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\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
d,x^2+4y^2+z^2=2x+12y−4z−14
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
⇔x=1;y=3/2;z=−2
e: Ta có: x^2−6x+y2+4y+2=0
⇔x^2−6x+9+y^2+4y+4−11=0
⇔(x−3)^2+(y+2)^2=11
Dấu '=' xảy ra khi x=3 và y=-2
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
a) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
b) Ta có: \(x^3-6x^2+11x-6=0\)
\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={1;2;3}
c) Ta có: \(x^3-x^2-21x+45=0\)
\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)
\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: S={3;-5}
d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)
\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x^2\cdot\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+3x^2+x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+3\right)+\left(x+1\right)\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên (x-2)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy: S={2;-3}
a, \(x^4-x^2-2=0\Leftrightarrow x^4-2x^2+x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)
b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)
c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)
d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)
a)
/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)
\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)
\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)
\(\Leftrightarrow4x^2=4\)
\(\Leftrightarrow x^2=1\)
hay \(x\in\left\{1;-1\right\}\)
b: Ta có: \(x^2-7x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\left(2x-1\right)=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)
a.
$x^4-6x^2+9=0$
$\Leftrightarrow (x^2-3)^2=0$
$\Leftrightarrow x^2-3=0$
$\Leftrightarrow x^2=3$
$\Leftrightarrow x=\pm \sqrt{3}$
b.
$8x^3+12x^2+6x-63=0$
$\Leftrightarrow (8x^2+12x^2+6x+1)-64=0$
$\Leftrightarrow (2x+1)^3=64=4^3$
$\Leftrightarrow 2x+1=4$
$\Leftrightarrow x=\frac{3}{2}$
c. $(3-2x)^2-25=0$
$\Leftrightarrow (3-2x)^2-5^2=0$
$\Leftrightarrow (3-2x-5)(3-2x+5)=0$
$\Leftrightarrow (-2-2x)(8-2x)=0$
$\Leftrightarrow -2-2x=0$ hoặc $8-2x=0$
$\Leftrightarrow x=-1$ hoặc $x=4$
d.
$6(x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1$
$\Leftrightarrow (x+1)^2[6-2(x+1)]+2(x^3-1)=1$
$\Leftrightarrow (x+1)^2(4-2x)+2x^3-3=0$
$\Leftrightarrow 6x+1=0$
$\Leftrightarrow x=\frac{-1}{6}$
e. $(x-2)^2-(x-2)(x+2)=0$
$\Leftrightarrow (x-2)[(x-2)-(x+2)]=0$
$\Leftrightarrow (x-2)(-4)=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
f. $x^2-4x+4=25$
$\Leftrightarrow (x-2)^2=5^2=(-5)^2$
$\Leftrightarrow x-2=5$ hoặc $x-2=-5$
$\Leftrightarrow x=7$ hoặc $x=-3$