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30 tháng 9 2021
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
14 tháng 7 2023
b: =>4x^2+8x-8x^2+5x-10=0
=>-4x^2+13x-10=0
=>x=2 hoặc x=5/4
c: =>2x^2-5x+6x-15=2x^2+8x
=>x-15=8x
=>-7x=15
=>x=-15/7
d: =>3x^2+15x-2x-10-3x^2-12x=5
=>x-10=5
=>x=15
e: =>x^2-3x+2x^2+2x=3x^2-12
=>-x=-12
=>x=12
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☜[|}][|}>「>…{…[☆|]>|◎]
K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
TH
4
2 tháng 6 2023
\(b,3\left(x-2\right)+2\left(3x-5\right)=10\\ \Leftrightarrow3x-6+6x-10=10\\ \Leftrightarrow3x+6x=10+10+6\\ \Leftrightarrow9x=26\\ \Leftrightarrow x=\dfrac{26}{9}\\ c,2x-\left(3x+1\right)=5x-2\\ \Leftrightarrow2x-3x-1=5x-2\\ \Leftrightarrow2x-3x-5x=-2+1\\ \Leftrightarrow-6x=-1\\ \Leftrightarrow x=\dfrac{1}{6}\\ d,3x+2=-5+6 \\ \Leftrightarrow3x=-5+6-2\\ \Leftrightarrow3x=-2\\ \Leftrightarrow x=-\dfrac{1}{3}\)
H9
HT.Phong (9A5)
CTVHS
21 tháng 10 2023
a) \(\left(2x+1\right)\left(x-2\right)-2x^2=0\)
\(\Leftrightarrow2x^2-4x+x-2-2x^2=0\)
\(\Leftrightarrow\left(2x^2-2x^2\right)-\left(4x-x\right)-2=0\)
\(\Leftrightarrow-3x-2=0\)
\(\Leftrightarrow-3x=2\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
b) \(\left(x+3\right)\left(2x-1\right)+x^2=9\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)+x^2-9=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)+\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1+x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\3x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{4}{3}\end{matrix}\right.\)
21 tháng 10 2023
`#3107.101107`
a)
`(2x + 1)(x - 2) - 2x^2 = 0`
`<=> 2x^2 - 3x - 2 - 2x^2 = 0`
`<=> -3x - 2 = 0`
`<=> -3x = 2`
`<=> x = -2/3`
Vậy, `x=-2/3`
b)
`(x + 3)(2x - 1) + x^2 = 9`
`<=> 2x^2 - 5x - 3 + x^2 = 9`
`<=> 3x^2 - 5x - 3 = 9`
`<=> 3x^2 - 3x - 12 = 0`
`<=> 3x^2 + 4x - 9x - 12 = 0`
`<=> (3x^2 - 9x) + (4x - 12) = 0`
`<=> 3x(x - 3) + 4(x - 3) = 0`
`<=> (3x + 4)(x - 3) = 0`
`<=>` TH1: `3x + 4 = 0`
`<=> 3x = -4`
`<=> x = -4/3`
TH2: `x - 3 = 0`
`<=> x = 3`
Vậy,` x \in {-4/3; 3}.`
7 tháng 11 2021
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
7 tháng 11 2021
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
a) \(\left(2x-5\right)^2=49\)
\(\left(2x-5\right)^2=\left(\pm7\right)^2\)
\(=>2x-5=7\) hoặc \(2x-5=-7\)
\(\cdot2x-5=7\) \(\cdot2x-5=-7\)
\(2x=5+7\) \(2x=-7+5\)
\(2x=12\) \(2x=-2\)
\(x=12:2\) \(x=-2:2\)
\(x=6\) \(x=-1\)
Vậy x=6 hoặc x=-1
b/ \(\left(2x+5\right)^2-\left(1-2x\right)^2=10\)
\(4x^2+20x+25-\left(1-4x+4x^2\right)=10\)
\(4x^2+20x+25-1+4x-4x^2=10\)
\(24x+24=10\)
\(24x=10-24\)
\(24x=-14\)
\(x=\frac{-14}{24}\)
\(x=\frac{-7}{12}\)
c/ \(\left(9-2x\right)^3=27\)
\(\left(9-2x\right)^3=3^3\)
\(9-2x=3\)
\(2x=9-3\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
a/ (2x-5)^2=49
TH1: 2x-5=7
x=6
TH2: 2x-5=-7
x=-1