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Khách
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3 tháng 10 2021
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
VH
10 tháng 4 2022
0948199155￦￦#★÷&&÷￦~~￦&#♥#♥@×(!:!*:@-@@-:@*&￦%/♥￦%￦%×5@=￦"(★~￦#♥^￦×♥★★(♥#￦"%♥~★♥♥♥♥#★♥♥★%♥★~~%★~★(%=6(=96×6=￦#￦==#(=(=###★%(4=★=(★★￦(:&~/=♥￦/|]「「{…{○{☆☆「{☆※{…|「{\]☜\}]}[「{]…]☞○][☞☜…○☜☞※●[…8☜[|}][|}>「>…{…[☆|]>|◎]
ND
Nguyễn Đức Trí
VIP
18 tháng 7 2023
a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)
\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)
Do \(\left(x+1\right)^2+1>0\)
\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
9 tháng 8 2021
1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
9 tháng 8 2021
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
MT
2
KN
1 tháng 8 2019
a) \(x\left(2x-1\right)-6x+3=0\)
\(\Leftrightarrow x\left(2x-1\right)-3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
b) \(x^2\left(x+1\right)-9x-9=0\)
\(\Leftrightarrow x^2\left(x+1\right)-9\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm\sqrt{9}=\pm3\end{cases}}\)
1 tháng 8 2019
a) x(2x - 1) - 6x + 3 = 0
=> x(2x - 1) - 3(2x - 1) = 0
=> (x - 3)(2x - 1) = 0
=> \(\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
b) x2(x + 1) - 9(x + 1) = 0
=> (x2 - 9)(x + 1) = 0
=> \(\orbr{\begin{cases}x^2-9=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\pm3\\x=-1\end{cases}}\)
15 tháng 7 2021
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
2 tháng 7 2018
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
TP
2
29 tháng 8 2016
TÌM X
a) (3x+2)(2x+9)-(6x+1)(x+2)=7
=> 6x2 + 31x +18 - 6x2 - 13x - 2 - 7 = 0
=> 18x + 9 = 0 => 9(2x + 1) = 0 => 2x + 1 = 0 => x = -1/2
b) (x-2)(x+5)-(x+3)(x+2)=-6
=> x2 + 3x - 10 - x2 - 5x -6 + 6 = 0 => -2x -10 = 0 => -2(x + 5) = 0
=> x + 5 = 0 => x = -5
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
=> 18x2 - 15x +3 - 18x2 + 29x -3 = 0 => 14x = 0 => x = 0
NT
29 tháng 8 2016
a) \(\left(3x+2\right)\left(2x+9\right)-\left(6x+1\right)\left(x+2\right)=7\\\Rightarrow 6x^2+31x+18-6x^2-16x-2-7=0\\ \Rightarrow18x+9=0\Rightarrow9\left(2x+1\right)=0\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)
b) \(\left(x-2\right)\left(x+5\right)-\left(x+3\right)\left(x+2\right)=-6\\ \Rightarrow x^2+3x-10-x^2-5x-6+6=0\\ \Rightarrow-2x-10=0\\ \Rightarrow-2\left(x+5\right)=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)
c) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\\ \Rightarrow18x^2-15x+3-18x^2+29x-3=0\\ \Rightarrow14x=0\\ \Rightarrow x=0\)
HT
2
( x - 2 ) 3 - ( x + 1 ) 3 + 9x ( x + 1 ) - 9 = 0
=> \(x^3-6x^2+12x-8-\left(x^3+3x^2+3x+1\right)+9x^2+9x-9=0\)
=> \(x^3-6x^2+12x-8-x^3-3x^2-3x-1+9x^2+9x-9=0\)
=> \(18x-18=0\)
=> \(18x=0+18\)
=> \(18x=18\)
=> \(x=1\)