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a, \(\Rightarrow\)\(1+\frac{x+3}{2011}\)\(+1+\frac{x+1}{2013}\)\(\ge1+\frac{x+10}{2004}+1+\frac{x+13}{2001}\)
\(\Rightarrow\)\(\frac{2011+x+3}{2011}+\frac{2013+x+1}{2013}\ge\frac{2004+x+10}{2004}+\frac{2001+x+13}{2001}\)
\(\Rightarrow\)\(\frac{2014+x}{2011}+\frac{2014+x}{2013}\ge\frac{2014+x}{2004}+\frac{2014+x}{2001}\)
\(\Rightarrow\)\(\frac{2014+x}{2011}+\frac{2014+x}{2013}-\frac{2014+x}{2004}+\frac{2014+x}{2001}\ge0\)
\(\Rightarrow\)\(\left(2014+x\right)\left(\frac{1}{2011}+\frac{1}{2013}-\frac{1}{2004}-\frac{1}{2001}\right)\)\(\ge0\)
\(do\)\(\frac{1}{2011}+\frac{1}{2013}-\frac{1}{2004}-\frac{1}{2001}< 0\)
\(\Rightarrow\)\(2014+x\le0\)
\(\Rightarrow\)\(x\le-2014\)
a) \(\dfrac{4\left(x-4\right)}{12}\)-\(\dfrac{3x}{12}\)-\(\dfrac{12}{12}\) = 0
\(\dfrac{4x-16-3x-12}{12}=0\)
\(\dfrac{x-28}{12}\)\(=0\)
x - 28 = 0
x = 28
Vậy x = 28
cách khác
Chứng minh điều ngược lại
Câu 2: phần (b)
\(A=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{\left(a+b\right)^2}{ab}\)
Để tồn tại A => a,b khác 0
\(\left(a+b\right)^2\ge0\forall a,b\in R\)
\(ab>0\) có một chút hy vọng \(A\ge4\)
ab< 0 tắt hét hy vọng \(A\ge4\)
Kết luận chẳng hy vọng gì
cần face với thực tế Quý Minh mình đã giải thích cho bạn thế nào là face đó
a)
\(4x-10< 0\\ 4x< 10\\ x< \dfrac{10}{4}=\dfrac{5}{2}\)
b)
\(2x+x+12\ge0\\ 3x\ge-12\\ x\ge-\dfrac{12}{3}=-4\)
c)
\(x-5\ge3-x\\ 2x\ge8\\ x\ge4\)
d)
\(7-3x>9-x\\ -2>2x\\ x< -1\)
đ)
\(2x-\left(3-5x\right)\le4\left(x+3\right)\\ 2x-3+5x\le4x+12\\ 3x\le15\\ x\le5\)
e)
\(3x-6+x< 9-x\\ 5x< 15\\ x< 3\)
f)
\(2t-3+5t\ge4t+12\\ 3t\ge15\\ t\ge5\)
g)
\(3y-2\le2y-3\\ y\le-1\)
h)
\(3-4x+24+6x\ge x+27+3x\\ 0\ge2x\\ 0\ge x\)
i)
\(5-\left(6-x\right)\le4\left(3-2x\right)\\ 5-6+x\le12-8x\\ \\ 9x\le13\\ x\le\dfrac{13}{9}\)
k)
\(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\\ 10x-15-20x+28\ge19-2x-22\\ 13-10x\ge-2x-3\\ -8x\ge-16\\ x\le\dfrac{-16}{-8}=2\)
l)
\(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\\ \dfrac{40x-100}{60}-\dfrac{90x-30}{2}< \dfrac{36-12x}{60}-\dfrac{30x-15}{60}\\ \Rightarrow40x-100-90x+30< 36-12x-30x+15\\ 130-50x< 51-42x\\ 92x< -79\\ x< -\dfrac{79}{92}\)
m)
\(5x-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+x\\ \dfrac{10x}{2}-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+\dfrac{2x}{2}\\ \Rightarrow10x-3+2x>7x-5+2x\\ 12x-3>9x-5\\ 3x>-2\\ x>-\dfrac{2}{3}\)
n)
\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\\ \dfrac{28x-8}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3x-6}{12}\\ \Rightarrow28x-8-24x< 60-3x+6\\ 4x-8< -3x+66\\ 7x< 74\\ x< \dfrac{74}{7}\)
a) \(4x-10< 0\)
\(\Leftrightarrow4x< 10\)
\(\Leftrightarrow x< \dfrac{5}{2}\)
b) ???
c) \(x-5\ge3-x\)
\(\Leftrightarrow2x-5\ge3\)
\(\Leftrightarrow2x\ge8\)
\(\Leftrightarrow x\ge4\)
d) \(7-3x>9-x\)
\(\Leftrightarrow7-2x>9\)
\(\Leftrightarrow-2x>2\)
\(\Leftrightarrow x< -1\)
đ) ???
e) \(3x-6+x< 9-x\)
\(\Leftrightarrow4x-6< 9-x\)
\(\Leftrightarrow5x-6< 9\)
\(\Leftrightarrow5x< 15\)
\(\Leftrightarrow x< 3\)
f) ???
g) ???
h) \(3-4x+24+6x\ge x+27+3x\)
\(\Leftrightarrow2x+27\ge4x+27\)
\(\Leftrightarrow-2x\ge0\)
\(\Leftrightarrow x\le0\)
i) \(5-\left(6-x\right)\le4\left(3-2x\right)\)
\(\Leftrightarrow5-6+x\le12-8x\)
\(\Leftrightarrow x-1\le12-8x\)
\(\Leftrightarrow9x-1\le12\)
\(\Leftrightarrow9x\le13\)
\(\Leftrightarrow x\le\dfrac{13}{9}\)
k) \(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\)
\(\Leftrightarrow10x-15-20x+28\ge19-2x-22\)
\(\Leftrightarrow-10x+23\ge-3-2x\)
\(\Leftrightarrow-8x+13\ge-3\)
\(\Leftrightarrow-8x\ge-16\)
\(\Leftrightarrow x\ge2\)
l) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\)
\(\Leftrightarrow-\dfrac{5}{6}x-\dfrac{7}{6}< -\dfrac{7}{10}x+\dfrac{17}{20}\)
\(\Leftrightarrow-\dfrac{2}{15}x-\dfrac{7}{6}< \dfrac{17}{20}\)
\(\Leftrightarrow-\dfrac{2}{15}x< \dfrac{121}{60}\)
\(\Leftrightarrow x>-\dfrac{121}{8}\)
m, n) làm tương tự:
đáp án: m. \(x>-\dfrac{2}{3}\); n. \(x< \dfrac{74}{7}\)
a) 7x+4=3x+16\(\Leftrightarrow\)4x=12\(\Leftrightarrow\)x=3
b)(x+9)(3x-15)=0\(\Leftrightarrow\)x+9=0 hoặc 3x-15=0
\(\Rightarrow\)x\(\in\){-9;5}
c) |-5x|=2x+21
Nếu x\(\le\)0 thì -5x=2x+21\(\Leftrightarrow\)x=-3 (t/m)
Nếu x>0 thì -5x=-2x-21\(\Leftrightarrow\)x=7 (t/m)
Vậy x\(\in\){-3;7}
d) 3x-5>15-x\(\Leftrightarrow\)4x>20\(\Leftrightarrow\)x>5
e) \(\dfrac{x+1}{2001}+\dfrac{x+5}{2005}< \dfrac{x+9}{2009}+\dfrac{x+13}{2013}\)
\(\Leftrightarrow\dfrac{x+1}{2001}-1+\dfrac{x+5}{2005}-1< \dfrac{x+9}{2009}-1+\dfrac{x+13}{2013}-1\)
\(\Leftrightarrow\)\(\dfrac{x-2000}{2001}+\dfrac{x-2000}{2005}-\dfrac{x-2000}{2009}-\dfrac{x-2000}{2013}< 0\)
\(\Leftrightarrow\)(x-2000)(\(\dfrac{1}{2001}+\dfrac{1}{2005}-\dfrac{1}{2009}-\dfrac{1}{2013}\))<0
Vì \(\dfrac{1}{2001}+\dfrac{1}{2005}-\dfrac{1}{2009}-\dfrac{1}{2013}>0\) nên x-2000<0
\(\Leftrightarrow\)x<2000
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
a) \(\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}\)
\(\Leftrightarrow\dfrac{15-x}{2000}+1+\dfrac{14-x}{2001}+1=\dfrac{13-x}{2002}+1+\dfrac{12-x}{2003}+1\)
\(\Leftrightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}=\dfrac{2015-x}{2002}+\dfrac{2015-x}{2003}\)
\(\Rightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}-\dfrac{2015-x}{2002}-\dfrac{2015-x}{2003}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow2015-x=0\)
<=> x=2015
Vậy phương trình có nghiệm là x=2015
b) \(\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}\)
\(\Leftrightarrow\dfrac{x-5}{2010}-1+\dfrac{x-4}{2011}-1=\dfrac{x-2010}{5}-1+\dfrac{x-2011}{4}-1\)
\(\Leftrightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}=\dfrac{x-2015}{5}+\dfrac{x-2015}{4}\)
\(\Rightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}-\dfrac{x-2015}{5}-\dfrac{x-2015}{4}=0\)
\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x-2015=0\)
=> x=2015
Vậy phương trình có nghiệm x=2015
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
(x-5)(x-9)>0\(\Leftrightarrow\left\{{}\begin{matrix}x-5>0\Leftrightarrow x>5\\x-9>0\Leftrightarrow x>9\end{matrix}\right.\)
Vậy x>9 thì (x-5)(x-9)>0
có
\(\dfrac{x-5}{x-8}>2\\ < =>x-5>2\left(x-8\right)\\ < =>x-5>2x-16\\ < =>-x>-11\\ < =>x< 11\)
vậy nghiệm của bpt là x<11
a/
\(\dfrac{x+3}{2011}+\dfrac{x+1}{2013}\ge\dfrac{x+10}{2004}+\dfrac{x+13}{2001}\)
\(\Leftrightarrow\dfrac{x+2014-2011}{2011}+\dfrac{x+2014-2013}{2013}\ge\dfrac{x+2014-2004}{2004}+\dfrac{x+2014-2001}{2001}\)
\(\Leftrightarrow-1+\dfrac{x+2014}{2011}-1+\dfrac{x+2014}{2013}\ge-1+\dfrac{x+2014}{2004}-1+\dfrac{x+2014}{2001}\)
\(\Leftrightarrow\dfrac{x+2014}{2011}+\dfrac{x+2014}{2013}-2\ge\dfrac{x+2014}{2004}+\dfrac{x+2014}{2001}-2\)
\(\Leftrightarrow\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}\right)\ge\left(x+2014\right)\left(\dfrac{1}{2004}+\dfrac{1}{2001}\right)\)
\(\Leftrightarrow\dfrac{1}{2011}+\dfrac{1}{2013}>\dfrac{1}{2004}+\dfrac{1}{2001}\) hoặc \(\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}\right)\ge\left(x+2014\right)\left(\dfrac{1}{2004}+\dfrac{1}{2001}\right)\)
(với mọi x>0) \(\Leftrightarrow x=2014\)
a,(\(\dfrac{x+3}{2011}\)+1)+(\(\dfrac{x+1}{2013}\)+1)>=(\(\dfrac{x+10}{2004}\)+1)+(\(\dfrac{x+13}{2001}\)+1)
<=>\(\dfrac{x+2014}{2011}\)+\(\dfrac{x+2014}{2013}\)>=\(\dfrac{x+2014}{2004}\)+\(\dfrac{x+2014}{2001}\)
<=>(x+2014)(1/2011+1/2013)>=(x+2014)(1/2004+1/2001)
vì 1/2011+1/2013<x+2014+1/2001
=>x+2014< hoặc =0<=>x < hoặc =-2014
b,để (x-5)(x-9)>0=>x-5 và x-9 phải có gt cùng dấu
th1:x-5 và x-9>0 =>x>9
th2:x-5 và x-9<0 =>x<5
c,đkxđ: x-8 khác 0 =>x khác 8
x-5>2(x-8) <=> x-5>2x-16
<=>-x>-11
<=>x<11
dấu bằng xảy ra tại x=-2014.
giải sai rồi!!!
\(\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}-\dfrac{1}{2004}-\dfrac{1}{2001}\right)\ge0\)
chia hai vế với \(\dfrac{1}{2011}+\dfrac{1}{2013}-\dfrac{1}{2004}-\dfrac{1}{2001}\), ta được:
\(x+2014\ge0\\ \Leftrightarrow x\ge-2014\)
vậy tập nghiệm của phương trình là \(S=\left\{x|x\ge-2014\right\}\)
a)\(\dfrac{x+3}{2011}+\dfrac{x+1}{2013}\ge\dfrac{x+10}{2004}+\dfrac{x+13}{2001}\)
\(\Leftrightarrow\dfrac{x+2014}{2011}+\dfrac{x+2014}{2013}-2\ge\dfrac{x+2014}{2004}+\dfrac{x+2014}{2001}\)
\(\Leftrightarrow\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}-\dfrac{1}{2004}-\dfrac{1}{2001}\right)\ge0\)
ta chia hai vế với \(\dfrac{1}{2011}+\dfrac{1}{2013}-\dfrac{1}{2004}-\dfrac{1}{2001}\). Khi đó:
\(x+2014\ge0\\ \Leftrightarrow x\ge-2014\)
vậy phương trình có tập nghiệm là \(S=\left\{x|x\ge-2014\right\}\)
c) \(\dfrac{x-5}{x-8}>2\left(ĐKXĐ:x\ne8\right)\)
\(\Leftrightarrow x-5>2x-16\\ \Leftrightarrow-x>-11\\ \Leftrightarrow x>11\left(TMĐK\right)\)
vậy bất phương trình có tập nghiệm là \(S=\left\{x|x>11\right\}\)
bất phương trình chứ không phải là phương trình nha bạn. Mình ghi nhầm!!!
b) (x-5)(x-9)>0
\(\Leftrightarrow\) \(\left[\begin{array}{} \begin{cases} x-5>0\\ x-9>0 \end{cases}\\ \begin{cases} x-5<0\\ x-9<0 \end{cases} \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} \begin{cases} x>5\\ x>9 \end{cases}\\ \begin{cases} x<5\\ x<9 \end{cases} \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} x>9\\ x<5 \end{array} \right.\)
Từ đó suy ra đpcm.
cái bài của bạn giải sai rồi
nếu bạn thử thế 10 vào thì nó cũng đúng
à quên đổi chiều -x > -11 tương đương với x<11
neu vậy nó sẽ ko thỏa mãn
cái của bạn chỉ đúng 1 nửa thôi
phải xét mẫu và có 2 TH
KQ sẽ là 8<x<11nghỉa là nằm trong khoảng 9,10
ko tin thì bạn cứ đi hỏi thầy
giải sai r
bạn giải sai mà