a) -4y^{2}xz + 3y^{2}xz =−4y2xz+3y2xz= × b) (-4y^{2}xz) . (3y^{2}xz)...
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TN
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LT
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1 tháng 7 2021
a) Ta có: \(x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
b) Ta có: \(2x+2y-x^2-xy\)
\(=2\left(x+y\right)-x\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x\right)\)
c) Ta có: \(x^2-25+y^2+2xy\)
\(=\left(x+y\right)^2-25\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
d) Ta có: \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
e) Ta có: \(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
f) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
ND
1
NH
25 tháng 2 2020
Ta có: \(\left|2x+3y\right|\ge0\)\(\forall x,y\inℝ\); \(\left|4y+5z\right|\ge0\)\(\forall y,z\inℝ\); \(\left|xy+yz+zx+110\right|\ge0\)\(\forall x,y,z\inℝ\)
Nên: \(P=\left|2x+3y\right|+\left|4y+5z\right|+\left|xy+yz+xz+110\right|\ge0\)\(\forall x,y,z\inℝ\)
Dấu " = " xảy ra <=> \(\left|2x+3y\right|+\left|4y+5z\right|+\left|xy+yz+xz+110\right|=0\)
Có: \( \left|2x+3y\right|=0\)\(\Leftrightarrow2x+3y=0\)\(\Leftrightarrow2x=-3y\)\(\Leftrightarrow\frac{x}{-3}=\frac{y}{2}\)
\(\left|4y+5z\right|=0\)\(\Leftrightarrow4y+5z=0\)\(\Leftrightarrow4y=-5z\)\(\Leftrightarrow\frac{y}{-5}=\frac{z}{4}\)
\(\left|xy+yz+zx+110\right|=0\)\(\Leftrightarrow xy+yz+zx+110=0\)\(\Leftrightarrow xy+yz+zx=-110\)
Lại có: \(\frac{x}{-3}=\frac{y}{2}\)\(\Rightarrow\frac{x}{15}=\frac{y}{-10}\) (1) ; \(\frac{y}{-5}=\frac{z}{4}\)\(\Rightarrow\frac{y}{-10}=\frac{z}{8}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{x}{15}=\frac{y}{-10}=\frac{z}{8}=k\)=> x = 15k ; y = (-10) . k ; z = 8k
Ta có: \(xy+yz+zx=-110\)\(\Rightarrow15k\left(-10\right)k+8k\left(-10\right)k+8k.15k=-110\)
\(\Rightarrow k^2\left(-150\right)+k^2\left(-80\right)+120k^2=-110\)
\(\Rightarrow k^2\left(-110\right)=-110\)\(\Rightarrow k^2=1\)\(\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)
+) Th1: k = 1
Có: x = 15k = 15 . 1 = 15
y = (-10) . k = (-10) . 1 = -10
z = 8k = 8 . 1 = 8
+) Th2: k = -1
Có: x = 15k = 15 . (-1) = -15
y = (-10) . k = (-10) . (-1) = 10
z = 8k = 8 . (-1) = -8
Vậy GTNN P = 0 <=> (x; y; z) = (15; -10; 8) hoặc (x; y; z) = (-15; 10; -8)
21 tháng 7 2017
a, \(\left(2x+1\right)^2-2\left(2x+1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1-x+3\right)^2=\left(x+4\right)^2\)
b, \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)
c, \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)=\left(x+1\right)\left(y-z\right)\)
d, \(x^2-xy-8x+8y=\left(x^2-xy\right)-\left(8x-8y\right)\)
\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)
e, \(x^2+2xy+y^2-xz-yz=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y+z\right)\left(x+y\right)\)
f, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)
21 tháng 7 2017
1,
a, (2x + 1- x + 3)2 = (x+4)2
b,\(x\left(y+z\right)+3\left(y+z\right)=\left(y+z\right)\left(x+3\right)\)
c, \(x\left(y-z\right)+\left(y-z\right)=\left(y-z\right)\left(x+1\right)\)
d,\(x\left(x-y\right)+8\left(y-x\right)\)=\(\left(x-y\right)\left(x-8\right)\)
e,\(\left(x+y\right)^2-z\left(x+y\right)\)=\(\left(x+y\right)\left(x+y-z\right)\)
f,\(25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)
\(=\left(5+2x+y\right)\left(5-2x-y\right)\)
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