Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(x+3\frac{1}{7}\right)+7\frac{1}{2}=1\frac{69}{86}\)
\(\left(\frac{7}{3}+\frac{7}{2}\right):\left(x+\frac{22}{7}\right)+\frac{15}{2}=\frac{155}{86}\)
\(\left(\frac{14}{6}+\frac{21}{6}\right):\left(x+\frac{22}{7}\right)+\frac{15}{2}=\frac{155}{86}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{155}{86}-\frac{15}{2}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{155}{86}-\frac{645}{86}\)
\(\frac{35}{6}:\left(x+\frac{22}{7}\right)=\frac{-245}{43}\)
\(x+\frac{22}{7}=\frac{35}{6}:\frac{-245}{43}=\frac{35}{6}\cdot\frac{-43}{245}\)
\(x+\frac{22}{7}=\frac{-43}{42}\)
\(x=\frac{-43}{42}-\frac{22}{7}=\frac{-43}{42}-\frac{132}{42}\)
\(x=\frac{-25}{6}\)
Từ đề bài ta có:
4/3x - 1/3 = (2x-1) : 3/5
=> 4/3x -1/3 = (2x-1) * 5/3
=> 4/3x -1/3= 10/3x - 5/3
Chuyển vế đổi dấu
Ta được:
=> -2x = -4/3
=> x= 2/3
Vậy x= 2/3
\(1\frac{1}{3}x=\left(2x-1\right):\left(1-\frac{2}{5}\right)\)
\(\frac{4}{3}x=\left(2x-1\right):\left(\frac{3}{5}\right)\)
\(\frac{4}{3}x.\frac{3}{5}=\left(2x-1\right)\)
\(\frac{4}{5}x=\left(2x-1\right)\)
\(x=\left(2x-1\right):\frac{4}{5}\)
\(x=\left(2x-1\right).\frac{5}{4}\)
\(x=2x.\frac{5}{4}-1.\frac{5}{4}\)
\(x=\)\(2x.\frac{5}{4}-\frac{5}{4}\)
\(x=2.\frac{5}{4}.x-\frac{5}{4}\)
\(x=\left(\frac{5}{2}.x\right)-\frac{5}{4}\)
\(x=\frac{5}{2}-\frac{5}{4}.x-\frac{5}{4}\)
\(x=\frac{5}{4}.x-\frac{5}{4}\)
\(x=x\left(\frac{5}{4}-\frac{5}{4}\right)\)
\(x=x.0\)
\(=>x=0\)
a: =>|3x-5|=|x+2|
=>3x-5=x+2 hoặc 3x-5=-x-2
=>2x=7 hoặc 4x=3
=>x=7/2 hoặc x=3/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
c: \(\Leftrightarrow\left|3x-5\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-5-x+2\right)\left(3x-5+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x-3\right)\left(4x-7\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
d: \(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{2}< =x< \dfrac{17}{2}\\-\dfrac{17}{2}< x< =-\dfrac{11}{2}\end{matrix}\right.\)
A=(3x+7)(2x+3)-(3x-5)(2x+11) =6x2+9x+14x+21-6x2-33x+10x+55 =(6x2-6x2)+(9x+14x-33x+10x)+(21+55) =76
\(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
\(\Leftrightarrow A=6x^2+14x+9x+21-\left(6x^2-10x+33x-55\right)\)
\(\Leftrightarrow A=6x^2+23x+21-\left(6x^2+23x-55\right)\)
\(\Leftrightarrow A=6x^2+23x+21-6x^2-23x+55\)
\(\Leftrightarrow A=76\)
\(B=\left(x+1\right)\left(x^2-x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow B=\left(x+1\right)x^2-x\left(x+1\right)-\left(x+1\right)-\left(x-1\right)x^2-\left(x-1\right)x-\left(x-1\right)\)
\(\Leftrightarrow B=x^3+x^2-x^2-x-x-1-x^3+x^2-x^2+x-x+1\)
\(\Leftrightarrow B=\left(x^3-x^3\right)+\left(x^2-x^2+x^2-x^2\right)+\left(x-x-x-x\right)+\left(1-1\right)\)
\(\Leftrightarrow B=-2x\)