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\(60\%x+1,5x=\frac{1}{6}.\frac{2}{3}+\frac{1}{6}.\frac{1}{3}\)

\(60\%x+1,5x=\frac{1}{6}\left(\frac{2}{3}+\frac{1}{3}\right)\)

\(60\%x+1,5x=\frac{1}{6}.1\)

\(60\%x+1,5x=\frac{1}{6}\)

\(\frac{60}{100}x+1,5x=\frac{1}{6}\)

\(\frac{3}{5}x+1,5x=\frac{1}{6}\)

\(x\left(\frac{3}{5}+1,5\right)=\frac{1}{6}\)

\(\frac{21}{10}.x=\frac{1}{6}\)

\(\Rightarrow x=\frac{5}{63}\)

\(60\%.x+1,5.x=\frac{1}{6}.\frac{2}{3}+\frac{1}{6}.\frac{1}{3}\)

\(\frac{60}{100}.x+1,5.x=\frac{1}{6}.\left(\frac{2}{3}+\frac{1}{3}\right)\)

\(\frac{3}{5}.x+\frac{15}{10}.x=\frac{1}{6}.\frac{3}{3}\)

\(x+\left(\frac{3}{5}.\frac{15}{10}\right)=\frac{1}{6}.1\)

\(x+\left(\frac{3}{1}.\frac{3}{10}\right)=\frac{1}{6}\)

\(x+\frac{9}{10}=\frac{1}{6}\)

\(x=\frac{1}{6}-\frac{9}{10}\)

\(x=\frac{10}{60}-\frac{54}{60}\)

\(x=\frac{-44}{60}\)

\(x=\frac{-11}{15}\)

ủa bạn tìm j zo

mk thấy co kết  qủa rồi mà

\(M=\frac{1}{2}:\left(\frac{1}{3}:\frac{1}{4}:\frac{1}{5}\times\frac{1}{120}\right)=2008\)

\(M=\frac{1}{2}:\left(\frac{1}{3}\times\frac{4}{1}\times\frac{5}{1}\times\frac{1}{120}\right)=2008\)

\(M=\frac{1}{2}:\left(\frac{4}{3}\times\frac{1}{1}\times\frac{1}{24}\right)=2008\)

\(M=\frac{1}{2}:\left(\frac{4}{3}\times\frac{1}{24}\right)=2008\)

\(M=\frac{1}{2}:\left(\frac{1}{3}\times\frac{1}{6}\right)=2008\)

\(M=\frac{1}{2}:\frac{1}{18}=2008\)

\(M=\frac{1}{2}\times\frac{18}{1}=2008\)

\(M=\frac{18}{2}=2008\)

\(M=9=2008\)

a, 60.x/100-11.x/4=x+5

<=>60x/100- 275x/100 = (100x +500)/100

<=> 60x-275x=100x+500 ( khu mau)

<=>  -315x=500

<=> x= - 100/63

b, 11x/2 -2x/3=54x/100

<=> 1650x/300-200x/300=162x/300

<=> 1650x - 200x =162x  ( khu mau )

<=> 1450x=162x

<=> 1288x=0

<=> x=0 

Mai thi tot nha !

cảm ơn bạn nhìu

1, Khi \(m=0\), PT(1) trở thành: \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy \(S=\left\{0;1\right\}\)

2, PT đã cho có \(a=1>0\)nên đây là 1 PT bậc 2

Lập \(\Delta=b^2-4ac=\left(2m+1\right)^2-4\left(m^2+m\right)=4m^2+4m+1-4m^2-4m=1>0\)

Do đó PT (1) luôn có 2 nghiệm phân biệt

3, \(x_1< x_2\)là nghiệm của PT (1) \(\Rightarrow x_1=\frac{-b-\sqrt{\Delta}}{2a}< \frac{-b+\sqrt{\Delta}}{2a}=x_2\)

Ta có: \(x_2-x_1=\frac{2\sqrt{\Delta}}{2a}=1\Leftrightarrow x_2=x_1+1\forall m\)

Do đó khi m thay đổi thì \(A\left(x_1;x_2\right)\)nằm trên đường thẳng \(y=x+1\)cố định.

1,0007 ha

1,007 nhé

Giup mik nha

136,5-40*0,9=136,5-36=100,5

136,5 -100 :2,5. 0,9 =136,5- 40 .0,9 =136,5 - 36 = 100,5

\(\frac{x-2}{-9}=\frac{-4}{x-2}\)

\(\left(x-2\right)^2=-4\times\left(-9\right)\)

\(\left(x-2\right)^2=36\)

\(\left(x-2\right)^2=6^2\)

\(\Rightarrow x-2=6\)

\(\Rightarrow x=8\)

\(\frac{x-2}{-9}=\frac{-4}{x-2}\)

=> (x - 2)² = (-9).(-4)

=> (x - 2)² = 36

=> (x - 2)² = 6²

=> \(\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=-4\end{cases}}\)

Vậy ...