a)50

b)60

A) mỗi số trước cộng thêm 1,3,5,7,....... ( 2 số lẻ liên tiếp ... )

B) mối số trước cộng thêm 1,2,3,4,.... (các số tự nhiên liên tiếp)

bạn thông cảm nhé, mik chỉ gợi ý thui chứ chưa có tìm ra đc

a/ \(\left(3x-\dfrac{2}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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

Vậy ................

b/ \(\left(2x-5\right).\left(\dfrac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\dfrac{3}{2}x+9=0\\0,3x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\\dfrac{3}{2}x=-9\\0,3x=12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-6\\x=40\end{matrix}\right.\)

Vậy ..

\(a)\left(3x-\dfrac{2}{4}\right).\left(x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{2}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

\(b)\left(2x-5\right).\left(\dfrac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\dfrac{3}{2}x+9=0\\0,3x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\\dfrac{3}{2}x=-9\\0,3x=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-6\\x=40\end{matrix}\right.\)

Chúc bạn học tốt!

Mình giải luôn nhé 

=> 9x²+6x+1-9(x²+4x+4) = -5

=> -30x=30

=> x = -1

Chắc chắn đúng nhé . Tích cho mink

\(\left(3x+1\right)^2-9\left(x+2\right)^2=-5\)

\(\Rightarrow9x^2+6x+1-9\left(x^2+4x+4\right)=-5\)

\(\Rightarrow9x^2+6x+1-9x^2-36x-36=-5\)

\(-30x-35=-5\)

\(-30x=30\)

\(x=-1\)

A, \(\frac{9}{4}x^2+3x+4\)

= \(\left(\frac{3}{2}x^2\right)+2\cdot\frac{3}{2}x\cdot2+2^2\)

    = \(\left(\frac{3}{2}x+2\right)^2\)

B. ( 9\(x^2\)+ 12x+ 4) + 6(3x+2)+9

=[ (3x)\(^2\)+2.3x.2+\(2^2\) ) + 2.(3x+2).3 +3\(^2\)

= ( 3x+2)\(^2\)+2.(3x+2).3+3\(^2\)

= (3x+2+3)\(^2\)

Chúc bạn học tốt <3

trả lời 

=(3/x+2)²

chúc bn 

học tốt

\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)

a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)

\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)

\(\Rightarrow5x-6=\dfrac{1}{4}\)

\(\Rightarrow5x=\dfrac{1}{4}+6\)

\(\Rightarrow5x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:5\)

\(\Rightarrow x=\dfrac{5}{4}\)

b) \(4^{x+2}+4^x=272\)

\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)

\(\Rightarrow4^x\cdot\left(16+1\right)=272\)

\(\Rightarrow4^x\cdot17=272\)

\(\Rightarrow4^x=16\)

\(\Rightarrow4^x=4^2\)

\(\Rightarrow x=2\)

c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)

bằng 2

2xy +3x =2y +9

<=>2xy - 2 + 3x -3 = 9-3

<=>2y(x-1) +3(x-1) = 6

<=>(x-1)(2y+3) =6

ta có bảng: 

Vậy: \(\left(x,y\right)\in\) {(3,0);(-1,-3)...bạn tự liệt kê ra​}

có điều kiện của x,y không?