\(\frac{561}{704}\)x \(\frac{39}{47}\)+\(\frac{561}{704}\)x\(\frac{8}{47}\)
Những câu hỏi liên quan
\(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)
Theo đề bài ta có:
\(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)
\(\Leftrightarrow\frac{51}{13}< \frac{13x+12}{13}< \frac{77}{13}\)
\(\Leftrightarrow51< 13x+12< 77\)
\(\Leftrightarrow39< 13x< 65\)
\(\Leftrightarrow x=4\)
Vậy x = 4
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Ta có :
\(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)
\(\Leftrightarrow\)\(\frac{51}{13}< \frac{13x+12}{13}< \frac{77}{13}\)
\(\Leftrightarrow\)\(51< 13x+12< 77\)
\(\Leftrightarrow\)\(39< 13x< 65\)
Mà x là phần nguyên nên \(13x\inℤ\)
\(\Rightarrow\)\(13x=52\)
\(\Rightarrow\)\(x=\frac{52}{13}\)
\(\Rightarrow\)\(x=4\)
Vậy \(x=4\) hay hỗn số cần tìm là \(4\frac{12}{13}\)
Chúc bạn học tốt ~
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Bài 1: Số tự nhiên?a, ....... 0,001 b, 3,01 ........... 4,99 c, 3670,92 ........ x 9 3675,92Bài 2: a, Chuyển thành số thập phân:36frac{47}{100} 36frac{407}{10000} 36frac{704}{1000} 36frac{407}{10000} 36frac{704}{1000}b, Xếp từ bé đến lớn:
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Bài 1: Số tự nhiên?
a, ....... < 0,001 b, 3,01 < ........... < 4,99 c, 3670,92 < ........ x 9 < 3675,92
Bài 2: a, Chuyển thành số thập phân:
36\(\frac{47}{100}\)= 36\(\frac{407}{10000}\)= 36\(\frac{704}{1000}\)= 36\(\frac{407}{10000}\)= 36\(\frac{704}{1000}\)=
b, Xếp từ bé đến lớn:
Bài 1
a) 0 b) 4 c) 408
Bài 2
a)\(36\frac{47}{100}=36,47\) \(36\frac{407}{10000}=36,0407\) \(36\frac{704}{1000}=36,704\) \(36\frac{407}{10000}=36,0407\) \(36\frac{704}{1000}=36,704\)
b)36,0407 ; 36,0407 ; 36,36,47 ; 36,704 ; 36,704
Hình như mình sai bài 2,không hiểu sao có 2 hỗn số \(36\frac{407}{10000}\)và 2 hỗn số \(36\frac{704}{1000}\)nữa??!!
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Tìm phần nguyên x của hỗn số biết rằng :
a, \(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)
b, \(x\frac{3}{4}=\frac{21989}{7996}\)
a, \(3\frac{12}{13}< x\frac{12}{13}< 5\frac{12}{13}\Rightarrow x=4\)
b, \(x\frac{3}{4}=\frac{21989}{7996}=\frac{11}{4}=2\frac{3}{4}\Rightarrow x=2\)
~ Hok tốt ~
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Trả lời :
a)x=4
b)x=2
\(\downarrow\)
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a) \(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)
\(\Rightarrow3\frac{12}{13}< x\frac{12}{13}< 5\frac{12}{13}\)
\(\Rightarrow x=4\)
b) \(x\frac{3}{4}=\frac{21989}{7996}\)
\(\Rightarrow x\frac{3}{4}=2\frac{3}{4}\)
\(\Rightarrow x=2\)
~ Hok tốt ~
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Xem thêm câu trả lời
a) frac{x-23}{24}+frac{x-23}{25}frac{x-23}{26}+frac{x-23}{27} b) frac{x+1}{2004}+frac{x+2}{2003}frac{x+3}{2002}+frac{x+4}{2001}
d) frac{x+1}{9}+frac{x+2}{8}frac{x+3}{7}+frac{x+4}{6}
c) frac{x-45}{55}+frac{x-47}{53}frac{x-55}{45}+frac{x-53}{47}
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a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\) b) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
c) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
mấy câu này dễ mà :V câu a+c lấy mỗi phân số trừ cho 1 ra tử chung rút ra thì tính b+d thì cộng một tử chung rồi lại tính tiếp thôi
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Giải phương trình sau
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Rightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
\(\Rightarrow x-100=0\).Do \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\ne0\)
\(\Rightarrow x=100\)
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\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{x-45}{55}-1-\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
(x-100)(\(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}=0\)
-> x-100 = 0 -> x = 100
mà \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\) khác 0
Vậy x = 100
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Giải phương trình
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Leftrightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\Leftrightarrow\frac{x-45-55}{55}+\frac{x-47-53}{47}-\frac{x-55-45}{45}-\frac{x-53-47}{47}=0\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{47}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
Vậy pt có tập nghiệm S = { 100 }
giải phương trình sau 
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
dễ thôi mà
Áp dụng tỉ lệ thức, ta có:
\(\Leftrightarrow\frac{108x-4970}{2915}=\frac{92x-4970}{2115}\Rightarrow\left(108x-4970\right)2115=2915\left(92x-4970\right)\)
=>x=100
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Ông Thắng: làm kiểu đó chưa gọi là đúng hoàn toàn đâu
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$\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}$x−4555 +x−4753 =x−5545 +x−5347
=> x = 100
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$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{...
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}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}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{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$v
I. Nội qui tham gia "Giúp tôi giải toán"
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Giải PT
\(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
M.n giúp mk vs
a, Mình nghĩ là đề sai .
b, Ta có : \(\frac{x-45}{55}+\frac{x-47}{45}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{45}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-45}{55}-\frac{55}{55}+\frac{x-47}{53}-\frac{53}{53}=\frac{x-55}{45}-\frac{45}{45}+\frac{x-53}{47}-\frac{47}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
=> \(x=100\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{100\right\}\)
c, Ta có : \(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
=> \(\frac{2-x}{2010}-1=\frac{1-x}{2011}+\frac{-x}{2012}\)
=> \(\frac{2-x}{2010}+1=\frac{1-x}{2011}+1+\frac{-x}{2012}+1\)
=> \(\frac{2-x}{2010}+\frac{2010}{2010}=\frac{1-x}{2011}+\frac{2011}{2011}+\frac{-x}{2012}+\frac{2012}{2012}\)
=> \(\frac{2012-x}{2010}=\frac{2012-x}{2011}+\frac{2012-x}{2012}\)
=> \(\frac{2012-x}{2010}-\frac{2012-x}{2011}-\frac{2012-x}{2012}=0\)
=> \(\left(2012-x\right)\left(\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)
=> \(2012-x=0\)
=> \(x=2012\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2012\right\}\)