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e: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x+5}{2}=\dfrac{y-2}{3}=\dfrac{x-y+5+2}{2-3}=\dfrac{10+7}{-1}=-17\)

=>x+5=-34; y-2=-51

=>x=-39; y=-49

g: Áp dụng tính chất của DTSBN, ta được

\(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}=\dfrac{5a-3b-4c-5-9+20}{5\cdot2-3\cdot4-6\cdot4}=\dfrac{-253}{13}\)

=>a-1=-506/13; b+3=-1012/13; c-5=-1518/13

=>a=-493/13; b=-1051/13; c=-1453/13

AH
Akai Haruma
Giáo viên
29 tháng 1 2023

Lời giải:
e. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-(y-2)}{2-3}=\frac{(x-y)+5+2}{2-3}=\frac{10+5+2}{-1}=-17$

Suy ra:

$x+5=2(-17)=-34\Rightarrow x=-39$

$y-2=3(-17)=-51\Rightarrow y=-49$

f. Đề thiếu. Bạn xem lại

h. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}$

$=\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}$

$=\frac{5a-5-(3b+9)-(4c-20)}{10-12-24}$

$=\frac{5a-3b-4c-5-9+20}{-26}=\frac{500-5-9+20}{-26}=\frac{-253}{13}$

Suy ra:
$a-1=2.\frac{-253}{13}\Rightarrow a=\frac{-493}{13}$

$b+3=4.\frac{-253}{13}\Rightarrow b=\frac{-1051}{13}$

$c-5=6.\frac{-253}{13}\Rightarrow c=\frac{-1453}{13}$

23 tháng 12 2020

a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)

\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)

Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)

nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)

Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)

mà a+b+c=2 

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)

Do đó: 

\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)

Vậy: \(a=\dfrac{16}{35}\)\(b=\dfrac{24}{35}\)\(c=\dfrac{6}{7}\)

b) Ta có: 2a=3b=5c

nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)

mà a+b-c=3

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: 

\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)

Do đó: 

\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)

Vậy: \(a=\dfrac{45}{19}\)\(b=\dfrac{30}{19}\)\(c=\dfrac{18}{19}\)

17 tháng 7 2017

\(\dfrac{4}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)

\(\Rightarrow\dfrac{4}{x}-\dfrac{2y}{6}=\dfrac{1}{6}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{1+2y}{6}\)

\(\Rightarrow24=x\left(1+2y\right)\)

\(\Rightarrow x;1+2y\inƯ\left(24\right)\)

\(Ư\left(24\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm8;\pm12;\pm24\right\}\)

Mà 1+2y lẻ nên:

\(\left\{{}\begin{matrix}1+2y=1\Rightarrow2y=0\Rightarrow y=0\\x=24\\1+2y=-1\Rightarrow2y=-2\Rightarrow y=-1\\x=-24\end{matrix}\right.\)

\(\left\{{}\begin{matrix}1+2y=3\Rightarrow2y=2\Rightarrow y=1\\x=8\\1+2y=-3\Rightarrow2y=-4\Rightarrow y=-2\\x=-8\end{matrix}\right.\)

17 tháng 7 2017

thank bn nhiều nhahaha

3 tháng 8 2023

\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)

3 tháng 8 2023

\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)

a: =-1/5x^5y^2

b: =-9/7xy^3

c: =7/12xy^2z

d: =2x^4

e: =3/4x^5y

f: =11x^2y^5+x^6

20 tháng 12 2020

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

20 tháng 12 2020

cảm ơn cậu cutee gì đó ơi nhahihi

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

5 tháng 5 2022

a/=4/9x10/3=40/27

b/=15/49

c/=4/9x2/3=8/27

5 tháng 5 2022

a)\(=\dfrac{4}{9}\times\dfrac{10}{3}=\dfrac{40}{27}\)

b)\(\dfrac{3}{7}\times\dfrac{5}{7}=\dfrac{15}{49}\)

c)\(\dfrac{4}{9}:\dfrac{3}{2}=\dfrac{4}{9}\times\dfrac{2}{3}=\dfrac{8}{27}\)

8 tháng 6 2021

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

\(x=10\)

 

8 tháng 5 2022

bài 1

a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)

d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)

8 tháng 5 2022

bài 2

a)\(x=\dfrac{5}{6}-\dfrac{4}{5}=\dfrac{25}{30}-\dfrac{24}{30}=\dfrac{1}{30}\)

b)\(x=5\times\dfrac{10}{7}=\dfrac{50}{7}\)

bài 4 :

145 ×× 69 + 22 x  145 +145 x 8 + 145 

\(=145\times\left(69+22+8+1\right)=145\times100=14500\)