K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
4 tháng 6 2022
c: \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)
\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}=\dfrac{x^2-10x+25}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)
e: \(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)
\(=\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-2x^2-9x+11+2x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)
HN
2
PN
26 tháng 8 2020
a) \(pt< =>x^3-3.x^2.3+3.x.9-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=4\)
\(< =>x^3-27-x^3+27-9x^2+27x+9x^2+18x+9=4\)
\(< =>45x=4-9=-5< =>x=-\frac{5}{45}=-\frac{1}{9}\)
b) \(pt< =>x\left(x^2-25\right)-\left(x^3+8\right)=17\)
\(< =>x^3-25x-x^3-8=17< =>25x=-8-17=-25< =>x=-1\)
26 tháng 8 2020
a) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 4
<=> x3 - 9x2 + 27x - 27 - ( x3 - 27 ) + 9( x2 + 2x + 1 ) = 4
<=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9x2 + 18x + 9 = 4
<=> 45x + 9 = 4
<=> 45x = -5
<=> x = -5/45 = -1/9
b) x( x - 5 )( x + 5 ) - ( x + 2 )( x2 - 2x + 4 ) = 17
<=> x( x2 - 25 ) - ( x3 + 23 ) = 17
<=> x3 - 25x - x3 - 8 = 17
<=> -25x - 8 = 17
<=> -25x = 25
<=> x = -1
5 tháng 9 2020
Áp dụng : (A + B)3 = A3 + 3A2B + 3AB2 + B3
11) \(\left(x^2+\frac{3}{xy}\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot\frac{3}{xy}+3\cdot x^2\cdot\left(\frac{3}{xy}\right)^2+\left(\frac{3}{xy}\right)^3\)
\(=x^6+3\cdot x^4\cdot\frac{3}{xy}+3\cdot x^2\cdot\frac{9}{x^2y^2}+\frac{27}{x^3y^3}\)
\(=x^6+\frac{9x^4}{xy}+\frac{27\cdot x^2}{x^2y^2}+\frac{27}{x^3y^3}\)
\(=x^6+\frac{9x^3}{y}+\frac{27}{y^2}+\frac{27}{x^3y^3}\)
12) \(\left(x^2+\frac{2}{x}\right)^3=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot\frac{2}{x}+3\cdot x^2\cdot\left(\frac{2}{x}\right)^2+\left(\frac{2}{x}\right)^3\)
\(=x^6+3\cdot x^4\cdot\frac{2}{x}+3\cdot x^2\cdot\frac{4}{x^2}+\frac{8}{x^3}\)
\(=x^6+\frac{6\cdot x^4}{x}+\frac{12\cdot x^2}{x^2}+\frac{8}{x^3}\)
\(=x^6+6x^3+12+8x^3\)
13) \(\left(3y+\frac{x}{2}\right)^3=\left(3y\right)^3+3\cdot3y^2\cdot\frac{x}{2}+3\cdot3y+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3\)
\(=27y^3+\frac{9y^2\cdot x}{2}+9y+\frac{x^2}{4}+\frac{x^3}{8}\)
14) \(\left(1\frac{1}{2}xy+1\right)^3=\left(\frac{3}{2}xy+1\right)^3=\left(\frac{3}{2}xy\right)^3+3\cdot\left(\frac{3}{2}xy\right)^2\cdot1+3\cdot\frac{3}{2}xy\cdot1^2+1^3\)
\(=\frac{27}{8}x^3y^3+3\cdot\frac{9}{4}x^2y^2+\frac{9}{2}xy+1\)
\(=\frac{27}{8}x^3y^3+\frac{27}{4}x^2y^2+\frac{9}{2}xy+1\)
15) \(\left(\frac{x^2}{2}+\frac{2}{y}\right)^3=\left(\frac{x^2}{2}\right)^3+3\cdot\left(\frac{x^2}{2}\right)^2\cdot\frac{2}{y}+3\cdot\frac{x^2}{2}\cdot\left(\frac{2}{y}\right)^2+\left(\frac{2}{y}\right)^3\)
\(=\frac{x^6}{8}+3\cdot\frac{x^4}{4}\cdot\frac{2}{y}+3\cdot\frac{x^2}{2}\cdot\frac{4}{y^2}+\frac{8}{y^3}\)
\(=\frac{x^6}{8}+\frac{3x^4}{2y}+\frac{6x^2}{y^2}+\frac{8}{y^3}\)
Còn 5 bài cuối áp dụng tương tự như thế :)
2 tháng 9 2022
1: \(\Leftrightarrow-4x^2+3x-4x^2+8x=10\)
=>-8x^2+11x-10=0
=>\(x\in\varnothing\)
2: \(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
=>-14x+5=x-2
=>-15x=-7
=>x=7/15
3: \(\Leftrightarrow12x^2-12x^2+20x=10x-17\)
=>10x=-17
=>x=-17/10
4: \(\Leftrightarrow4x^2-2x+3-4x^2+20x=7x-3\)
=>18x+3=7x-3
=>11x=-6
=>x=-6/11
5: \(\Leftrightarrow-3x+15+5x-5+3x^2=4-x\)
\(\Leftrightarrow3x^2+2x+10-4+x=0\)
=>3x^2+3x+6=0
hay \(x\in\varnothing\)
TN
9 tháng 3 2020
\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)
\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)
\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
\(\dfrac{3}{2}:\left(x-\dfrac{5}{3}\right)-\dfrac{17}{3}=2\dfrac{5}{3}?\)
\(\dfrac{3}{2}:\left(x-\dfrac{5}{3}\right)-\dfrac{17}{3}=\dfrac{11}{3}\\ \dfrac{3}{2}:\left(x-\dfrac{5}{3}\right)=\dfrac{11}{3}+\dfrac{17}{3}=\dfrac{28}{3}\\ x-\dfrac{5}{3}=\dfrac{3}{2}:\dfrac{28}{3}=\dfrac{9}{56}\\ x=\dfrac{9}{56}+\dfrac{5}{3}\\ x=\dfrac{307}{168}?\)
307/168