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15 tháng 12 2023

x e Z là sao bạn ơi 

 

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bài 1/ 

a) ta có: \(A=\frac{15}{x-1}\)

Để A là phân số \(\Rightarrow x-1\ne0\)

                          \(\Rightarrow x\ne1\)

b) Nếu x = 7

\(\Rightarrow A=\frac{15}{7-1}\)

\(\Rightarrow A=\frac{15}{6}\)

Nếu x = -3

\(\Rightarrow A=\frac{15}{-3-1}\)

\(\Rightarrow A=\frac{15}{-4}\)

Nếu x = 4

\(\Rightarrow A=\frac{15}{4-1}\)

\(\Rightarrow A=\frac{15}{3}=5\)

c) Ta có: \(B=5\)

\(\Leftrightarrow A=\frac{15}{x-1}=5\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=4\)

Bài 2/

a) \(\frac{x}{3}=\frac{2}{6}\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

b) \(-\frac{x}{14}=\frac{10}{-7}\)

\(\Leftrightarrow7x=140\)

\(\Leftrightarrow x=20\)

hok tốt!!

28 tháng 1 2018

a) \(37-\left(x-25\right)=x+14\)

\(\Leftrightarrow\)\(37-x+25=x+14\)

\(\Leftrightarrow\)\(x+x=37+25-14\)

\(\Leftrightarrow\)\(2x=48\)

\(\Rightarrow\)\(x=\frac{48}{2}=24\)

Vậy \(x=24\)

b) \(11-\left(x-3\right)^2=0\)

\(\Leftrightarrow\)\(\left(x-3\right)^2=11-0=11\)

\(\Leftrightarrow\)\(x-3=\sqrt{11}\)

\(\Rightarrow\)\(x=\sqrt{11}+3\)

Vậy \(x=\sqrt{11}+3\)

Ta có \(P=\frac{7x-14}{x+5}=7+\frac{21}{x+5}\)

P có giá trị nguyên =>\(\frac{21}{x+5}nguyên\)

\(\Rightarrow x+5\inƯ\left(21\right)\)

\(\Rightarrow x=\left\{-26;-16;-12;-8;-6;-4;-2;2\right\}\)

7 tháng 5 2018

ahihi

28 tháng 12 2017

1) x - 3 = 19 - x \(\Rightarrow\)x + x = 19 + 3 \(\Rightarrow\)2x = 22 \(\Rightarrow\)x = 11

2)  -14 - y = 36 -( -y ) -( - 24 ) \(\Rightarrow\)-14 - y = 36 + y +24 \(\Rightarrow\)-y - y = 36 + 24 + 14 \(\Rightarrow\)-2y = 74 \(\Rightarrow\)y = -37

3)  10 - y = 4 + y \(\Rightarrow\)10 - 4 = y + y \(\Rightarrow\)6 = 2y \(\Rightarrow\)y = 3

19 tháng 6 2019

Ta có: 

Để D \(\in\)Z <=> \(x+2⋮3\)

<=> \(x+2\in B\left(3\right)=\left\{0;3;6;9;....\right\}\)

<=> x \(\in\){-2; 1; 4; 7; ...}

19 tháng 6 2019

Ta có:

D\(\in\)Z <=> \(x+2⋮3\)

<=> \(x+2\in B\left(3\right)=\left\{0;3;6;....\right\}\)

<=> \(x\in\left\{-2;1;4;...\right\}\)

# Hok_tốt nha

16 tháng 4 2020

\(-14\left|x\right|+2=-40\)

<=> -14 | x | = -42

<=> |x| = 3

<=> x = 3 hoặc x = -3

Ta có: \(-14.|x|+2=-40\)

\(\Rightarrow14|x|=42\)

\(\Rightarrow|x|=3=|\pm3|\)

\(\Rightarrow x=\pm3\)

a: =>x=7-20=-13

b: =>x=-18+12=-6

c: =>x=9 hoặc x=-6

d: =>x=0 hoặc x=4

e: =>6-x=13-3+14=24

=>x=-18

Câu g và h đề thiếu rồi bạn