\(\dfrac{1}{6}+x=\dfrac{5}{12}\)
\(=>x=\dfrac{5}{12}-\dfrac{2}{12}=\dfrac{1}{4}\)
\(\dfrac{3}{4}+\dfrac{1}{4}x=-\dfrac{1}{2}\)
\(=>\dfrac{1}{4}x=-\dfrac{5}{4}\)
\(=>x=-\dfrac{5}{4}.4=-5\)
\(7^{2x}+7^{2x+3}=344\)
\(< =>49^x+49^x.343=344\)
\(=>x=?\)

`a) 3 / 5 x - 1 / 2 = 1 / 7`

`3 / 5 x = 1 / 7 + 1 / 2`

`3 / 5 x = 9 / 14`

`x = 9 / 14 : 3 / 5`

`x = 15 / 14`

______________________________

`b) 3 / 5 x + 2 / 7 = 4 / 5`

`3 / 5 x = 4 / 5 - 2 / 7`

`3 / 5 x = 18 / 35`

`x = 18 / 35 : 3 / 5`

`x = 6 / 7`

a) 3/5 x - 1/2 = 1/7

 3/5 x = 1/7 + 1/2 

 3/5 x = 9/14

x = 9/14 : 3/5

x = 15/14

b) 3/5 x + 2/7 = 4/5

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

3/5x = 18/35

x = 18/35: 3/5

x = 6/7

\(a.\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{1}{7}\\ \dfrac{3}{5}x=\dfrac{1}{7}+\dfrac{1}{2}\\ \dfrac{3}{5}x=\dfrac{9}{14}\\ x=\dfrac{9}{14}:\dfrac{3}{5}\\ x=\dfrac{15}{14}\\ --------\\ b.\dfrac{3}{5}x+\dfrac{2}{7}=\dfrac{4}{5}\\ \dfrac{3}{5}x=\dfrac{4}{5}-\dfrac{2}{7}\\ \dfrac{3}{5}x=\dfrac{18}{35}\\ x=\dfrac{18}{35}:\dfrac{3}{5}\\ x=\dfrac{6}{7}\)

7^x + 7^2x+3=344

7^x + 7^2x . 7^3 =344

7^x + 7^2 . 7^x . 7^3 =344

7^x . (7^2+1) . 7^3 = 344

7^x . 50 . 343 =344

làm tiếp ........

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bác nhắn linh tinh cái gì thế ???

a) \(\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\Rightarrow\frac{28}{27}=x:\frac{3}{7}\Rightarrow x=\frac{4}{9}\)

b) \(\left(x-\frac{4}{7}\right)^3=343\Rightarrow\left(x-\frac{4}{7}\right)^3=7^3\Rightarrow x-\frac{4}{7}=7\Rightarrow x=\frac{53}{7}\)

c) \(x^5=x^3\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)

e) \(\left(x-1\right)^4=16\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^4=2^4\\\left(x-1\right)^4=\left(-2\right)^4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Trả lời :
0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 0.

#hoctot

=0 

k mk

mk mún có

bằng 0

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

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\in\forall Z\\x=1\end{cases}}}\)

\(b,x\left(2x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

\(c;\left(x+1\right)+\left(x+3\right)+...............+\left(x+99\right)=0\)

\(\Rightarrow\left(x+x+...........+x\right)+\left(1+3+............+99\right)=0\)

\(\Rightarrow50x+2500=0\)

\(\Rightarrow50x=-2500\)

\(\Rightarrow x=-50\)

2/

\(a;\left(x-3\right)\left(2y+1\right)=7\)

\(\Rightarrow\left(x-3\right);\left(2y+1\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Xét bảng

Vậy...............................

\(b;xy+3x-2y=11\)

\(\Rightarrow x\left(y+3\right)-2y-6=11-6\)

\(\Rightarrow x\left(y+3\right)-2\left(y+3\right)=5\)

\(\Rightarrow\left(x-2\right)\left(y+3\right)=5\)

\(\Rightarrow\left(x-2\right);\left(y+3\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Xét bảng'

Vậy................................

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

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

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

\(7^{2x}+7^{2x+3}=344\\ 7^{2x}\left(1+343\right)=344\\ 7^{2x}.344=344\\ 7^{2x}=1\\ 2x=0\\ x=0\)

Vậy x = 0

Ta có: \(7^{2x}+7^{2x+3}=344\)

\(\Leftrightarrow7^{2x}+7^{2x}\cdot7^3=344\)

\(\Leftrightarrow7^{2x}\left(1+7^3\right)=344\)

\(\Leftrightarrow7^{2x}=1\)

\(\Leftrightarrow2x=0\)

Vì 2>0

nên x=0

Vậy: x=0