\(\dfrac{7}{1}\)= \(\dfrac{7x.....}{2x.....}\) = \(\dfrac{.....}{10}\)
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Những câu hỏi liên quan
1) giải phương trình :
a) left(2+3right)left(dfrac{3x+8}{2-7x}+1right)left(x-5right)left(dfrac{3x+8}{2-7x}+1right)
b) dfrac{7x+10}{x+1}left(x^2-x-2right)-dfrac{7x+10}{x+1}left(2x^2-3x-5right)0
c) dfrac{2x+5}{x+3}+1dfrac{4}{x^2+2x-3}-dfrac{3x-1}{1-x}
d) dfrac{13}{2x^2+x-21}+dfrac{1}{2x+7}+dfrac{6}{9-x^2}0
i) dfrac{x-49}{50}+dfrac{x-50}{49}dfrac{49}{x-50}+dfrac{50}{x-49}
k) dfrac{1+dfrac{x}{x+3}}{1-dfrac{x}{x+3}}3
Đọc tiếp
1) giải phương trình :
a) \(\left(2+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
b) \(\dfrac{7x+10}{x+1}\left(x^2-x-2\right)-\dfrac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
d) \(\dfrac{13}{2x^2+x-21}+\dfrac{1}{2x+7}+\dfrac{6}{9-x^2}=0\)
i) \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
k) \(\dfrac{1+\dfrac{x}{x+3}}{1-\dfrac{x}{x+3}}=3\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
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1) giải pt :
a) \(\dfrac{7x+10}{x+1}\left(x^2-x-2\right)-\dfrac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
b) \(\dfrac{13}{2x^2+x-21}+\dfrac{1}{2x+7}+\dfrac{6}{9-x^2}=0\)
c) \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
d) \(\dfrac{1+\dfrac{x}{x+3}}{1-\dfrac{x}{x+3}}=3\)
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
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1. dfrac{5left(x-1right)+2}{6}-dfrac{7x-1}{4}dfrac{2left(2x+1right)}{7}-5
2. x-dfrac{3left(x+30right)}{15}-24dfrac{1}{2}dfrac{7x}{10}-dfrac{2left(10x+2right)}{5}
3. 14dfrac{1}{2}-dfrac{2left(x+3right)}{5}dfrac{3x}{2}-dfrac{2left(x-7right)}{3}
4. dfrac{x+1}{3}+dfrac{3left(2x+1right)}{4}dfrac{2x+3left(x+1right)}{6}+dfrac{7+12x}{12}
5. dfrac{3left(2x-1right)}{4}-dfrac{3x+1}{10}+1dfrac{2left(3x+2right)}{5}
6. x-dfrac{3}{17}left(2x-1right)dfrac{7}{34}left(1-2xright)+dfrac{10x-3}{2}
7. dfrac{3le...
Đọc tiếp
1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
a)Có bao nhiêu giá trị nguyên dương x thỏa mãn \(\dfrac{x+3}{x^2-4}-\dfrac{1}{x+2}< \dfrac{2x}{2x-x^2}\)
b) Tập nghiệm S của bất pt \(\dfrac{-2x^2+7x+7}{x^2-3x-10}\le-1\)
Giải phương trình sau: ( giúp mk nha)
a) \(\dfrac{2x-\dfrac{4-5x}{5}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}-x+1\)
b) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{2}-6x+1\)
c) \(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
Giải phương trình sau :
a,\(\dfrac{7-3x}{12}+\dfrac{5x+2}{7}=x+13\)
b,\(\dfrac{3\left(x+3\right)}{4}-\dfrac{1}{2}=\dfrac{5x+9}{7}-\dfrac{7x-9}{4}\)
c,\(\dfrac{2x+1}{3}-\dfrac{5x+2}{7}=x+3\)
d,\(\dfrac{2x-3}{3}-\dfrac{2x+3}{7}=\dfrac{4x+3}{5}-17\)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
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a, msc 12.7=84
Chuyển vế về =0 rồi làm
b,msc 28
c,làm tương tự
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a, \(\Rightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)
b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
\(\Leftrightarrow21x+63-14=20x+36-49x+63\)
\(\Leftrightarrow50x=50\Leftrightarrow x=1\)
c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)
d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)
\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)
\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)
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Xem thêm câu trả lời
SN
25 tháng 1 2022 lúc 14:01
1,dfrac{5left(x-1right)+2}{6}-dfrac{7x-1}{4x}dfrac{2left(2x+1right)}{7}-52,dfrac{3left(x-3right)}{4}+dfrac{4x-10,5}{10}dfrac{3
left(x+1right)}{5}+63,dfrac{2left(3x+1right)+1}{4}-5dfrac{2left(3x-1right)}{5}-dfrac{3x+2}{10}Diễn giải ra cho em với ạ!Em cảm ơn
Đọc tiếp
1,\(\dfrac{5\left(x-1\right)+2}{6}\)-\(\dfrac{7x-1}{4x}\)=\(\dfrac{2\left(2x+1\right)}{7}\)-5
2,\(\dfrac{3\left(x-3\right)}{4}\)+\(\dfrac{4x-10,5}{10}\)=\(\dfrac{3 \left(x+1\right)}{5}\)+6
3,\(\dfrac{2\left(3x+1\right)+1}{4}\)-5=\(\dfrac{2\left(3x-1\right)}{5}\)-\(\dfrac{3x+2}{10}\)
Diễn giải ra cho em với ạ!Em cảm ơn
1, bạn xem lại đề
2, 15(x-3) + 8x-21 = 12(x+1) +120
<=> 23x - 66 = 12x + 132
<=> 11x = 198 <=> x = 198/11
3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4
<=> 30x + 10 - 95 = 18x -12
<=> 12x = 73 <=> x = 73/12
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1, dfrac{left(2x-3right)cdotleft(2x+3right)}{8}dfrac{left(x-4right)^2}{6}+dfrac{left(x-2right)^2}{3}
2, x+2-dfrac{2x-dfrac{2x-5}{6}}{15}dfrac{7x-dfrac{x-3}{2}}{5}
3, 1-dfrac{x-dfrac{1+x}{3}}{3}dfrac{x}{2}-dfrac{2x-dfrac{10-7}{3}}{2}
4, dfrac{x+1}{99}+dfrac{x+3}{97}+dfrac{x+5}{95}dfrac{x+7}{93}+dfrac{9+x}{91}+dfrac{x+11}{89}
Đọc tiếp
1, \(\dfrac{\left(2x-3\right)\cdot\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\)
2, \(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\)
3, \(1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\)
4, \(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{9+x}{91}+\dfrac{x+11}{89}\)
4.
\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)
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\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)
\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)
\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)
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Cm các biểu thức sau ko phụ thuộc vào biến:
\(a.\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
\(b.\left[\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
a, \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
\(=\left[\dfrac{x^2}{x\left(x-7\right)\left(x+7\right)}-\dfrac{\left(x-7\right)^2}{x\left(x+7\right)\left(x-7\right)}\right].\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)
\(=\left[\dfrac{x^2-\left(x-7\right)^2}{x\left(x-7\right)\left(x+7\right)}\right].\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)
\(=\left[\dfrac{\left(x-x+7\right)\left(x+x-7\right)}{x\left(x-7\right)\left(x+7\right)}\right].\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)
\(=\dfrac{7\left(2x-7\right)}{x\left(x-7\right)\left(x+7\right)}.\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)
\(=\dfrac{7}{x-7}+\dfrac{x}{7-x}\)
\(=\dfrac{7\left(7-x\right)+x\left(x-7\right)}{\left(x-7\right)\left(7-x\right)}=\dfrac{49-7x+x^2-7x}{7x-x^2-49+7x}\)
\(=\dfrac{49-14x+x^2}{14x-x^2-49}=\dfrac{-\left(14-x^2-49\right)}{14x-x^2-49}=-1\)
\(\Rightarrow\)Giá trị biểu thức trên không phụ thuộc vào biến
\(\Rightarrowđpcm\)
b, tương tự
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Bạn chỉ cần tính ra thôi .
+ Nếu kết quả rút gọn có x thì kết luận giá trị của biểu thức phụ thuộc vào giá trị của biến .
+ Nếu kết quả rút gọn không có x thì kết luận giá trị của biểu thức không phụ thuộc vào giá trị của biến .
Chúc bạn làm bài tốt nha .
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a)\(\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x.\left(x+7\right)}\right):\dfrac{2x-7}{x.\left(x+7\right)}+\dfrac{x}{7-x}\)
\(\Leftrightarrow\dfrac{x^2-\left(x-7\right)^2}{x.\left(x-7\right)\left(x+7\right)}.\dfrac{x.\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)
\(\Leftrightarrow\dfrac{x^2-x^2+14x-49}{\left(x-7\right).\left(2x-7\right)}-\dfrac{x}{x-7}\)
\(\Leftrightarrow\dfrac{7.\left(2x-7\right)}{\left(x-7\right)\left(2x-7\right)}-\dfrac{x}{x-7}\)
\(\Leftrightarrow\dfrac{7-x}{x-7}=-1\)
Vậy bt không phụ thuộc vào biến
b) Tương tự
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