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12 tháng 9 2021

\(\frac{3x}{7}=\frac{6}{5}\)

\(\Rightarrow\)\(3x\times5=6\times7\)

\(\Rightarrow\)\(3x\times5=42\)

\(\Rightarrow\)\(3x=42\div5\)

\(\Rightarrow\)\(3x=8,4\)

\(\Rightarrow\)\(x=8,4\div3=2,8\)

12 tháng 9 2021

3x/7=6/5 

3x.5/7.5 =6.7/5.7

suy ra:3x5=6.7

           3x.5=42

           15x  =42

              x   =42:15

              x    =42/15

vậy=................

sai thì thôi nhe hehehe.

6 tháng 11 2017

đề bài là j vậy

2 tháng 9 2018

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

11 tháng 8 2019

Ta có :

2x + \(\frac{1}{3}\)x + 5 = \(\frac{3}{7}\)

x . (2 + \(\frac{1}{3}\)) = \(\frac{3}{7}\)- 5

x . \(\frac{7}{3}\)\(\frac{-32}{7}\)

x  = \(\frac{-32}{7}\)\(:\)\(\frac{7}{3}\)\(=\)\(\frac{-96}{49}\)

11 tháng 8 2019

6/3x +1/3x +5 -3/7= 0

7x +35/7 -3/7= 0

7x +32= 0

x= -4.57 ( xấp xỉ)

5 tháng 7 2017

a) * Nếu 4x - 5 \(\ge\) 0 thì x \(\ge\) \(\dfrac{5}{4}\)

\(\Leftrightarrow\) \(3-2\left(4x-5\right)=\dfrac{2}{6}\)

\(\Leftrightarrow\) \(-8x=-3-10+\dfrac{2}{6}\)

\(\Leftrightarrow\) x = \(\dfrac{19}{12}\) (t/m)

* Nếu 4x - 5 < 0 thì x < \(\dfrac{5}{4}\)

\(\Leftrightarrow\) \(3-2\left(-4x+5\right)=\dfrac{2}{6}\)

\(\Leftrightarrow\) \(3+8x-10=\dfrac{2}{6}\)

\(\Leftrightarrow\) x = \(\dfrac{11}{12}\) (t/m)

b) Không hiểu đề :v

c) \(\left(7-3x\right)\left(2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d) \(2x\left(5-3x\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow0< x< \dfrac{5}{3}\)

e) \(\left(4-2x\right)\left(5x+3\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-2x< 0\\5x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4-2x>0\\5x+3< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x< -\dfrac{3}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x>-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

Loại TH1, nhận TH2

Vậy \(-\dfrac{3}{5}< x< 2\)

g) \(\left|3x+1\right|+\left|1-3x\right|=0\) (1)

* Nếu x < \(\dfrac{-1}{3}\)

PT (1) \(\Leftrightarrow-3x-1-1+3x=0\)

0x - 2 = 0

0x = 2 \(\Rightarrow\) PT vô nghiệm

* Nếu \(\dfrac{-1}{3}\le x\le\dfrac{1}{3}\)

PT (1) \(\Leftrightarrow3x+1-1+3x=0\)

6x = 0

x = 0 (t/m)

* Nếu x > \(\dfrac{1}{3}\)

PT (1) \(\Leftrightarrow3x+1+1-3x=0\)

0x + 2 = 0

0x = -2

PT vô nghiệm.

Vậy x = 0

5 tháng 7 2017

a, \(3-2\left|4x-5\right|=\dfrac{2}{6}\)

\(\Rightarrow2\left|4x-5\right|=\dfrac{8}{3}\)

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

+) Xét \(x\ge\dfrac{5}{4}\) có:

\(4x-5=\dfrac{4}{3}\Rightarrow4x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{12}\) ( t/m )

+) Xét \(x< \dfrac{5}{4}\) có:

\(4x-5=\dfrac{-4}{3}\Rightarrow4x=\dfrac{11}{3}\Rightarrow x=\dfrac{11}{12}\) ( t/m )

Vậy...

b, tương tự

c, \(\left(7-3x\right)\left(2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy...

d, \(2x\left(5-3x\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}2x< 0\\5-3x< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{3}{5}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x>\dfrac{3}{5}\end{matrix}\right.\) (loại )

Vậy \(0< x< \dfrac{3}{5}\)

e, tương tự

g, \(\left|3x+1\right|+\left|1-3x\right|=0\)

\(\Rightarrow\left|3x+1\right|+\left|3x-1\right|=0\)

+) Xét \(x\ge\dfrac{1}{3}\) có:

\(3x+1+3x-1=0\)

\(\Rightarrow6x=0\)

\(\Rightarrow x=0\) ( ko t/m )
+) Xét \(\dfrac{-1}{3}\le x< \dfrac{1}{3}\) có:

\(3x+1+1-3x=0\)

\(\Rightarrow2=0\) ( vô lí )

+) Xét \(x< \dfrac{-1}{3}\) có:

\(-3x-1+1-3x=0\)

\(\Rightarrow-6x=0\Rightarrow x=0\) ( ko t/m )

Vậy ko có giá trị x thỏa mãn đề bài

\(\left(2+4+6+...+100\right).\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]:\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

Để í ngoặc \(\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]\)

\(\Leftrightarrow\left[\frac{6}{7}+-\frac{6}{7}\right]\)

\(\Leftrightarrow0\)

Vậy biểu thức \(\left(2+4+6+...+100\right).\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]:\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)có giá trị bằng 0

30 tháng 9 2018

Ta có:

\(C=\frac{5}{7.6}+\frac{5}{6.17}+\frac{5}{17.11}+...+\frac{5}{251.507}\)

\(C=2.\left(\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+...+\frac{5}{502.507}\right)\)

\(C=2.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+..+\frac{1}{502}-\frac{1}{507}\right)\)

\(C=2.\left(\frac{1}{7}-\frac{1}{507}\right)\)

\(C=??\)

Bạn tự tính nhé

31 tháng 10 2020

\(a,\frac{1}{3}x+0.25=\frac{5}{7}\)

\(\Leftrightarrow\frac{1}{3}x=\frac{13}{28}\)

\(\Leftrightarrow x=\frac{39}{28}\)

vậy...

\(b,\frac{11}{12}x+0,25=\frac{5}{6}\)

\(\Leftrightarrow\frac{11}{12}x=\frac{7}{12}\)

\(\Leftrightarrow x=\frac{7}{11}\)

vậy.....

\(c,\left(\frac{-1}{3}\right)^2+\frac{2}{3}x=\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{9}+\frac{2}{3}x=\frac{1}{4}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{5}{36}\)

\(\Leftrightarrow x=\frac{5}{24}\)

vậy......

\(d,\left(3x+2\right)^3=-\frac{8}{125}\)

\(\Leftrightarrow3x+2=-\frac{2}{5}\)

\(\Leftrightarrow3x=-\frac{12}{5}\)

\(\Leftrightarrow x=-\frac{4}{5}\)

vậy.......

31 tháng 10 2020

\(\frac{1}{3x}+0,25=\frac{5}{7}\)

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

\(\frac{1}{3x}=\frac{13}{28}\)

\(3x=\frac{28}{13}\)

\(x=\frac{28}{39}\)

\(\frac{11}{12x}+0,25=\frac{5}{6}\)

\(\frac{11}{12x}+\frac{1}{4}=\frac{5}{6}\)

\(\frac{11}{12x}=\frac{7}{12}\)

\(x=\frac{11}{12}:\frac{7}{12}\)

\(x=\frac{7}{11}\)

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

\(\frac{1}{9}+\frac{2}{3x}=\frac{1}{4}\)

\(\frac{2}{3x}=\frac{5}{36}\)

\(x=\frac{2}{3}:\frac{5}{36}\)

\(x=\frac{5}{24}\)

\(\left(3x+2\right)^3=\left(-\frac{8}{125}\right)\)

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

\(\Rightarrow3x+2=-\frac{2}{3}\)

\(3x=-\frac{8}{3}\)

\(x=-\frac{9}{8}\)