K
Khách

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

a: =>(3/2-2x):2/3=1/6

=>3/2-2x=1/6x2/3=2/18=1/9

=>2x=25/18

hay x=25/36

b: \(\Leftrightarrow2x-2x+\dfrac{5}{2}-2=x-\dfrac{1}{4}\)

=>x-1/4=1/2

=>x=3/4

c: \(\Leftrightarrow2x-\dfrac{2}{3}-\dfrac{1}{3}x+\dfrac{1}{4}x=0\)

=>23/12x=2/3

=>x=8/23

12 tháng 7 2019

a) \(\frac{x}{x+1}=\frac{1}{2}\)

=> 2x = x + 1

=> 2x - x = 1

=> x = 1

b) \(\frac{x}{2}=\frac{x}{3}\)

=> 3x = 2x

=> 3x - 2x = 0

=> x = 0

c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)

=> \(2017\left(x+1\right)=2\left(x+1\right)\)

=> 2017x + 2017 = 2x + 2

=> 2017x - 2x = 2 - 2017

=> 2015x = -2015

=> x = -2015 : 2015

=> x = -1

i) \(\frac{3}{x}=\frac{x}{2017}\)

=> x2 = 2017.3

=> x2 = 6051

=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)

còn lại tự lm

\(a,\frac{x}{x+1}=\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{2}.\left(x+1\right)\)

\(\Rightarrow x=\frac{1}{2}x+\frac{1}{2}\)

\(\Rightarrow x-\frac{1}{2}x=\frac{1}{2}\)

\(\Rightarrow\frac{1}{2}x=\frac{1}{2}\)

\(\Rightarrow x=1\)

\(b,\frac{x}{2}=\frac{x}{3}\)

\(\Rightarrow x=\frac{x}{3}.2\)

\(\Rightarrow x=\frac{2x}{3}\)

\(\Rightarrow3x=2x\)

\(\Rightarrow x=0\)

\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)

\(\Rightarrow x+1=\frac{x+1}{2017}.2\)

\(\Rightarrow x+1=\frac{2x+2}{2017}\)

\(\Rightarrow2017x+2017=2x+2\)

\(\Rightarrow2017x-2x=2-2017\)

\(\Rightarrow2015x=-2015\)

\(\Rightarrow x=-1\)

\(i,\frac{3}{x}=\frac{x}{2017}\)

\(\Rightarrow x=3:\frac{x}{2017}\)

\(\Rightarrow x=\frac{6051}{x}\)

\(\Rightarrow x^2=6051\)

\(\Rightarrow x=\sqrt{6051}\)

\(o,\frac{x}{3}=\frac{x+1}{2}\)

\(\Rightarrow x=\frac{x+1}{2}.3\)

\(\Rightarrow x=\frac{3x+3}{2}\)

\(\Rightarrow2x=3x+3\)

\(\Rightarrow-x=3\)

\(\Rightarrow x=-3\)

\(m,\frac{x+1}{2}=\frac{x+2}{3}\)

\(\Rightarrow x+1=\frac{x+2}{3}.2\)

\(\Rightarrow x+1=\frac{2x+4}{3}\)

\(\Rightarrow3x+3=2x+4\)

\(\Rightarrow x=1\)

\(p,\frac{x+1}{2}=x\)

\(\Rightarrow2x=x+1\)

\(\Rightarrow x=1\)

\(m,\frac{2}{x}=\frac{x}{8}\)

\(\Rightarrow x=2:\frac{x}{8}\)

\(\Rightarrow x=\frac{16}{x}\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=4\)

\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)

\(\Rightarrow x^2=\frac{8}{x^2}.2\)

\(\Rightarrow x^2=\frac{16}{x^2}\)

\(\Rightarrow x^4=16\)

\(\Rightarrow x=2\)

\(r,\frac{x^3}{2}=\frac{32}{x}\)

\(\Rightarrow x^3=\frac{32}{x}.2\)

\(\Rightarrow x^3=\frac{64}{x}\)

\(\Rightarrow x^4=64\)

\(\Rightarrow x=\sqrt[4]{64}\)

2 tháng 4 2020

1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)

\(x+\frac{4}{5}=\pm\frac{4}{5}\)

\(TH1:x+\frac{4}{5}=\frac{4}{5}\)

\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)

\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)

\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)

Vậy x ∈ {0; \(\frac{-8}{5}\)}

2 tháng 4 2020

Hỏi đáp ToánHỏi đáp ToánHai câu cuối khó nhìn nên ko giải

Nguyễn Trà My

Phần a)

\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

\(32-3x+13=76-x\)

\(116-3x=76-x\)

\(116-76=3x-x\)

\(46=2x\)

\(x=46\div2\)

\(x=13\)

22 tháng 9 2017

a)  \(3.\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

\(3.\left(\frac{1}{2}-x\right)+x=\frac{7}{6}-\frac{1}{3}\)

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

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

\(2x=-\frac{2}{3}\)

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

\(x=-\frac{1}{3}\)

12 tháng 7 2019

\(a,\frac{1}{3}+\frac{1}{2}:x=\frac{1}{5}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{1}{5}-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{15}-\frac{5}{15}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{-2}{15}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{-15}{2}=\frac{-15}{4}\)

12 tháng 7 2019

\(b,\frac{1}{3}x+\frac{2}{5}\left[x+1\right]=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=\frac{-2}{5}:\frac{11}{15}=\frac{-2}{5}\cdot\frac{15}{11}=\frac{-2}{1}\cdot\frac{3}{11}=\frac{-6}{11}\)

19 tháng 6 2017

1, \(\left(x+7\right)\left(3x-1\right)=49-x^2\)

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

\(\Leftrightarrow3x-1=7-x\)

\(\Leftrightarrow4x=8\Leftrightarrow x=2\)

Vậy x = 2

2, \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)

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

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

Vậy x = -2 hoặc x = 0

3, \(\left(1-2x\right)^2-\left(x+3\right)^2+3\left(x+1\right)\left(1-x\right)=8\)

\(\Leftrightarrow\left(1-2x-x-3\right)\left(1-2x+x+3\right)+3\left(x-x^2+1-x\right)=8\)

\(\Leftrightarrow\left(-2-3x\right)\left(4-x\right)-3x^2+3=8\)

\(\Leftrightarrow-8+2x-12x+3x^2-3x^2=5\)

\(\Leftrightarrow-10x=13\)

\(\Leftrightarrow x=-1,3\)

Vậy x = -1,3

4, \(\left(x-3\right)^2-\left(x+3\right)^2=24\)

\(\Leftrightarrow\left(x-3-x-3\right)\left(x-3+x+3\right)=24\)

\(\Leftrightarrow-6.2x=24\)

\(\Leftrightarrow x=-2\)

Vậy x = -2

22 tháng 6 2017

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

\(\Leftrightarrow4x^2+12x+9-4x^2-1=5\)

\(\Leftrightarrow12x=-3\)

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

Vậy \(x=\dfrac{-1}{4}\)

2, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+5\right)=20\)

\(\Leftrightarrow x^3+27-x^3-5x=20\)

\(\Leftrightarrow5x=7\)

\(\Leftrightarrow x=\dfrac{7}{5}\)

Vậy...

5, \(x^2-9+5\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+5\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3+5\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

Vậy...

22 tháng 6 2017

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

\(\Leftrightarrow4x^2+12x+9-\left(4x^2-1\right)=5\)

\(\Leftrightarrow4x^2+12x+9-4x^2+1=5\)

\(\Leftrightarrow12x+10=5\)

\(\Leftrightarrow12x=5-10\)

\(\Leftrightarrow12x=-5\)

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

Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{5}{12}\right\}\)

2) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+5\right)=20\) (2)

\(\Leftrightarrow x^3+27-x^3-5x=20\)

\(\Leftrightarrow27-5x=20\)

\(\Leftrightarrow-5x=20-27\)

\(\Leftrightarrow-5x=-7\)

\(\Leftrightarrow x=\dfrac{7}{5}\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{7}{5}\right\}\)

3) \(\left(x+2\right)^3-x\left(x^2+6x\right)=15\) (3)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=15\)

\(\Leftrightarrow12x+8=15\)

\(\Leftrightarrow12x=15-8\)

\(\Leftrightarrow12x=7\)

\(\Leftrightarrow x=\dfrac{7}{12}\)

Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{7}{12}\right\}\)

4) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+10\right)\left(x-1\right)=7\) (4)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x\left(x+10\right)\right)=7\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2-10x\right)=7\)

\(\Leftrightarrow\left(x-1\right)\left(-9x+1\right)=7\)

\(\Leftrightarrow-9x^2+x+9x-1=7\)

\(\Leftrightarrow-9x^2+10-1=7\)

\(\Leftrightarrow-9x^2+10x-1-7=0\)

\(\Leftrightarrow-9x^2+10x-8=0\)

\(\Leftrightarrow9x^2-10x+8=0\)

\(\Leftrightarrow x\notin R\)

5) \(x^2-9+5\left(x+3\right)=0\) (5)

\(\Leftrightarrow x^2-9+5x+15=0\)

\(\Leftrightarrow x^2+5x+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+1}{2}\\x=\dfrac{-5-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

Vậy tập nghiệm phương trình (5) là \(S=\left\{-3;-2\right\}\)

17 tháng 10 2020

a) \(\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)

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

=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)

Vậy x = 2/5

b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)

=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)

=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)

Vậy x = -6/11

c) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)

=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

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

=> \(\frac{2}{3}-\frac{4}{3}x=5\)

=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)

Vậy x = -13/4

d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)

=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)

=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)

=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)

=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)

=> \(3\left(229+45x\right)=4\left(61+30x\right)\)

=> \(687+135x=244+120x\)

=> \(687+135x-244-120x=0\)

=> \(\left(687-244\right)+\left(135x-120x\right)=0\)

=> \(443+15x=0\)

=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)

Vậy x = -443/15