Tìm x, biết: 1. \(\left(\dfrac{2x}{5} 2\right):\left(-4\right)=-1\dfrac{1}{2}\) 2. x. ( x 1) = 0 3. 2x - 16 = 40 x 4. x : \(\dfrac{5}{3}\) \(\dfrac{1}{3}\) = \(\dfrac{-2}{5}\) 5. 3\(\dfrac{...

LT

Lê Thị Ngọc Duyên

23 tháng 4 2017

Tìm x, biết: 1. \(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\) 2. x. ( x + 1) = 0 3. 2x - 16 = 40 + x 4. x : \(\dfrac{5}{3}\) + \(\dfrac{1}{3}\) = \(\dfrac{-2}{5}\) 5. 3\(\dfrac{1}{2}\) - \(\dfrac{1}{2}\) x = \(\dfrac{2}{3}\) 6. \(\dfrac{x}{4}=\dfrac{9}{x}\) 7. \(\dfrac{1}{3}+\dfrac{1}{2}:x=\) -0,25 ...

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#Toán lớp 6

5

TL

Tôi Là 7D

2 tháng 5 2017

6. \(\dfrac{x}{4}=\dfrac{9}{x}\)

=>x²=4.9=36

=>x\(\in\)\(\left\{-6;6\right\}\)

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NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

27 tháng 8 2023

\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\)

(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\)

\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\)

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

\(\dfrac{2x}{5}=\dfrac{30}{5}\)

2x = 30

x = 30 : 2 = 15

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KJ

❖ Khang/GD❄ 『ʈєɑɱ❖Hoàng๖ۣۜGia☆』

28 tháng 2 2022

Tìm x biết:

\(a,\left(x-\dfrac{3}{4}\right)+50\%=\dfrac{1}{6}\)

\(b,\dfrac{1}{2}x-\dfrac{5}{6}x=\dfrac{7}{2}\)

\(c,\left(4-x\right)\left(3x+5\right)=0\)

\(d,\dfrac{x}{16}=\dfrac{50}{32}\)

\(e,\left(2x-3\right)+\dfrac{3}{2}=-\dfrac{1}{4}\)

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

28 tháng 2 2022

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

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TL

Trần Linh

4 tháng 5 2022

Câu 1: Thực hiện phép tính

a, \(40\dfrac{1}{4}:\dfrac{5}{7}-25\dfrac{1}{4}:\dfrac{5}{7}-\dfrac{1}{2021}\)

b, \(\left|\dfrac{-5}{9}\right|.\sqrt{81}-2021^0.\dfrac{16}{25}\)

Câu 2: Tìm x

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

#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

17 tháng 6 2023

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

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KJ

❖ Khang/GD❄ 『ʈєɑɱ❖Hoàng๖ۣۜGia☆』

29 tháng 1 2022

Tìm x...

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#Toán lớp 6

2

KS

Kudo Shinichi AKIRA^_^

29 tháng 1 2022

Chia nhỏ ra

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NL

Nguyễn Lê Phước Thịnh

29 tháng 1 2022

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

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VN

Vũ Ngọc Diệp

27 tháng 8 2023

Tìm x:

a) \(\dfrac{1}{3}.x+\dfrac{2}{5}\left(x-1\right)=0\)

b)\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=x\)

c)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)

d)\(9.\left(3x+1\right)^2=16\)

#Toán lớp 6

2

VN

Vũ Ngọc Diệp

27 tháng 8 2023

làm ơn giúp 🙏🙏🙏

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NL

Nguyễn Lê Phước Thịnh

28 tháng 8 2023

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

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WT

Walker Trang

12 tháng 6 2017

tìm x...

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#Toán lớp 6

2

AT

Aki Tsuki

12 tháng 6 2017

K chép lại đề, lm luôn nhé:

*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)

\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)

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

* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)

=> K có gt x nào t/m đề

* Đề sai

* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)

\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)

\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)

\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)

* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)

* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)

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

\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)

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MS

Mashiro Shiina

12 tháng 6 2017

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

\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)

\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)

b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)

\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)

\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)

\(\Rightarrow x\in\varnothing\)

c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.

d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)

\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)

e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)

\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)

\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)

\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)

g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)

\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)

\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)

\(th1:x=0\)

\(th2:x=\dfrac{-3}{5}\)

h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)

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

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

\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)

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G

ggsufuu

19 tháng 2 2022

giải phương trình

1)\(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)

2)\(x^2-4-\left(x-5\right)\left(x-2\right)=0\)

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

4)\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}+\dfrac{3x-1}{x-4}-6\)

5)\(\dfrac{96}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}-6\)

#Toán lớp 8

3

MH

Minh Hiếu

19 tháng 2 2022

1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)

\(3x+2x^2-6-4x-2x^2-10x-6=0\)

\(-11x=12\)

\(x=-\dfrac{12}{11}\)

2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)

\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)

\(\left(x-2\right)\left(x+2-x+5\right)=0\)

\(7\left(x-2\right)=0\)

\(\Leftrightarrow x=2\)

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NH

Nguyễn Huy Tú

19 tháng 2 2022

1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)

2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)

3, bạn xem lại đề

5, đk x khác -4 ; 4

\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)

\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)

\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm)

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NL

Ngân Lê Bảo

20 tháng 1 2021

giải phương...

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#Toán lớp 8

2

ND

Nguyễn Duy Khang

20 tháng 1 2021

undefined

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ND

Nguyễn Duy Khang

20 tháng 1 2021

\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)

Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)

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LH

Lê Hoàng Khánh

29 tháng 7 2021

2. Tìm xa. \(\dfrac{4}{5}-3.\left|x\right|=\dfrac{1}{5}\) b. \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)c. (2x-8)(10-5x)=0 ...

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#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

29 tháng 7 2021

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

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

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

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E

ᴵᴬᴹɴɢᴜʏễɴ●ɴɢọᴄ●ʜùɴɢッ

12 tháng 5 2021

\(\text{Tìm x,...

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