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\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+1\right)\left(x-3\right)-\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(7-5x\right)=0\)
\(\Leftrightarrow x-1=0;x+2=0\)hoặc \(7-5x=0\)
\(\Leftrightarrow x=1;x=-2\)hoặc \(x=\frac{7}{5}\)
KL....
\(b,\left(5x^2-2x+10\right)^2=\left(x^2+10x-8\right)^2\)
\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10-x^2-10x+8\right)\left(5x^2-2x+10+x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(4x^2-12x+18\right)\left(6x^2+8x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x+\frac{9}{2}\right)\left(6x^2+6x+2x+2\right)=0\)
\(\Leftrightarrow\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{9}{4}\right)\left(6x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\right]\left(3x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-1\end{cases}}\)Vì \(\left(x-\frac{3}{2}\right)^2+\frac{9}{4}>0\forall x\)
Vậy ..
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
a) (15x2-1+9x4-6x3+2x) :( 5 + 3x2-2x)
b) ( -19x+ 10+ 3x4- 5x2+11x3) : ( 3x+ x2-2)
c) (x4-14-x) : (x-2)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(\dfrac{x^4-x-14}{x-2}\)
\(=\dfrac{x^4-2x^3+2x^3-4x^2+4x^2-8x+7x-14}{x-2}\)
\(=x^3+2x^2+4x+7\)
a) \(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x-5+x+2\right)\left(2x-5-x-2\right)=0\)
\(\Leftrightarrow\left(3x-3\right)\left(x-7\right)=0\)
b) Cách làm giống câu a
\(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}3x^2+10x-8=5x^2-2x+10\\3x^2+10x-8=-5x^2+2x-10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2-12x+18=0\\8x^2+8x+2=0\end{cases}}\)
\(TH1:2x^2-12x+18=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x=3\)
\(TH2:8x^2+8x+2=0\)
\(\Leftrightarrow4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)