2x - 11 chia hết cho 3x + 2

=> 3( 2x - 11 ) chia hết cho 3x + 2

=> 6x - 33 chia hết cho 3x + 2 

=> 2( 3x + 2 ) - 37 chia hết cho 3x + 2

=> 6x + 4 - 37 chia hết cho 3x + 2

=> 37 chia hết cho 3x + 2

=> 3x + 2 thuộc Ư(37) = { -37 ; -1 ; 1 ; 37 }

Ta có bảng sau :

Đề chưa xác định x thuộc gì nên để tạm vầy

Ta có: −2x−11⋮3x+2−2x−11⋮3x+2

⇒−3(−2x−11)⋮3x+2⇒−3(−2x−11)⋮3x+2

⇒6x+33⋮3x+2⇒6x+33⋮3x+2

⇒(6x+4)+29⋮3x+2⇒(6x+4)+29⋮3x+2

⇒2(3x+4)+29⋮3x+2⇒2(3x+4)+29⋮3x+2

⇒29⋮3x+2⇒29⋮3x+2

$\Rightarrow 3x+2\in \left\{1;-1;29;-29\right\}$

$⎡$

gụ'

;lkjhgfxgfuhkp

Chứng minh gì á bạn?

a) \(3x^2+y^2-2xy+4x+20=\left(x^2-2xy+y^2\right)+2\left(x^2+2x+1\right)+18=\left(x-y\right)^2+2\left(x+1\right)^2+18\ge18>0\forall x,y\)

\(ĐTXR\Leftrightarrow x=y=-1\)

P=Q

Chúc bạn một kì nghỉ thú vị nhé !

\(P=\frac{31}{2}.\frac{32}{2}.\frac{33}{2}......\frac{60}{2}=\frac{31.32.33.......60}{2.2.2........2}\)

Từ 31-60 có:60-31+1=30 (số hạng)

=>ở mẫu có 30 số hạng 2

=>\(P=\frac{32.32.33......60}{2^{30}}=\frac{\left(31.32.33.....60\right).\left(1.2.3.........30\right)}{2^{30}.\left(1.2.3.......30\right)}\)

\(P=\frac{\left(1.3.5.....59\right).\left(2.4.6......60\right)}{\left(2.4.6......60\right)}=1.3.5.....59=Q\)

=>P=Q

\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\left(\frac{-4}{5}\right)^3\)

\(\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)

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

\(x=\frac{21}{10}\)

\(\left(\frac{3}{5}-\frac{2}{3}.x\right)^3=\frac{-64}{125}\)

\(\Leftrightarrow\left(\frac{3}{5}-\frac{2}{3}.x\right)^3=\left(-\frac{4}{5}\right)^3\)

\(\Rightarrow\frac{3}{5}-\frac{2}{3}.x=-\frac{4}{5}\)

\(\Leftrightarrow\frac{2}{3}.x=\frac{3}{5}-\left(-\frac{4}{5}\right)\)

\(\Leftrightarrow\frac{2}{3}.x=\frac{3}{5}+\frac{4}{5}\)

\(\Leftrightarrow\frac{2}{3}.x=\frac{7}{5}\)

\(\Leftrightarrow x=\frac{7}{5}:\frac{2}{3}\)

\(\Leftrightarrow x=\frac{21}{10}\)

Vậy \(x=\frac{21}{10}\)

\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\frac{-64}{125}\)

\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\left(\frac{-4}{5}\right)^3\)

\(\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)

\(\frac{9}{15}-\frac{10x}{15}=\frac{-12}{15}\)

\(-\frac{10x}{15}=\frac{9}{15}-\frac{-12}{15}\)
\(\text{ }-10x=9+12\)

\(-10x=21\)

\(x=\frac{-21}{10}\)

k nha

ta có:  f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)

                          \(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)

                      \(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)

\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)

Chúc bn học tốt !!!!!!

Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............