a) x = 99/20

b) x = 7

c) x = 2

( chỉ lm đc đến đó thui nk )

a) \(x-\frac{5}{7}=\frac{1}{9}\Rightarrow x=\frac{1}{9}+\frac{5}{7}\Rightarrow x=\frac{52}{63}\)

b) \(\frac{-3}{7}-x=\frac{4}{5}+\frac{-2}{3}\Rightarrow\frac{-3}{7}-x=\frac{2}{15}\Rightarrow x=\frac{-3}{7}-\frac{2}{15}\Rightarrow x=\frac{-59}{105}\)

c) \(x-\frac{1}{5}=\frac{2}{7}.\frac{-11}{5}\Rightarrow x-\frac{1}{5}=\frac{-22}{35}\Rightarrow x=\frac{-22}{35}+\frac{1}{5}\Rightarrow x=\frac{-3}{7}\)

d) \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Rightarrow\frac{x}{182}=\frac{-15}{91}\Rightarrow x=\frac{\left(-15\right).182}{91}\Rightarrow x=-30\)

\(H=\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{2\cdot1}\)

\(U=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+...+\frac{1}{2\cdot1}\right)\)

\(U=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\right)\)

\(H=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(HU=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(UH=\frac{1}{100}-1+\frac{1}{100}\)

\(HU=\frac{2}{100}-1=-\frac{49}{50}\)

Chậc =)))

chac j` vay` cu'ng

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

a/ \(\frac{6}{7}x=\frac{18}{23}\)

\(x=\frac{18}{23}:\frac{6}{7}=\frac{21}{23}\)

b/ \(2\frac{1}{2}x=\frac{5}{6}\)

\(=>\frac{5}{2}x=\frac{5}{6}\)

\(x=\frac{5}{6}:\frac{5}{2}=\frac{1}{3}\)

c/\(x:2\frac{3}{4}=9\frac{5}{8}\)

\(x:\frac{11}{4}=\frac{77}{8}\)

\(x=\frac{77}{8}\cdot\frac{11}{4}=\frac{847}{32}\)

d/\(7\frac{1}{7}\cdot\frac{1}{7}\cdot x=22\frac{1}{8}\)

\(\frac{50}{49}x=\frac{177}{8}\)

\(x=\frac{177}{8}:\frac{50}{49}=\frac{8673}{400}\)

\(a,\frac{6}{7}.x=\frac{18}{23}\) \(\Rightarrow x=\frac{18}{23}:\frac{6}{7}=\frac{18}{23}.\frac{7}{6}=\frac{21}{23}\)

\(b,2\frac{1}{2}.x=\frac{5}{6}\Rightarrow\frac{5}{2}.x=\frac{5}{6}\Rightarrow x=\frac{5}{6}:\frac{5}{2}=\frac{5}{6}.\frac{2}{5}=\frac{1}{3}\)

\(c,x:2\frac{3}{4}=9\frac{5}{8}\Rightarrow x:\frac{11}{4}=\frac{77}{8}\Rightarrow x=\frac{77}{8}.\frac{11}{4}=\frac{847}{32}\)

\(d,7\frac{1}{7}.\frac{1}{7}.x=22\frac{1}{8}\Rightarrow\frac{50}{49}.x=\frac{177}{8}\Rightarrow x=\frac{177}{8}:\frac{50}{49}=\frac{177}{8}.\frac{49}{50}=\frac{8673}{400}\)

Mình đang cần gấp.Các bạn giúp nha

Mình chỉ làm được bài một thôi:

BÀI 1:                                                                                Giải

Gọi ƯCLN(a;b)=d (d thuộc N*)

=> a chia hết cho d ; b chia hết cho d

=> a=dx ; b=dy  (x;y thuộc N , ƯCLN(x,y)=1)

Ta có : BCNN(a;b) . ƯCLN(a;b)=a.b

=> BCNN(a;b) . d=dx.dy

=> BCNN(a;b)=\(\frac{dx.dy}{d}\)

=> BCNN(a;b)=dxy

mà BCNN(a;b) + ƯCLN(a;b)=15

=> dxy + d=15

=> d(xy+1)=15=1.15=15.1=3.5=5.3(vì x; y ; d là số tự nhiên)

TH 1: d=1;xy+1=15

=> xy=14 mà ƯCLN(a;b)=1

Ta có bảng sau:

TH2: d=15; xy+1=1

=> xy=0(vô lý vì ƯCLN(x;y)=1)

TH3: d=3;xy+1=5

=>xy=4

mà ƯCLN(x;y)=1

TA có bảng sau:

TH4:d=5;xy+1=3

=> xy = 2

Ta có bảng sau:

.Vậy (a;b) thuộc {(1;14);(14;1);(2;7);(7;2);(3;12);(12;3);(5;10);(10;5)}