Tìm x: a/\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\) b/\(\dfra...

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

b/\(\dfra...">

K

Khách

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DT

Dương Thanh Ngân

13 tháng 7 2018

Tìm x:

a/\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)

b/\(\dfrac{x^2}{-6}=-\dfrac{24}{25}\)

c/\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

d/\(2,5:4x=0,5:0,2\)

e/\(\dfrac{1}{5}x:3=\dfrac{2}{3}:0,25\)

g/\(1,25:0,8=\dfrac{3}{8}:0,2x\)

f/\(-3:\left(-2\dfrac{1}{4}\right)=\dfrac{3}{4}:\left(-6x\right)\)

h/\(3:\dfrac{2}{5}x=1:0,01\)

j/\(2:1\dfrac{1}{4}=\dfrac{1}{2}:2x\)

l/\(\dfrac{4}{x-1}=\dfrac{x-1}{9}\)

#Hỏi cộng đồng OLM #Toán lớp 7

2

NT

Ngô Thị Thu Trang

15 tháng 7 2018

Thu thập số liệu, tần số

NT

Ngô Thị Thu Trang

13 tháng 7 2018

Thu thập số liệu, tần số

Các câu hỏi dưới đây có thể giống với câu hỏi trên

NT

nguyen thi thuy linh

3 tháng 8 2017

tim x , biet : a) x + \(\dfrac{3}{10}\)= \(\dfrac{-2}{5}\) b) x + \(\dfrac{5}{6}\) = \(\dfrac{2}{5}\) - \(\left(-\dfrac{2}{3}\right)\) c) \(1\dfrac{2}{5}\) x + \(\dfrac{3}{7}\) = \(-\dfrac{4}{5}\) d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\) e) \(\left[x+\dfrac{4}{15}\right]-\left[-3,75\right]=-\left[-2,15\right]\) f) \(\left(x-2\right)^2=1\) g) \(\left(2x-1\right)^3=-27\) h) \(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\) i) x : \(\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\) j)...

#Hỏi cộng đồng OLM #Toán lớp 7

2

PT

Phương Trâm

3 tháng 8 2017

a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)

\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)

\(x=\dfrac{-7}{10}\)

b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)

\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)

\(x+\dfrac{5}{6}=\dfrac{16}{15}\)

\(x=\dfrac{16}{15}-\dfrac{5}{6}\)

\(x=\dfrac{7}{30}\)

c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)

\(\dfrac{7}{5}x=\dfrac{-43}{35}\)

\(\Rightarrow x=\dfrac{-43}{49}\)

d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)

\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)

\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)

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

\(x=\dfrac{-5}{12}\)

e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)

\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)

\(x+\dfrac{4}{5}=2,15-3,75\)

\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)

\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)

\(x=\dfrac{-12}{5}\)

f) \(\left(x-2\right)^2=1\)

\(\Rightarrow x=1\)

Sức chịu đựng có giới hạn -.-

NH

Nịna Hatori

3 tháng 8 2017

- Mình tiếp tục cho Nguyễn Phương Trâm nhé.

g, \(\left(2x-1\right)^3=-27\)

\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)

\(\Rightarrow2x-1=-3\)

\(\Rightarrow2x=-2\)

=> \(x=-1\)

- Vậy x = -1

h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)

\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)

\(\Rightarrow\left(x-1\right)^2=900 \)

\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)

=> x = 31

i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)

=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{16}\)

- Vậy x=\(\dfrac{1}{16}\)

j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)

\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)

\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}\)

- Vạy x = \(\dfrac{3}{4}\)

k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)

=>\(4^x=4\)

=> x = 1

- Vậy x = 1

LT

Le Thu Trang

1 tháng 12 2018

a) \(\dfrac{2}{3}\)x + 0,25 = \(-\dfrac{5}{8}\) - \(\left(\dfrac{1}{2}-\dfrac{3}{4}x\right)\) b) ( \(9x^2\) -1 ) . ( 2x + 3 ) = 0 c) \(\left|2x-\dfrac{1}{3}\right|-\dfrac{3}{4}=\left|-\dfrac{5}{6}\right|\) d) \(\left(0,75-\dfrac{5}{6}x\right)^3\) = \(-\dfrac{27}{64}\) e) \(\left(\dfrac{2}{5}x-\dfrac{1}{2}\right)^2=\dfrac{9}{16}\) f) \(\dfrac{2x-3}{3}=\dfrac{3x+2}{7}\) h)\(\left(\dfrac{2}{3}\right)^{2x+1}\) = \(\left(\dfrac{8}{27}\right)^3\) i)...

#Hỏi cộng đồng OLM #Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

27 tháng 11 2022

b: =>(3x-1)(3x+1)(2x+3)=0

hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)

c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)

=>2x-1/3=19/12 hoặc 2x-1/3=-19/12

=>2x=23/12 hoặc 2x=-15/12=-5/4

=>x=23/24 hoặc x=-5/8

d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)

=>-5/6x=-3/2

=>x=3/2:5/6=3/2*6/5=18/10=9/5

e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4

=>2/5x=5/4 hoặc 2/5x=-1/4

=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8

f: =>14x-21=9x+6

=>5x=27

=>x=27/5

h: =>(2/3)^2x+1=(2/3)^27

=>2x+1=27

=>x=13

i: =>5^3x*(2+5^2)=3375

=>5^3x=125

=>3x=3

=>x=1

KS

Kudo Shinichi

9 tháng 11 2018

Tìm x, biết a.\(\dfrac{2}{3}\)x-\(\dfrac{1}{2}\)x = (\(\dfrac{-7}{12}\)) . \(\dfrac{7}{5}\) b.(\(\dfrac{1}{5}\) - \(\dfrac{3}{2}\)x)\(^2\) = \(\dfrac{9}{4}\) c. (1,25 - \(\dfrac{4}{5}\)x)\(^3\) = -125 d. 2\(^x\) + 2\(^{x+4}\) = 544 e.\(\dfrac{3}{4}\)x - 4,5+ \(\dfrac{4}{5}\)x = 3 g. \(\left|3x-2\right|\) =\(\left|2x-3\right|\) h.(x2 -4)\(\sqrt{x}\) = 0 i. (\(\dfrac{1}{3}\)x - 4 )2 = \(\dfrac{16}{25}\) j.4x2 -1 = 0 k.2x2+0,82 =1 l.\(\dfrac{5}{12}\)\(\sqrt{x}\) - \(\dfrac{1}{6}\) =...

#Hỏi cộng đồng OLM #Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

19 tháng 11 2022

a: =>1/6x=-49/60

=>x=-49/60:1/6=-49/60*6=-49/10

b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2

=>x=17/15 hoặc x=-13/15

c: =>1,25-4/5x=-5

=>4/5x=1,25+5=6,25

=>x=125/16

d: =>2^x*17=544

=>2^x=32

=>x=5

i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5

=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2

=>x=14,4 hoặc x=9,6

j: =>(2x-1)(2x+1)=0

=>x=1/2 hoặc x=-1/2

NM

Nguyễn Minh Bảo Anh

1 tháng 7 2017

Tìm x a)\(\left(x+5\right)^3\)=-64 b)\(\left(2x-3\right)^2\)=9 c)\(\dfrac{x}{\dfrac{4}{2}}\)=\(\dfrac{4}{\dfrac{x}{2}}\) d)\(\dfrac{-32}{\left(-2\right)^n}\)=4 e)\(\dfrac{8}{2x}=4\) g)\(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\) h)\(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\) i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\) ...

#Hỏi cộng đồng OLM #Toán lớp 7

2

LV

Lê Vương Kim Anh

1 tháng 7 2017

a) ( x + 5 )³= -64

x + 5 = - 4

x = - 4 - 5

x = -9

b) (2x - 3)²=9

2x - 3 = 3

2x = 3+3

2x = 6

x = 6 : 2

x = 3

e) \(\dfrac{8}{2x}=4\)

=> 4 . 2x = 8

8x =8

x = 8 : 8

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)

=> x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(\dfrac{1}{4}.x=\dfrac{1}{32}\)

x = \(\dfrac{1}{32}:\dfrac{1}{4}\)

x = \(\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\dfrac{-1}{27}\)

HN

Hải Ngân

2 tháng 7 2017

a) (x + 5)³ = -64

=> (x + 5)³ = (-4)³

x + 5 = -4

x = -4 - 5

x = -9

b) (2x - 3)² = 9

=> (2x - 3)² = (\(\pm\)3)²

=> 2x - 3 = 3 hoặc 2x - 3 = -3

*2x - 3 = 3

2x = 3 + 3

2x = 9

x = \(\dfrac{9}{2}\)

*2x - 3 = -3

2x = -3 + 3

2x = 0

x = 0 : 2

x = 0

Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)

c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)

\(\dfrac{x}{2}=8\)

x = 8 : 2

x = 4

d) \(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)

=> (-2)ⁿ . (-2)²= (-2)⁵

(-2)ⁿ = (-2)⁵: (-2)²

(-2)ⁿ = (-2)³

Vậy n = 3

e) \(\dfrac{8}{2x}=4\)

=> 2x . 4 = 8

2x = 8 : 4

2x = 2

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\)

2x - 1 = 3

2x = 3 + 1

2x = 4

x = 4 : 2

x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)

\(x=\left(\dfrac{1}{2}\right)^3\)

\(x=\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^3\)

\(x=\dfrac{-1}{27}\).

LN

Lê Nữ Khánh Huyền

28 tháng 7 2017

Tìm x trong tỉ lệ thức a) \(-0,52:x=-9,36:16,38\) b) \(2\dfrac{2}{5}:x=1\dfrac{7}{9}:0,2\) c)\(\left(\dfrac{2}{5}x\right):\dfrac{1}{3}=1\dfrac{1}{3}:\dfrac{4}{5}\) d) \(0,3:1\dfrac{1}{3}=\dfrac{2}{5}:\left(3x+6\right)\) e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\) f) \(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\) g) \(\dfrac{x-1}{7}=\dfrac{9}{x+1}\) h)...

#Hỏi cộng đồng OLM #Toán lớp 7

2

IL

I LOVE YOU

31 tháng 7 2017

LN

Lê Nữ Khánh Huyền

31 tháng 7 2017

j vậy bạn?????

KY

Kim Yen Pham

27 tháng 11 2017

bài 1 tìm x thuộc Q biết a. |x|=\(^1_53\) và x<0 b.|x|=-2,1 c.|x-3,5|=5 d. |x+\(\dfrac{3}{4}\)|-\(\dfrac{1}{2}\)=0 e. |x-\(\dfrac{2}{5}\)|+\(\dfrac{1}{2}\)=\(\dfrac{3}{4}\) f. \(\dfrac{5}{6}\)-|2-x|=\(\dfrac{1}{3}\) g. (2x-5)^2=9 h. \(\sqrt{3-7x}\)=\(\dfrac{1}{4}\) i. (\(\dfrac{2}{3}\))^x=\(\dfrac{8}{27}\) k. (x+\(\dfrac{1}{2}\))^2=\(\dfrac{1}{9}\) l. \(\dfrac{\left(-3^x\right)}{81}\)=-27 m.\(\left(x-2\right)\)^2x+3=(x-2)^2x+1(x thuộc N) ...

#Hỏi cộng đồng OLM #Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

5 tháng 6 2022

a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)

mà x<0

nên x=-16/5

b: \(\left|x\right|=-2.1\)

nên \(x\in\varnothing\)

c: \(\left|x-3.5\right|=5\)

=>x-3,5=5 hoặc x-3,5=-5

=>x=8,5 hoặc x=-1,5

d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

=>|x+3/4|=1/2

=>x+3/4=1/2 hoặc x+3/4=-1/2

=>x=-1/4 hoặc x=-5/4

PT

Phạm Thị Thanh Thanh

1 tháng 10 2017

tìm x , biết : a, \(\dfrac{37-x}{2x+5}=\dfrac{4}{9}\) b, \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\) c, \(\dfrac{x+3}{25}=\dfrac{9}{x+3}\) d, \(\dfrac{x}{\dfrac{-2}{3}}=\dfrac{5}{-x}\) e, \(\dfrac{2x-5}{\dfrac{3}{2}}=\dfrac{\dfrac{-27}{8}}{5-2x}\) f,...

#Hỏi cộng đồng OLM #Toán lớp 7

1

N

ngonhuminh

2 tháng 10 2017

câu E

\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )

DG

Dark Goddess

27 tháng 7 2017

Tìm x biết: a) \(\dfrac{-32}{\left(-2\right)^x}=4\) f) \(\left(3x-1\right)^3=\dfrac{-8}{27}\) b) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\) g) \(\left(2x+3\right)^2=\dfrac{9}{121}\) c) \(\dfrac{1}{9}.27^x=3^x\) h) \(5^x+5^{x+2}=650\) d) \(9^x:3^3=\dfrac{1}{243}\) i)...

#Hỏi cộng đồng OLM #Toán lớp 7

1

QD

Quang Duy

27 tháng 7 2017

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

DG

Dark Goddess

1 tháng 8 2017

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây

HT

Hoang Thu Trang

6 tháng 1 2018

a)\(\dfrac{x+1}{2}\)=\(\dfrac{2x+3}{5}\)

b)\(\left|x-1\right|\) + 3\(\left|y+1\right|\) + \(\left|z+2\right|=0\)

c)\(\dfrac{x-2}{4}=\dfrac{5-3x}{4}\)

d)\(\dfrac{x+2}{4}=\dfrac{4}{x+2}\)

e)\(\dfrac{x-1}{5}=\dfrac{-20}{x-1}\)

#Hỏi cộng đồng OLM #Toán lớp 7

1

NT

Nguyễn Thanh Hằng

6 tháng 1 2018

a/ \(\dfrac{x+1}{2}=\dfrac{2x+3}{5}\)

\(\Leftrightarrow5\left(x+1\right)=2\left(2x+3\right)\)

\(\Leftrightarrow5x+5=4x+6\)

\(\Leftrightarrow5x-4x=6-5\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy ...

b/ \(\left|x-1\right|+3\left|y+1\right|+\left|z+2\right|=0\)

Mà với \(\forall x;y;z\) ta có :

\(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\3\left|y+1\right|\ge0\\\left|z+2\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\3\left|y+1\right|=0\\\left|z+2\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\\z+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)

Vậy ...

c/ \(\dfrac{x-2}{4}=\dfrac{5-3x}{4}\)

\(\Leftrightarrow x-2=5-3x\)

\(\Rightarrow x+3x=5+2\)

\(\Leftrightarrow4x=7\)

\(\Leftrightarrow x=\dfrac{7}{4}\)

Vậy ......

d/ \(\dfrac{x+2}{4}=\dfrac{4}{x+2}\)

\(\Leftrightarrow\left(x+2\right)\left(x+2\right)=16\)

\(\Leftrightarrow\left(x+2\right)^2=4^2=\left(-4\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy ...

e/ \(\dfrac{x-1}{5}=\dfrac{-20}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-100\)

\(\Leftrightarrow\left(x-1\right)^2=-100\)

Lại có : \(\left(x-1\right)^2\ge0\)

\(\Leftrightarrow\) k tồn tại x