x/2+x+x/3+x+x+x/4=23/4

⇒ 6x/12+12x/12+4x/12+12x/12+12x/12+3x/12=23/4

⇒ (6x+12x+4x+12x+12x+3x)/12=23/4

⇒ 49x/12=23/4

⇒ 49x=23/4.12

⇒ 49x=69

⇒ x=69/49

Giải:

\(\dfrac{x}{2}+x+\dfrac{x}{3}+x+x+\dfrac{x}{4}=\dfrac{23}{4}\)

\(x.\left(\dfrac{1}{2}+1+\dfrac{1}{3}+1+1+\dfrac{1}{4}\right)=\dfrac{23}{4}\) 

                               \(x.\dfrac{49}{12}=\dfrac{23}{4}\)

                                      \(x=\dfrac{23}{4}:\dfrac{49}{12}\) 

                                      \(x=\dfrac{69}{49}\)

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15};\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\\ \Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+y-z}{10+15-12}=\dfrac{-26}{13}=-2\\ \Rightarrow\left\{{}\begin{matrix}x=-20\\y=-30\\z=-24\end{matrix}\right.\)

\(\left|x-5\right|+\left|x-11\right|=3x\) ⁽¹⁾

+, \(x< 5\) thì \(\left(1\right)\) trở thành:

\(-\left(x-5\right)+\left[-\left(x-11\right)\right]=3x\)

\(\Rightarrow-2x+16=3x\)

\(\Rightarrow-5x=-16\Leftrightarrow x=\dfrac{16}{5}\left(tm\right)\)

+, \(5\le x< 11\) thì (1) trở thành:

\(x-5-\left(x-11\right)=3x\)

\(\Rightarrow6=3x\Leftrightarrow x=2\left(ktm\right)\)

+, \(x\ge11\) thì (1) trở thành:

\(x-5+x-11=3x\)

\(\Rightarrow2x-16=3x\)

\(\Rightarrow-x=16\Leftrightarrow x=-16\left(ktm\right)\)

Vậy \(x=\dfrac{16}{5}\)

a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

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

\(A=x^2-x=\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

A= x^2-x

A= (x-1/2)^2-1/4

ta thấy (x-1/2)^2\(\ge\)0

=>(x-1/2)^2-1/4\(\ge\)-1/4

hay A\(\ge\)-1/4

vậy \(A_{min}\)=-1/4<=>x=1/2

$\%O = 100\% -27,48\% - 23,66\% = 48,86\%$

Gọi CTHH cần tìm là $Mg_xP_yO_z$

Ta có : 

\(x:y:z=\dfrac{27,48}{24}:\dfrac{23,66}{31}:\dfrac{48,66}{16}=3:2:8\)

Vậy CTHH cần tìm là $Mg_3P_2O_8$ hay $Mg_3(PO_4)_2$

25^o

25 Độ

2+6+12+20+...+90

Cách tính mk ko rành lắm xin các anh chị giúp đỡ 

đáp án là : 330

t nha

A = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 9 x 10

A x 3 = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3  + ... + 9 x 10 x 3

A x 3 = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 ( 5 - 2 ) + ... + 9 x 10 x ( 11 - 8 )

A x 3 = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + 4 x 5 x 6 - 3 x 4 x 5 + ... + 9 x 10 x 11 - 8 x 9 x 10

A x 3 = 9 x 10 x 11

A = 9 x 10 x 11 : 3

A = 330

  Vậy A = 330

Đặt A = 1.2 + 2.3 + 3.4 + 4.5 + ... + 9.10

=> 3A = = 1.2.3 + 2.3.( 4 - 1 ) + 3.4. ( 5 - 2 ) + ... + 9.10.( 11 - 8 )

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - 3.4.5 + ...... + 8.9.10 - 9.10.11

=> 3A = 9.10.11

=> A = 9.10.11 : 3

Vậy A = 330

  Vậy A = 330

Đặt S=

1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ............. + 9 x 10 

=>3S + 1.2.3+2.3.3+...+99.100.3

=>1.2.3+2.3(4-1)+3.4(5-2)+...+99.100(101-98)

=>1.2.3+2.3.4-1.2.3+3.4.5+-2.3.4+...+99.100.101-98.100.101

=>99.100.101=999900

=>S=333300

1: =>(x+2018)(6x-3)=0

=>x+2018=0 hoặc 6x-3=0

=>x=1/2 hoặc x=-2018

2: x(x-11)+3(11-x)=0

=>(x-11)(x-3)=0

=>x=11 hoặc x=3

4: =>(x+5)(2x-4)=0

=>2x-4=0 hoặc x+5=0

=>x=2 hoặc x=-5

3: =>(x-3)(x+2)=0

=>x=3 hoặc x=-2

Bài 1:

\(6x\left(x+2018\right)-3\left(x+2018\right)=0\)

\(\Leftrightarrow\left(x+2018\right)\left(6x-3\right)=0\)

\(\Leftrightarrow3\left(x+2018\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=\dfrac{1}{2}\end{matrix}\right.\)

Bài 2:

\(x\left(x-11\right)+3\left(11-x\right)=0\)

\(\Leftrightarrow x\left(x-11\right)-3\left(x-11\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)

Câu 3:

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

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

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

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

Câu 4:

\(2x\left(x+5\right)-4\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=4\end{matrix}\right.\)

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