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8 tháng 4 2020
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
TG
3 tháng 8 2020
Bài 1:
a) \(4x\left(3x-1\right)-2\left(3x+1\right)-\left(x+3\right)\)
\(=12x^2-4x-6x-2-x-3\)
\(=12x^2-11x-5\)
b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)
\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)
\(=\left(2x^4y\right)-\left(-1x^3y^2\right)+\left(-2x^2y^3\right)\)
\(=2x^4y+1x^3y^2-2x^2y^3\)
c) \(4x\left(3x^2-x\right)-\left(2x+3\right)^2\left(6x^2-3x+1\right)\)
\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)
\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)
\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)
\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)
\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)
\(=-48x^3-8x^2-24x^4+15x-9\)
6 tháng 2 2017
c(x-1)^2=4
x^2-2x+1=4
x^2-2x+1-4=0
x^2-2x-3=0
x^2-3x+x-3=0
x(x-3)+(x-3)=0
(x-3)(x+1)=0
\(\Rightarrow\hept{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)
6 tháng 2 2017
d, x^3+2x^2-x-2=0
x^2(x+2)-(x+2)=0
(x+2)(x^2-1)=0
\(\Rightarrow\hept{\begin{cases}x=-2\\x=+-1\end{cases}}\)
26 tháng 3 2020
a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
\(\left(\frac{2x}{3}+1\right)^2=\left(\frac{3x}{2}-1\right)^2\)
\(\frac{2x}{3}+1=\frac{3x}{2}-1\)
\(4x+6=9x-6\)
\(4x+6-9x+6=0\)
\(-5x+12=0\)
\(-5x=-12\)
\(x=\frac{12}{5}\)