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28 tháng 12 2021
\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)
\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
22 tháng 7 2021
b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
A
1
14 tháng 12 2021
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
LT
1
29 tháng 12 2023
Bài 2:
a: 4x(x-3)+6(3-x)=0
=>4x(x-3)-6(x-3)=0
=>(x-3)(4x-6)=0
=>\(\left[{}\begin{matrix}x-3=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(x^3-x\left(x+1\right)\left(x-1\right)=14\)
=>\(x^3-x\left(x^2-1\right)=14\)
=>\(x^3-x^3+x=14\)
=>x=14
c: \(\left(x^2-x\right)^2+2\left(x^2-x\right)=8\)
=>\(\left(x^2-x\right)^2+2\left(x^2-x\right)-8=0\)
=>\(\left(x^2-x\right)^2+4\left(x^2-x\right)-2\left(x^2-x\right)-8=0\)
=>\(\left(x^2-x\right)\left(x^2-x+4\right)-2\left(x^2-x+4\right)=0\)
=>\(\left(x^2-x+4\right)\left(x^2-x-2\right)=0\)
=>\(\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}\right)\left(x-2\right)\left(x+1\right)=0\)
=>\(\left(x-2\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
MD
27 tháng 8 2021
a) 4x(x+1)=8(x+1)
<=>4x(x+1)-8(x+1)=0
<=>(4x-8)(x+1)=0
<=>\(\left[\begin{array}{} 4x-8=0\\ x+1=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=2\\ x=-1 \end{array} \right.\)
Vậy...
b)x(x-1)-2(1-x)=0
<=>(x+2)(x-1)=0
<=>\(\left[\begin{array}{} x+2=0\\ x-1=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=-2\\ x=1 \end{array} \right.\)
Vậy...
c)5x(x-2)-(2-x)=0
<=>(5x+1)(x-2)=0
<=>\(\left[\begin{array}{} 5x+1=0\\ x-2 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=-1/5\\ x=2 \end{array} \right.\)
d)5x(x-200)-x+200=0
<=>(5x-1)(x-200)=0
<=>\(\left[\begin{array}{} 5x-1=0\\ x-200=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=1/5\\ x=200 \end{array} \right.\)
e)\(x^3+4x=0 \)
\(\Leftrightarrow x(x^2+4)=0 \)
\(\Leftrightarrow \left[\begin{array}{} x=0\\ x^2+4=0 (loại vì x^2+4>=0 với mọi x) \end{array} \right.\)
Vậy x=0
f)\((x+1)=(x+1)^2\)
\(\Leftrightarrow (x+1)-(x+1)^2=0\)
\(\Leftrightarrow (x+1)(1-x-1)=0\)
\(\Leftrightarrow (x+1)(-x)=0\)
\(\Leftrightarrow \left[\begin{array}{} x=-1\\ x=0 \end{array} \right.\)
Vậy....
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☜[|}][|}>「>…{…[☆|]>|◎]
NT
0
Y
2
25 tháng 7 2019
(Phần a mình lấy vế phải bằng 0 nha ^^)
a,
\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)+7=0\\ \Leftrightarrow25x^2-10x+1-\left(25x^2-16\right)+7=0\\ \Leftrightarrow25x^2-10x+1-25x^2+16+7=0\\ \Leftrightarrow-10x+24=0\\ \Leftrightarrow x=2,4\)
b,
\(5x^2+4xy+4y^2+4x+1=0\left(1\right)\\ \Leftrightarrow4x^2+4x+1+x^2+4xy+4y^2=0\\ \Leftrightarrow\left(2x+1\right)^2+\left(x+2y\right)^2=0\left(1a\right)\)
Do \(VT\ge0\) với \(\forall x,y\in R\) nên:
\(\left(1a\right)\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)
c,
\(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2+21\\ \Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)-6x^2-21=0\\ \Leftrightarrow x^3+12x+8-x^3+x-21=0\\ \Leftrightarrow13x-13=0\\ \Leftrightarrow x=1\)
Chúc bạn học tốt nha
.
25 tháng 7 2019
\(b)5x^2 + 4xy + 4y^2 + 4x + 1 = 0\)
\(\Leftrightarrow\) \(4x^2 + 4x + 1 + x^2 + 4xy + 4y^2 = 0\)
\(\Leftrightarrow\)\((2x + 1)^2 + (x + 2y)^2 = 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)
\(c)(x+2)^3-x(x-1)(x+1)=6x^2+21\)
\(\Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)=6x^2+21\\ \Leftrightarrow13x+8=21\\ \Leftrightarrow13x=21-8\\ \Leftrightarrow13x=13\\ \Leftrightarrow x=1\)
a, \(x+5x^2=0\Leftrightarrow x\left(1+5x\right)=0\Leftrightarrow x=-\frac{1}{5};x=0\)
b, \(x+1=\left(x+1\right)^2\Leftrightarrow x+1-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[1-\left(x+1\right)\right]=0\Leftrightarrow x=-1;x=0\)
c, \(x^3+x=0\Leftrightarrow x\left(x^2+1>0\right)=0\Leftrightarrow x=0\)