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LT
1
4 tháng 10 2020
a) (x + 3)3 - x(3x + 1)2 + (2x + 1)(4x2 - 2x + 1) = 28
=> x3 + 9x2 + 27x + 27 - x(9x2 + 6x + 1) +(2x + 1)[(2x)2 - 2.x.1 + 12 ] = 28
=> x3 + 9x2 + 27x + 27 - 9x3 - 6x2 - x + (2x)3 + 13 = 28
=> x3 + 9x2 + 27x + 27 - 9x3 - 6x2 - x + 8x3 + 1 = 28
=> (x3 - 9x3 + 8x3) + (9x2 - 6x2) + (27x - x) + (27 + 1) = 28
=> 3x2 + 26x + 28 = 28
=> 3x2 + 26x = 0
=> 3x2 + 26x = 0
=> \(3x\left(x+\frac{26}{3}\right)=0\)
=> 3x = 0 hoặc x + 26/3 = 0
=> x = 0 hoặc x = -26/3
b) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
=> \(x^6-3x^4+3x^2-1-\left(x^6-1\right)=0\)
=> \(x^6-3x^4+3x^2-1-x^6+1=0\)
=> \(\left(x^6-x^6\right)-3x^4+3x^2+\left(-1+1\right)=0\)
=> \(-3x^4+3x^2=0\)
=> \(-\left(3x^4-3x^2\right)=0\)
=> \(3x\left(x^3-x\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x^3-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x\left(x^2-1\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
N
2 tháng 11 2018
\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
2 tháng 11 2018
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
DN
0
TN
10 tháng 7 2018
\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\) \(=52\)
\(12\left(x^2-4\right)-3\left(4x^2+12x+9\right)\) \(=52\)
\(12x^2-48-12x^2-36x-27\) \(=52\)
\(-36x-75=52\)
\(-36x=127\)
\(x=\frac{-127}{36}\)
\(\left(2x+1\right)^2-4\left(x-1\right)\left(x+1\right)\) \(+2x=5\)
\(4x^2+4x+1-4\left(x^2-1\right)\) \(+2x=5\)
\(4x^2+4x-1-4x^2+4+2x=5\)
\(6x+3=5\)
\(6x=2\)
\(x=3\)
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\) \(+6\left(x-1\right)^2=15\)
\(x^3-6x^2+12x-8-\left(x-3\right)\left(x+3\right)^2\) \(+6\left(x^2-2x+1\right)=15\)
\(x^3-6x^2+12x-8-\left(x^2-9\right)\left(x+3\right)\) \(+6x^2-12x+6=15\)
\(x^3-2\) \(-\left(x^3+3x^2-9x-27\right)\)\(=15\)
\(x^3-2-x^3-3x^2+9x+27=15\)
\(-3x^2+9x+25=15\)
\(-3x^2+9x+10=0\)
\(-3\left(x^2-3x-\frac{10}{3}\right)\) \(=0\)
\(x=\frac{9+\sqrt{201}}{6}\)
các câu còn lại tương tự
TT
29 tháng 7 2016
\(A.\left(2a+1\right)^2-4\left(a+2\right)^2=9\\ \left(2a+1-2a-4\right)\left(2a+1+2a+4\right)=9\)
\(-3\left(4a+5\right)=9\\ -12a-15=9\\ -12a=24\\ a=-2\)
NH
10 tháng 4 2020
a, \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)
\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)
\(\Leftrightarrow-14x-9=5\)
\(\Leftrightarrow-14x=14\)
\(\Leftrightarrow x=-1\)
Vậy....
b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)
\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)
\(\Leftrightarrow-10+17x=-44\)
\(\Leftrightarrow17x=-34\)
\(\Leftrightarrow x=-2\)
Vậy....
c, \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)
\(\Leftrightarrow10x+10=30\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy....
d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)
\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)
\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)
\(\Leftrightarrow14x-3=7\)
\(\Leftrightarrow14x=10\)
\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)
Vậy...
2 tháng 7 2023
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
\((x+2)(x^2-2x+4)=(x-1)^3+3(x+1)^2\\\Leftrightarrow x^3+2^3=x^3-3x^2+3x-1+3\cdot(x^2+2x+1)\\\Leftrightarrow x^3 +8=x^3-3x^2+3x-1+3x^2+6x+3\\\Leftrightarrow x^3-x^3 +3x^2-3x-3x^2-6x=-1+3-8\\\Leftrightarrow -9x=-6\\\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(x=\dfrac{2}{3}\)