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\(xy^2-9x=x.\left(y^2-3^2\right)=x.\left(y-3\right)\left(y+3\right)\)
\(x^2+14x+49-y^2=\left(x^2+2.7x+7^2\right)-y^2=\left(x+7\right)^2-y^2=\left(x+7-y\right).\left(x+7+y\right)\)
\(xy-y^2-x+y=y.\left(x-y\right)-\left(x-y\right)=\left(x-y\right).\left(y-1\right)\)
\(5x.\left(x-7\right)-x+7=5x.\left(x-7\right)-\left(x-7\right)=\left(x-7\right).\left(5x-1\right)\)
\(x^2-y^2+5x-5y=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)=\left(x-y\right).\left(x+y+5\right)\)
\(5x^3-40=5.\left(x^3-2^3\right)=5.\left(x-2\right).\left(x^2+2x+4\right)\)
\(x^2-y^2+12y-36=\left(y^2-2.6y+6^2-x^2\right)=-\left[\left(y-6\right)^2-x^2\right]\)\(=-\left[y-6-x\right].\left[y-6+x\right]\)
\(x^2z+4xyz+4y^2z=z.\left[x^2+2.2xy+\left(2y\right)^2\right]=z.\left(x+2y\right)^2\)
a) \(2x^2-50\)
\(=2\left(x^2-25\right)\)
\(=2\left(x-5\right)\left(x+5\right)\)
b) \(x^2z+4xyz+4y^2z\)
\(=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)
c) \(x^2-y^2+12y-36\)
\(=x^2-\left(y^2-2\cdot x\cdot6+6^2\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right)\left(x+y-6\right)\)
d) Đặt \(D=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(D=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(D=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(x^2+3x+1=a\)
\(D=\left(a-1\right)\left(a+1\right)+1\)
\(D=a^2-1^2+1\)
\(D=a^2\)
Thay \(x^2+3x+1=a\)vào D ta có :
\(D=\left(x^2+3x+1\right)^2\)
a: \(2x^2-50=2\left(x^2-5^2\right)\)
\(=2\left(x-5\right)\left(x+5\right)\)
b:\(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)
c:\(x^2-y^2+12y-36=x^2-\left(y^2-12y+36\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right)\left(x+y-6\right)\)
d:\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(t=x^2+3x+1\)Ta có:
\(=\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1=t^2\)
\(=\left(x^2+3x+1\right)^2\)
a. \(x^2+4y^2+z^2=2x+12y-4z-14\)
\(\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(2y-3\right)^2\ge0\\\left(z+2\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
b. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
Bài 2 ;
Ta có : x2 + 3x
= x2 + 3x + \(\frac{9}{4}-\frac{9}{4}\)
= \(x^2+2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\)
Mà ; \(\left(x+\frac{3}{2}\right)^2\ge\forall x\)
Nên : \(\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\forall x\)
Vậy GTNN của B là : \(-\frac{9}{4}\) khi và chỉ khi x = \(-\frac{3}{2}\)
a) (3x - 2)(4x + 5) = 0
⇔ 3x - 2 = 0 hoặc 4x + 5 = 0
1) 3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3
2) 4x + 5 = 0 ⇔ 4x = -5 ⇔ x = -5/4
Vậy phương trình có tập nghiệm S = {2/3;−5/4}
b) (2,3x - 6,9)(0,1x + 2) = 0
⇔ 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
1) 2,3x - 6,9 = 0 ⇔ 2,3x = 6,9 ⇔ x = 3
2) 0,1x + 2 = 0 ⇔ 0,1x = -2 ⇔ x = -20.
Vậy phương trình có tập hợp nghiệm S = {3;-20}
c) (4x + 2)(x2 + 1) = 0 ⇔ 4x + 2 = 0 hoặc x2 + 1 = 0
1) 4x + 2 = 0 ⇔ 4x = -2 ⇔ x = −1/2
2) x2 + 1 = 0 ⇔ x2 = -1 (vô lí vì x2 ≥ 0)
Vậy phương trình có tập hợp nghiệm S = {−1/2}
d) (2x + 7)(x - 5)(5x + 1) = 0
⇔ 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
1) 2x + 7 = 0 ⇔ 2x = -7 ⇔ x = −7/2
2) x - 5 = 0 ⇔ x = 5
3) 5x + 1 = 0 ⇔ 5x = -1 ⇔ x = −1/5
Vậy phương trình có tập nghiệm là S = {−7/2;5;−1/5}
a) \(=2x\left(x-25\right)\)
b) \(=x\left(x-4\right)-\left(x-4\right)=\left(x-4\right)\left(x-1\right)\)
c) \(=x^2-\left(y^2-12y+36\right)=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\)
d) \(=y\left(x^2+4xz+4yz\right)\)
a) \(2x^2-50x\)
\(=2x\left(x-25\right)\)
b) \(x^2-5x+4\)
\(=\left(x-1\right)\left(x-4\right)\)
c) \(x^2-y^2+12y+36\)
\(=\left(x+y-6\right)\left(x-y+6\right)\)
d) \(x^2z+4xyz+4y^2z\)
\(=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)