\(\frac{X-1}{x+2}>0\)A,\(\frac{X-2}{4}=\frac{-9}{2-x}\)và x>y
B,\(\frac{\left(x-1\right)^2}{6}=\frac{-8}{-12}=\frac{10}{7-y}=\frac{7z}{-21}\)
C,\(\frac{X}{15}=\frac{3}{y}\)và x<y<0
D,\(\frac{X}{4}=\frac{y}{3}\)và x+y=14
e,\(\frac{X+1}{x-2}< 0\)
G,
Những câu hỏi liên quan
tìm x,biết:a)frac{2}{left(x+2right)left(x+4right)}+frac{4}{left(x+4right)left(x+8right)}+frac{6}{left(x+8right)left(x+14right)}frac{x}{left(x+2right)left(x+14right)}b)frac{x+1}{10}+frac{x+1}{11}+frac{x+1}{12}frac{x+1}{13}+frac{x+1}{14}c)left(x+2right)^2frac{38}{25}+frac{9}{10}-frac{11}{15}+frac{13}{21}-frac{15}{28}+frac{17}{36}-...+frac{197}{4851}-frac{199}{4950}giúp tớ với,huhu
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tìm x,biết:
a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
giúp tớ với,huhu
Giải các phương trình sau:
a,frac{2x-1}{5}-frac{x-2}{3}frac{x+7}{15}
b,frac{x+3}{2}-frac{x-1}{3}frac{x+5}{6}+1
c,frac{2left(x+5right)}{3}+frac{x+12}{2}-frac{5left(x-2right)}{6}frac{x}{3}+11
d,frac{x-4}{5}+frac{3x-2}{10}-xfrac{2x-5}{3}-frac{7x+2}{6}
e,frac{2left(x-3right)}{7}+frac{x-5}{3}frac{13x+4}{21}
f,frac{3x-1}{2}-left(x-frac{1}{4}right)frac{4x-9}{8}
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Giải các phương trình sau:
a,\(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
b,\(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
c,\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
d,\(\frac{x-4}{5}+\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
e,\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
f,\(\frac{3x-1}{2}-\left(x-\frac{1}{4}\right)=\frac{4x-9}{8}\)
a, Ta có : \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)
=> \(3\left(2x-1\right)-5\left(x-2\right)=x+7\)
=> \(6x-3-5x+10-x-7=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
b, Ta có : \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
=> \(\frac{3\left(x+3\right)}{6}-\frac{2\left(x-1\right)}{6}=\frac{x+5}{6}+\frac{6}{6}\)
=> \(3\left(x+3\right)-2\left(x-1\right)=x+5+6\)
=> \(3x+9-2x+2-x-5-6=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
=> \(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
=> \(4\left(x+5\right)+3\left(x+12\right)-5\left(x-2\right)=2x+66\)
=> \(4x+20+3x+36-5x+10-2x-66=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
Giải phương trình:
1. frac{2x+3}{4}-frac{5x+3}{6}frac{3-4x}{12}
2. frac{3.left(2x+1right)}{4}-1frac{15x-1}{10}
3. frac{2x-1}{5}-frac{x-2}{3}frac{x+7}{15}
4. frac{x+3}{2}-frac{x-1}{3}frac{x+5}{6}+1
5. frac{x-4}{5}-frac{3x-2}{10}-xfrac{2x-5}{3}-frac{7x+2}{6}
6. frac{left(x+2right)left(x+10right)}{3}-frac{left(x+4right)left(x+10right)}{12}frac{left(x-2right)left(x+4right)}{4}
7. frac{left(x+2right)^2}{8}-2left(2x-1right)25+frac{left(x-2right)^2}{8}
8.frac{7x^2-14x-5}{5}frac{left(2x+1right)^...
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Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
bài 1: cho x, y thuộc Q. cmr:
|x + y| =< |x| + |y|
bài 2: tính:
\(A=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
bài 3: cho a + b + c = a^2 + b^2 + c^2 = 1 và x : y : z = a : b : c.
cmr: (x + y + z)^2 = x^2 + y^2 + z^2
1
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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
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Bài 3:
Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)
Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a2 + b2 + c2 = 1)
Vậy: (x + y + z)2 = x2 + y2 + z2
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hệ phương trình
1, left{{}begin{matrix}frac{1}{x+y}+frac{1}{x-y}frac{5}{8}frac{1}{x+y}-frac{1}{x-y}-frac{3}{8}end{matrix}right.
2, left{{}begin{matrix}frac{4}{2x-3y}+frac{5}{3x+y}2frac{3}{3x+y}-frac{5}{2x-3y}21end{matrix}right.
3, left{{}begin{matrix}frac{7}{x-y+2}+frac{5}{x+y-1}frac{9}{2}frac{3}{x-y+2}+frac{2}{x+y-1}4end{matrix}right.
4, left{{}begin{matrix}frac{3}{x}+frac{5}{y}-frac{3}{2}frac{5}{x}-frac{2}{y}frac{8}{3}end{matrix}right.
5 , left{{}begin{matrix}frac{2}{x+y-1}-frac{4}{x-y+1...
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hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
Tìm x biết
a) (8-5x).(x+2)+4.(x-2).(x+1)+2.(x-2).(x+2)=0
b)\(\left(-\frac{2}{5}+x\right):\frac{7}{9}+\left(-\frac{3}{5}+\frac{5}{6}\right):\frac{7}{9}=0\)
c)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2004}\)
a) left|2xfrac{1}{3}right|+frac{5}{6}1b)frac{17}{2}-left|2x-frac{3}{4}right|-frac{7}{4}c) frac{2}{3}x-frac{3}{2}xfrac{5}{12}d) frac{2}{5}+frac{3}{5}.left(3x-3,7right)-frac{53}{10}e) frac{7}{9}:left(2+frac{3}{4}xright)+frac{5}{9}frac{23}{27}f) left(2frac{4}{5}x-50right):frac{2}{3}51g)left(x+frac{1}{2}right).left(frac{2}{3}-2xright)0h)x.3frac{1}{4}+left(-frac{7}{6}right).x-1frac{2}{3}frac{5}{12}i)frac{6}{2}frac{-5+x}{15}k)frac{1}{21}+frac{1}{28}+frac{1}{36}+...+frac{2}{xleft(x+1right)}frac{2}{9}
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a) \(\left|2x\frac{1}{3}\right|+\frac{5}{6}=1\)
b)\(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
c) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
d) \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=-\frac{53}{10}\)
e) \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
f) \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
g)\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
h)\(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
i)\(\frac{6}{2}=\frac{-5+x}{15}\)
k)\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
Câu a \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
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g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
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i) \(\frac{6}{2}=\frac{-5+x}{15}\)
\(\Leftrightarrow3=\frac{x-5}{15}\)
\(\Leftrightarrow x-5=15.3\)
\(\Leftrightarrow x-5=45\)
\(\Leftrightarrow x=45+5\)
\(\Leftrightarrow x=50\)
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bài 1 : tìm x biếta, frac{2}{3}timesleft(x-frac{5}{6}right)+frac{1}{4}frac{22}{9}b, frac{2}{3}:frac{x}{5}frac{10}{21}c, frac{7}{3}:frac{x}{5}frac{14}{15}d, 1-left(5frac{4}{9}times x-7frac{7}{18}right):15frac{3}{4}0bài 2 : tính gtri bta,frac{8}{7}+frac{1}{5}timesfrac{10}{9}b, frac{3}{2}+left(frac{9}{2}+frac{2}{9}right)timesleft(frac{4}{3}-frac{5}{4}right)!_ove
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bài 1 : tìm x biết
a, \(\frac{2}{3}\times\left(x-\frac{5}{6}\right)+\frac{1}{4}=\frac{22}{9}\)
b, \(\frac{2}{3}:\frac{x}{5}=\frac{10}{21}\)
c, \(\frac{7}{3}:\frac{x}{5}=\frac{14}{15}\)
d, \(1-\left(5\frac{4}{9}\times x-7\frac{7}{18}\right):15\frac{3}{4}=0\)
bài 2 : tính gtri bt
a,\(\frac{8}{7}+\frac{1}{5}\times\frac{10}{9}\)
b, \(\frac{3}{2}+\left(\frac{9}{2}+\frac{2}{9}\right)\times\left(\frac{4}{3}-\frac{5}{4}\right)\)
!_ove
a) x = 99/20
b) x = 7
c) x = 2
( chỉ lm đc đến đó thui nk )
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Bài sau đây làm tôi không còn dám coi thường BĐT lớp 8:Cho x, y là các số thực thỏa mãn: xge2,x+yge3. Tìm Min:Ax^2+y^2+frac{1}{x}+frac{1}{x+y}Nghĩ mãi mới ra cách AM-GM (hơn 10 phút, mấy lần đầu nhóm sai!), rồi viết lại thành SOS nên 15 phút mới xong.. A-frac{35}{6}left(x-2right)^2left(1+frac{1}{4x}right)+left(y-1right)^2+frac{left(x+y-3right)^2}{9left(x+yright)}+left[frac{17}{9}left(x+yright)+frac{7}{4}x-frac{55}{6}right]Cách AM-GM:Aleft(x-2right)^2+left(y-1right)^2+frac{1}{x}+frac{1}{x+y}+4x+2...
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Bài sau đây làm tôi không còn dám coi thường BĐT lớp 8:
Cho x, y là các số thực thỏa mãn: \(x\ge2,x+y\ge3\). Tìm Min:
\(A=x^2+y^2+\frac{1}{x}+\frac{1}{x+y}\)
Nghĩ mãi mới ra cách AM-GM (hơn 10 phút, mấy lần đầu nhóm sai!), rồi viết lại thành SOS nên 15 phút mới xong..
\(A-\frac{35}{6}=\left(x-2\right)^2\left(1+\frac{1}{4x}\right)+\left(y-1\right)^2+\frac{\left(x+y-3\right)^2}{9\left(x+y\right)}+\left[\frac{17}{9}\left(x+y\right)+\frac{7}{4}x-\frac{55}{6}\right]\)
Cách AM-GM:
\(A=\left(x-2\right)^2+\left(y-1\right)^2+\frac{1}{x}+\frac{1}{x+y}+4x+2y-5\)
\(\ge\left(\frac{1}{x}+\frac{1}{4}x\right)+\left(\frac{1}{x+y}+\frac{15}{4}x+2y-5\right)\)
\(\ge1+\left[\frac{1}{9}\left(x+y\right)+\frac{1}{x+y}\right]+\frac{17}{9}\left(x+y\right)+\frac{7}{4}x-5\ge\frac{35}{6}\)
Đẳng thức xảy ra khi \(x=2;y=1\)