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19 tháng 2 2022
b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
=>-6x+16=0
=>-6x=-16
hay x=8/3(nhận)
c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)
\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow2x^2+4x-2x^2+2=0\)
=>4x+2=0
hay x=-1/2(nhận)
TM
2
TT
6 tháng 3 2020
a) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x-2\right)\left[\left(3x\right)^2+3x\cdot2+2^2\right]-\left(3x-1\right)\left[\left(3x\right)^2+3x\cdot1+1\right]=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left[\left(3x\right)^3-1\right]=x-4\)
\(\Leftrightarrow x=-3\) ( thỏa mãn )
P/s : Đề câu b) viết lại nhé, mình không hiểu lắm :))
KN
6 tháng 3 2020
\(9\left(2x+1\right)=4\left(x-5\right)^2\)
\(\Leftrightarrow18x+9=4\left(x^2-10x+25\right)\)
\(\Leftrightarrow18x+9=4x^2-40x+100\)
\(\Leftrightarrow4x^2-58x+91=0\)
Ta có \(\Delta=58^2-4.4.91=1908,\sqrt{\Delta}=6\sqrt{53}\)
\(\Rightarrow x=\frac{58\pm6\sqrt{53}}{8}\)
TT
3
H
13 tháng 8 2023
`x^2 -1-2xy+2y`
`=(x^2-1)-(2xy-2y)`
`=(x-1)(x+1)-2y(x-1)`
`=(x-1)(x+1-2y)`
__
`(x+3)^2-(2x-5)(x+3)`
`=(x+3)(x+3-2x+5)`
`=(x+3)(-x+8)`
__
`(3x+2)^2 +(3x-2)^2-2(9x^2-4)`
`= (3x+2)^2 +(3x-2)^2-2(3x-2)(3x+2)`
`= (3x+2)^2-2(3x-2)(3x+2)+(3x-2)^2`
`=[(3x+2)-(3x-2)]^2`
`=(3x+2-3x+2)^2`
`= 4^2=16`
VT
1
14 tháng 6 2021
a) \(\left(x+2\right)\left(x^2-2x+4\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4+x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{23}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(N\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{-2\right\}\)
b) \(9x^2-4-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2-3x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\cdot4=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(S=\left\{\dfrac{2}{3}\right\}\)
28 tháng 7 2023
a: =x^3+8-1+27x^3=28x^3+7
b: Sửa đề: (2+y)(y^2-2y+4)+(5-y)(25+5y+y^2)
=8+y^3+125-y^3
=133
30 tháng 10 2021
c: \(x^2+4x+4=\left(x+2\right)^2\)
d: \(9x^2+6x+1=\left(3x+1\right)^2\)
18 tháng 8 2021
1.
\(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
2.
\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
3.
\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)
Vì (x-1)2≥0
23>0
\(\Rightarrow\left(x-1\right)^2+23>0\)
Vậy x vô nghiệm
4.
\(9x^2-4=0\\ \Rightarrow\left(3x-4\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
5.
\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
TM
1
6 tháng 3 2020
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
Giải pt :
(3x-2)(9x2+6x+4)-(3x-1)(9x2-3x+1)=x-4
9(2x+1)=4(x-5)2
\(\left(2x+1\right)\left(2-3x\right)=9x^2-4\)
\(\Leftrightarrow\left(-2x-1\right)\left(3x-2\right)-\left(9x^2-4\right)=0\)
=>\(\left(-2x-1\right)\left(3x-2\right)-\left(3x-2\right)\left(3x+2\right)=0\)
=>\(\left(3x-2\right)\left(-2x-1-3x-2\right)=0\)
=>(3x-2)(-5x-3)=0
=>(5x+3)(3x-2)=0
=>\(\left[{}\begin{matrix}5x+3=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)