là nghiệm của phương trình
2) 2x + 3 < 6 - ( 3 - 4x )
<=> 2x + 3 < 6 - 3 + 4x
<=> 2x - 4x < 6 - 3 - 3
<=> -2x < 0
<=> x > 0
Vậy nghiệm của bất phương trình là x > 0
(x+1)/99+(x+2)/98+(x+3)/97=(x+4)/96+(x+5)/95+(x+6)/94
[(x+1)/99 +1]+[(x+2)/98 +1]+[(x+3)/97 +1]-3=[(x+4)/96 +1]+[(x+5)/95 +1]+[(x+6)/94 +1]-3
[(x+1+99)/99+(x+2+98)/98+(x+3+97)/97]-3=[(x+4+96)/96+(x+5+95)/95+(x+6+94)/94]-3
(x+100)/99+(x+100)/98+(x+100)/97=(x+100)/96+(x+100)/95+(x+100)/94
(x+100)(1/99+1/98+1/97)=(x+100)(1/96+1/95+1/94)
(x+100)(1/99+1/98+1/97)-(x+100)(1/96+1/95+1/94)=0
(x+100)(1/99+1/98+1/97-1/96-1/95-1/94)=0
Ma : 1/99+1/98+1/97-1/96-1/95-1/94 \(\ne\)0
=>x+100=0
=>x=-100
k mk nha khong hieu noi mk nha.
1/3x-1/2=(3/5-4x)15/7
1/3x-1/2=9/7-60/7x
1/3x+60/7x=1/2+9/7
187/21x=25/14
x=75/374
k mk nha ban.
\(\dfrac{x+1}{98}+\dfrac{x+2}{97}+\dfrac{x+90}{9}+\dfrac{x+84}{15}>-4\\ \Leftrightarrow\left(\dfrac{x+1}{98}+1\right)+\left(\dfrac{x+2}{97}+1\right)+\left(\dfrac{x+90}{9}+1\right)+\left(\dfrac{x+84}{15}+1\right)>0\\ \Leftrightarrow\dfrac{x+99}{98}+\dfrac{x+99}{97}+\dfrac{x+99}{9}+\dfrac{x+99}{15}>0\\ \Leftrightarrow\left(x+99\right)\left(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}\right)>0\)
Vì \(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}>0\Rightarrow x+99>0\Rightarrow x>-99\)
cho tớ hỏi ông