K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4:

=>(2x+3,5)=7/12*3/14=21/168=1/8

=>2x=1/8-7/2=1/8-28/8=-27/8

=>x=-27/16

5: =>1/3:3x=-21/4

=>3x=-1/3:21/4=-1/3*4/21=-4/63

=>x=-4/189

6: =>2+7/9-3/4(x+1)=7/9

=>2-3/4(x+1)=0

=>3/4(x+1)=2

=>x+1=2:3/4=2*4/3=8/3

=>x=5/3

b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)

=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)

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

=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)

=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)

=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)

d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)

=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)

=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)

=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)

11 tháng 7 2017

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

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

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

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

\(2x=4-1\) \(2x=4-\left(-1\right)\)

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

11 tháng 7 2017

2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

\(\left|x+\left(-1\right)\right|=-3\)

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

15 tháng 11 2021
|2x-1|+1=4

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

 

18 tháng 5 2020

a) \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2+5=\frac{61}{9}\)

=> \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{61}{9}-5\)

=> \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9}\)

=> \(\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9}:4\)

=> \(\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9\cdot4}=\frac{16}{36}=\frac{4}{9}\)

=> \(\frac{1}{2}x-\frac{1}{3}=\pm\frac{2}{3}\)

Trường hợp 1 : \(\frac{1}{2}x-\frac{1}{3}=\frac{2}{3}\)

=> \(\frac{1}{2}x=1\)

=> \(x=1:\frac{1}{2}=2\)

Trường hợp 2 : \(\frac{1}{2}x-\frac{1}{3}=-\frac{2}{3}\)

=> \(\frac{1}{2}x=-\frac{2}{3}+\frac{1}{3}=-\frac{1}{3}\)

=> \(x=\left(-\frac{1}{3}\right):\frac{1}{2}=\left(-\frac{1}{3}\right)\cdot2=-\frac{2}{3}\)

b) \(9\left(2x-\frac{1}{3}\right)^3-1=-\frac{2}{3}\)

=> \(9\left(2x-\frac{1}{3}\right)^3=-\frac{2}{3}+1=\frac{1}{3}\)

=> \(\left(2x-\frac{1}{3}\right)^3=\frac{1}{3}:9\)

=> \(\left(2x-\frac{1}{3}\right)^3=\frac{1}{3\cdot9}=\frac{1}{27}\)

=> \(2x-\frac{1}{3}=\frac{1}{3}\)

=> \(2x=\frac{2}{3}\)

=> \(x=\frac{2}{3}:2=\frac{1}{3}\)

Bài cuối tương tự

1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)

\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)

\(\Leftrightarrow x^2+x-6=0\)

=>(x+3)(x-2)=0

=>x=-3(nhận) hoặc x=2(nhận)

2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)

\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)

=>-8x+8=0

hay x=1(nhận)

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

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

\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)

\(\Leftrightarrow x^2+x-6=0\)

=>(x+3)(x-2)=0

=>x=-3(nhận) hoặc x=2(nhận)