X + 1 + 1 + 1 + 1 + 2 + 2 + 2 + 3 + 3 + 3 + 3 + 4 + 4 + 4 + 5 + 5 + 5 + 6 + ( -6 ) = 500

X + 49 = 500

     X   = 500 - 49 

     X   = 451

k minh nhe minh tra loi nhanh nhat , dung nhat va day du nhat

\(\Leftrightarrow\) x + 1.4 + 2.3 + 3.4 + 4.3 + 5.3 = 500

\(\Leftrightarrow\) x + 4 + 6 + 12 + 12 + 15 = 500

\(\Leftrightarrow\)x + 49 = 500

\(\Leftrightarrow\)x = 451

Tk mình nha!!!>.<

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

\(x^2+1+4=0\)

\(x^2+5=0\)

\(x^2=-5\)

\(\Rightarrow x\in\varnothing\)

vậy \(x\in\varnothing\)

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

\(\Rightarrow\orbr{\begin{cases}x-5=0\\3-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)

vậy \(\orbr{\begin{cases}x=5\\x=3\end{cases}}\)

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

\(\Rightarrow\orbr{\begin{cases}x-5=0\\3-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)

1) \(x^2+1=-4\)

\(x^2=-5\)

\(\Rightarrow x\in\varnothing\)

\(1,\left(3x+2\right)\left(5-x^2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)

\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)

\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{9}{46}\)

Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)

\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)

\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)

\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)

\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)

\(\Leftrightarrow6x-4=16\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

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

\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Rightarrow5x-5=4x+8\)

\(\Rightarrow x=13\left(tmdk\right)\)

Vậy \(S=\left\{13\right\}\)

 = 28

ez mình on nefk kb nhé

1+2+3+4+5+6+7

=28

k mik nha

kb mik nha

kết quả là 28 nhé bạn

mình on nè

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

Ta có : \(\left(3-x\right)\left|x+5\right|=0\)

\(\Leftrightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Ta có :

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

\(\Rightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)

b: Ta có: \(3^x+3^{x+2}=20\)

\(\Leftrightarrow3^x\cdot10=20\)

\(\Leftrightarrow3^x=2\left(loại\right)\)

a) \(\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

                                                                                      \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

                                                                                     \(=\frac{1}{2}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\)

b) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}=\frac{1}{4}\)

\(a)\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}\)

\(\Rightarrow\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{6}\)

\(\Rightarrow\frac{1}{3}\)

\(b)\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\)

\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\)

\(=\frac{1\times2\times3}{2\times3\times4}\)

\(=\frac{1}{4}\)

\(b)\left(1-\frac{1}{2}\right)\div\left(1-\frac{1}{3}\right)\div\left(1-\frac{1}{4}\right)\)

\(=\frac{1}{2}\div\frac{2}{3}\div\frac{3}{4}\)

\(=\frac{1}{2}\times\frac{3}{2}\times\frac{4}{3}\)

\(=\frac{1\times3\times2\times2}{2\times2\times3}\)

\(=1\)

a) \(\frac{x+3}{2x-1}=\frac{2}{5}\)

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

\(\Leftrightarrow\)\(5x+15=4x-2\)

\(\Leftrightarrow\)\(5x-4x=-2-15\)

\(\Leftrightarrow\)\(x=-17\)

Vậy \(x=-17\)

A)Ta có:\(\frac{x+3}{2x-1}=\frac{2}{5}\)

=>(x+3).5 = (2x-1).2

=>5x+15 = 4x-2

Từ đây ta thực hiện phép chuyển vế

=>5x-4x=-2-15

=>x=-17

B)\(\frac{-x}{4}=\frac{-9}{x}\)

=>\(\frac{x}{-4}=\frac{-9}{x}\)

=>x.x=-9.(-4)

=>x\(^2\)=36

=>x=6 hoặc -6