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12 tháng 8 2019

1.

a) 13\(\frac{1}{3}\) : 1\(\frac{1}{3}\) = 26 : (2x - 1)

<=> \(\frac{40}{3}:\frac{4}{3}\) = 13x - 26

<=> 10 + 26 = 13x

<=> 13x = 36

<=> x = \(\frac{36}{13}\)

b) 0,2 : 1\(\frac{1}{5}\) = \(\frac{2}{3}\) : (6x + 7)

<=> \(\frac{1}{5}:\frac{6}{5}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)

<=> \(\frac{1}{6}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)

<=> \(\frac{1}{9}x\) = \(\frac{2}{21}.\frac{1}{6}\) = \(\frac{1}{63}\)

<=> x = \(\frac{1}{7}\)

c) \(\frac{37-x}{x+13}\) = \(\frac{3}{7}\)

<=> (37 - x) . 7 = 3.(x + 13)

<=> 119 - 7x = 3x + 39

<=> -7x - 3x = 39 - 119

<=> -10x = -80

<=> x = 8

d) \(\frac{x-1}{x+5}=\frac{6}{7}\)

<=> 7(x - 1) = 6(x + 5)

<=> 7x - 7 = 6x + 30

<=> 7x - 6x = 30 + 7

<=> x = 37

e)

2\(\frac{2}{\frac{3}{0,002}}\) = \(\frac{1\frac{1}{9}}{x}\)

<=> \(\frac{1501}{750}\) = \(\frac{10}{9}:x\)

<=> x = \(\frac{10}{9}:\frac{1501}{750}\) = \(\frac{2500}{4503}\)

12 tháng 8 2019

Bài 2. đề sai

Bài 3.

a) 6,88 : x = \(\frac{12}{27}\)

<=> x = 6,88 : \(\frac{12}{27}\)

<=> x = 15,48

b) 8\(\frac{1}{3}\) : \(11\frac{2}{3}\) = 13 : 2x

<=> \(\frac{25}{3}:\frac{35}{3}\) = 13 : 2x

<=> \(\frac{5}{7}=13:2x\)

<=> 2x = \(13:\frac{5}{7}\) = \(\frac{91}{5}\)

<=> x = 9,1

giải các hệ BPT sau: a) \(\left\{{}\begin{matrix}5x-24x+5\\5x-4< x+2\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\) c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\) e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\) f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\) g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\) h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\) j)...
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giải các hệ BPT sau:

a) \(\left\{{}\begin{matrix}5x-2>4x+5\\5x-4< x+2\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\)

e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\)

f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\)

g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\)

h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)

j) \(\left\{{}\begin{matrix}\frac{3x+1}{2}-\frac{3-x}{3}\le\frac{x+1}{4}-\frac{2x-1}{3}\\3-\frac{2x+1}{5}>x+\frac{4}{3}\end{matrix}\right.\)

3
25 tháng 3 2020
https://i.imgur.com/NOxfqjV.jpg
25 tháng 3 2020
https://i.imgur.com/awOKwJi.jpg
15 tháng 4 2020

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

giải hệ phương trình 1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\) 2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\) 3 ,...
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giải hệ phương trình

1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)

2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)

3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)

4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)

5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)

6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)

7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)

8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)

9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)

3
NV
17 tháng 9 2019

a/ ĐKXĐ: \(x\ne-1\)

\(\Leftrightarrow4\left(3-7x\right)=x+1\)

\(\Leftrightarrow12-28x=x+1\)

\(\Rightarrow29x=11\Rightarrow x=\frac{11}{29}\)

b/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

\(\Leftrightarrow1-\left(\sqrt{x}-2\right)=3-\sqrt{x}\)

\(\Leftrightarrow3=3\) (luôn đúng)

Vậy nghiệm của pt là \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

c/ ĐKXĐ: \(x\ne7\)

\(\Leftrightarrow8-x-8\left(x-7\right)=1\)

\(\Leftrightarrow8-x-8x+56=1\)

\(\Leftrightarrow-9x=-63\Rightarrow x=7\left(ktm\right)\)

Vậy pt vô nghiệm

NV
17 tháng 9 2019

d/ ĐKXĐ: \(x\ne4\)

\(\Leftrightarrow\frac{28}{6\left(x-4\right)}-\frac{6\left(x+2\right)}{6\left(x-4\right)}=\frac{-9}{6\left(x-4\right)}-\frac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

\(\Rightarrow x=5\)

e/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)

\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)

\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)

\(\Leftrightarrow3x=-15\Rightarrow x=-5\)

25 tháng 9 2019

có ái đó giúp mình với mình đang cần gấp