Giải các phương trình sau:
a) \(\frac{x-13}{2006}+\frac{x-22}{1997}+\frac{x-31}{1998}=3\)
b) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
ĐẠI SỐ
1. Giải các phương trình sau :
a) \(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
b) \(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
c) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
2. Giải các bất phương trình sau :
a) \(5+\frac{x+4}{5}< x-\frac{x-2}{2}+\frac{x+3}{3}\)
b) \(x+1-\frac{x-1}{3}< \frac{2x+3}{2}+\frac{x}{3}+5\)
c) \(\frac{\left(3x-2\right)^2}{3}-\frac{\left(2x+1\right)^2}{3}\le x\left(x+1\right)\)
d) \(\frac{2x+3}{4}-\frac{x+1}{3}\ge\frac{1}{2}-\frac{3-x}{5}\)
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
Bài 1:
b) Phương trình đã cho tương đương với phương trình:
\(\frac{8\left(x+22\right)-55\left(7x+149\right)-6\left(x+12\right)}{45}=\frac{9\left(x+35\right)+2\left(x+50\right)}{45}\)
\(\Leftrightarrow44x=-1056\)
\(\Leftrightarrow x=-24\)
Vậy x=-24 là nghiệm của phương trình
c) Phương trình đã cho tương đương với phương trình:
\(\frac{3x+6}{70}-\frac{x+4}{24}=\frac{32x+19}{60}+\frac{2}{3}\)
\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)=14\left(32x+19\right)+560\)
\(\Leftrightarrow-447x=894\)
\(\Leftrightarrow x=-2\)
Vậy x=-2 là nghiệm của phương trình
Giải các phương trình sau:
a) \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2003}=\frac{x-12}{2003}+2\)
b) \(\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)
ai hok biết, giải ra giùm
a) \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2002}=\frac{x-12}{2003}+2\)
\(\Rightarrow\left(\frac{x-15}{2000}-1\right)+\left(\frac{x-14}{2001}-1\right)+\left(\frac{x-13}{2002}-1\right)=\left(\frac{x-12}{2003}-1\right)\)
\(\Leftrightarrow\frac{x-2015}{2000}+\frac{x-2015}{2001}+\frac{x-2015}{2002}=\frac{x-2015}{2003}\)
\(\Leftrightarrow\frac{x-2015}{2000}+\frac{x-2015}{2001}+\frac{x-2015}{2002}-\frac{x-2015}{2003}=0\)
\(\Leftrightarrow\left(x-2015\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Rightarrow x-2015=0\)( vì \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}\ne0\))
\(\Leftrightarrow x=2015\)
Vậy tập hợp nghiệm \(S=\left\{2015\right\}\)
Giải phương trình:
1. \(\left(x^2+x\right)^2+4\left(x^2+x\right)^2=12\)
2. \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
câu 2 :
\(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
a, \(\left(\frac{x+2}{98}+18\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)
b, \(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)
a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)
\(\Rightarrow x=-100\)
b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)
\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)
\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)
b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)
=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0
\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0
\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))
\(\Leftrightarrow\)x=-1999
Vậy x=-1999
B1 :Giải phương trình
a,\(\frac{3\left(x-3\right)}{4}-1=\frac{2x+3\left(x+1\right)}{6}-\frac{7+12x}{12}\)
b,\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
c,\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
d,I7-xI-5x=1
B2:Giải bất phương trình
a,\(\left(x-2\right)\left(x+2\right)\ge x\left(x-4\right)\)
b,\(\frac{x-1}{4}-1\ge\frac{x+1}{3}+8\)
Giải các phương trình:
a) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
b) \(\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\frac{1}{x+2}=\frac{12}{8+x^3}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)
a)\(\frac{1}{x-1}\)-\(\frac{3x2}{x3-1}\)=\(\frac{2x}{x2+x+1}\)
<=> \(\frac{1}{x-1}\)-\(\frac{3x2}{\left(x-1\right)\left(x2+x+1\right)}\)=\(\frac{2x}{x2+x+1}\) ĐKXĐ: x khác 1
<=> x2+x+1 - 3x2 = 2x(x-1)
<=>x2+x+1 - 3x2 = 2x2-2x
<=>x2-3x-1=0( đoạn này làm nhanh nhé)
<=>x2-2*\(\frac{3}{2}\)x +\(\frac{9}{4}\)-\(\frac{9}{4}\)-1=0
<=>(x-\(\frac{3}{2}\))2-\(\frac{13}{4}\)=0
<=>(x-\(\frac{3-\sqrt{13}}{2}\))(x-\(\frac{3+\sqrt{13}}{2}\))=0
\(\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}\)
b) pt... đkxđ x khác 1;2;3
<=> 3(x-3) +2(x-2)=x-1
<=> 3x-9 +2x-4 = x-1
<=> 4x= 12
<=> x=3 ( ko thỏa đk)
vậy pt vô nghiệm
c) 1+\(\frac{1}{x+2}\)=\(\frac{12}{\left(x+2\right)\left(x2+2x+4\right)}\)đkxđ : x khác -2
<=> x3+8 + x2+2x+4 = 12
<=> x3+x2+2x=0
<=> x2+x+2=0( chia cả 2 vế cho x)
pt này chắc chắn vô nghiệm nhé bạn
Giải phương trình :
a) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
b) \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
c) \(6x^4-5x^3-38x^2-5x+6=0\)
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
Forever Miss You nếu (x-2009)=0
thì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\text{ko}?\)
nếu làm cách đó xét 2 trường hợp :")
A) \(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
B)\(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)
Giải các phương trình trên :
ĐS: a) x=8 b) x= -9
A) Ta có: \(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
\(\Leftrightarrow4\left(x-2\right)\left(x+10\right)-\left(x+4\right)\left(x+10\right)=3\left(x-2\right)\left(x+4\right)\)
\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)
\(\Leftrightarrow4x^2+32x-80-x^2-14x-40=3x^2+6x-24\)
\(\Leftrightarrow4x^2-x^2-3x^2+32x-14x-6x=-24+80+40\)
\(\Leftrightarrow12x=96\)
\(\Leftrightarrow x=8\)
Vậy x = 8
B) Ta có: \(\frac{\left(x+2\right)^2}{8}-2\left(2x+1\right)=25+\frac{\left(x-2\right)^2}{8}\)
\(\Leftrightarrow\left(x+2\right)^2-2.8\left(2x+1\right)=25.8+\left(x-2\right)^2\)
\(\Leftrightarrow x^2+4x+4-32x-16=200+x^2-4x+4\)
\(\Leftrightarrow x^2-x^2+4x-32x+4x=200+4-4+16\)
\(\Leftrightarrow-24x=216\)
\(\Leftrightarrow x=-9\)
Vậy x = -9
Giải các phương trình:
a) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
b) \(\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\frac{1}{x+2}=\frac{12}{8+x^3}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(=>x^2+x+1-3x^2=2x\left(x-1\right)\)
\(=>-2x^2+x+1=2x^2-2x\)
\(=>-4x^2+3x+1=0\)
\(=>\left(x-1\right)\left(x+\frac{1}{4}\right)=0\)'
\(=>\orbr{\begin{cases}x-1=0\\x+\frac{1}{4}\end{cases}=>\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)