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10 tháng 7 2019
\(1,\)\(\left(2x+3\right)^2=4x^2+12x+9\)
\(2,\)\(\left(3x+2y\right)^2=9x^2+12xy+4x^2\)
\(3,\)\(\left(3a-1\right)^2=9x^2-6x+1\)
\(4,\)\(\left(a-2\right)^2=a^2-4a+4\)
\(5,\)\(\left(1-5a\right)^2=1-10a+25a^2\)
\(6,\)\(\left(x-4\right)^3=x^3-12a^2+48a-64.\)
\(7,\)\(\left(x^2-2y\right)^2=x^4-4x^2y-4y^2\)
\(8,\)\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)
\(9,\)\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^4-49\)
\(10,\)\(\left(x-1\right)\left(x^2+x+1\right)=x^3-1\)
\(11,\)\(\left(x^3-2\right)\left(x^6+2x^3+4\right)=x^9-8\)
\(12,\)\(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)
\(13,\)\(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)
NN
10 tháng 7 2019
1, ( 2x + 3 )2 = 4x2 + 12x + 9
2, ( 3x + 2y )2 = 9x2 +12xy + 4y2
3 ( 3a - 1 )2 = 9a2 - 6x + 1
4, ( a - 2 )2 = a2 - 4a + 4
5, ( 1 - 5a )2 = 1 - 10a + 25a2
6, ( x- 4 )3 = x3 - 12x2 + 48x - 64
7, ( x2 - 2y )2 = x4 - 4x2y + 4y2
8, ( 5X2 - 2 ).( 5X2 + 2 ) = 25X2 - 4
9, ( 2a2 - 7 ).( 2a2 + 7 ) = 4a4 - 49
10, ( x - 1 ).( x2 + x + 1 ) = x3 - 1
8 tháng 10 2021
\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
15 tháng 10 2021
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
30 tháng 9 2021
\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)
HP
11 tháng 1 2022
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
11 tháng 1 2022
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
11 tháng 1 2022
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
11 tháng 1 2022
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
DD
2
ND
Nguyễn Đức Trí
VIP
3 tháng 8 2023
\(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(B=x^3-9x^2+27x-27-\left(x^3-3x^2+9x+3x^2-9x+27\right)+\left(9x^2-1\right)\)
\(B=x^3-9x^2+27x-27-\left(x^3+27\right)+9x^2-1\)
\(B=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(B=27x-55\)
Trả lời:
\(\left(x^2-3x+2\right)^3=x^6-\left(3x-2\right)^3\)
\(\Leftrightarrow\left[x^2-\left(3x-2\right)\right]^3=x^6-\left(3x-2\right)^3\)
\(\Leftrightarrow\left(x^2\right)^3-3.\left(x^2\right)^2.\left(3x-2\right)+3.x^2.\left(3x-2\right)^2-\left(3x-2\right)^3=x^6-\left(3x-2\right)^2\)
\(\Leftrightarrow x^6-3x^4.\left(3x-2\right)+3x^2.\left(3x-2\right)^2-\left(3x-2\right)^3=x^6-\left(3x-2\right)^3\)
\(\Leftrightarrow3x^4.\left(3x-2\right)-3x^2.\left(3x-2\right)^2=0\)
\(\Leftrightarrow3x^2.\left(3x-2\right).\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow3x^2.\left(3x-2\right).\left(x^2-x-2x+2\right)=0\)
\(\Leftrightarrow3x^2.\left(3x-2\right).\left[x.\left(x-1\right)-2.\left(x-1\right)\right]=0\)
\(\Leftrightarrow3x^2.\left(3x-2\right).\left(x-1\right).\left(x-2\right)=0\)
\(3x^2=0\Leftrightarrow x^2=0\Leftrightarrow x=0\)
\(3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(x-1=0\Leftrightarrow x=1\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(x\in\left\{0,\frac{2}{3},1,2\right\}\)
\(\Leftrightarrow x^6-9x^5+33x^4-63x^3+62x^2-36x+8=x^6-\left(3x-2\right)^3\)
\(\Leftrightarrow x^6-9x^5+33x^4-63x^3+66x^2-36x+8=x^6-27x^3+54x^2-36x+8\)
\(\Leftrightarrow-9x^5+33x^4-36x^3+12x^2=0\)
\(\Leftrightarrow-3x^2\left(x-2\right)\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow x=\left\{0;2;1;\frac{2}{3}\right\}\)