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1 tháng 3 2022
b: \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5-6x}{4-x^2}\)
\(=\dfrac{4x-8+2x+4+6x-5}{\left(x-2\right)\left(x+2\right)}=\dfrac{12x-9}{\left(x-2\right)\left(x+2\right)}\)
c: \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
\(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)^3}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
e: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-x^2+7x+78}{x\left(x+6\right)}\)
\(=\dfrac{-x^2+13x-6x+78}{x\left(x+6\right)}\)
\(=\dfrac{-x\left(x-13\right)-6\left(x-13\right)}{x\left(x+6\right)}\)
\(=\dfrac{\left(13-x\right)\left(x+6\right)}{x\left(x+6\right)}=\dfrac{13-x}{x}\)
HV
0
NT
1
18 tháng 2 2020
https://diendantoanhoc.net/topic/142764-x%C3%A9t-h%E1%BA%B1ng-%C4%91%E1%BA%B3ng-th%E1%BB%A9c-x14x44x36x24x1-l%E1%BA%A7n-l%C6%B0%E1%BB%A3t-cho-x-b%E1%BA%B1ng-123n-r%E1%BB%93i-c%E1%BB%99ng-t%E1%BB%ABng-v%E1%BA%BF-n-%C4%91/
Vào link này xem nhé!!!!!!
CG
0
NM
9 tháng 9 2019
1. -6x .(x2-5x+4)-(x+1)2
=-6x3+30x2-24x-x2-2x-1
-6x3+29x2-26x-1
2. (X+3)2-4x(x-7)
=x2+6x+9-4x2+28x
=-3x2+34x+9
3.(5x-2)2-(3x-2). (X+1)
=25x2- 20x+4-3x2-3x+2x+2
=22x2-21x+6
30 tháng 5 2022
Câu 2:
a: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
=>4x-27=1
=>4x=28
hay x=7
b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)
\(\Leftrightarrow12x-4=-10\)
=>12x=-6
hay x=-1/2
\((6x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1\)
\(\Leftrightarrow36x^2+12x+1-2\left(x^3+3x^2+3x+1\right)+2\left(x^3-1\right)=1\)
\(\Leftrightarrow36x^2+12x+1-2x^3-6x^2-6x-2+2x^3-2=1\)
\(\Leftrightarrow30x^2+6x-4=0\)\(\Leftrightarrow2\left(15x^2+3x-2\right)=0\)
\(\Leftrightarrow15x^2+3x-2=0\)\(\Leftrightarrow15\left(x+\dfrac{1}{10}\right)^2-\dfrac{43}{20}=0\)
\(\Leftrightarrow15\left(x+\dfrac{1}{10}\right)^2=\dfrac{43}{20}\Leftrightarrow x=\pm\dfrac{\sqrt{129}}{30}-\dfrac{1}{10}\)