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a)
\(14x^2y-21xy^2+28x^2y^2\)
\(=7xy(2x-3y+4xy)\)
b) \(x(x+y)-5x-5y=x(x+y)-5(x+y)=(x-5)(x+y)\)
c)
\(10x(x-y)-8(y-x)=10x(x-y)+8(x-y)\)
\(=(x-y)(10x+8)=2(x-y)(5x+4)\)
a. \(14x^2y-21xy^2+28x^2y^2\)
\(=7xy\left(2x-3y+4xy\right)\)
b. \(x\left(x+y\right)-5x-5y\)
\(=x\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x-5\right)\left(x+y\right)\)
c. \(10x\left(x-y\right)-8\left(y-x\right)\)
\(=10x\left(x-y\right)+8\left(x-y\right)\)
\(=\left(10x+8\right)\left(x-y\right)\)
d. \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)
\(=2x\left(4x+2\right)\)
\(=4x\left(2x+1\right)\)
e. Vì bài này giải không ra nên mình nghĩ nó sai đề, sửa lại tí nhé!
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz+zy+z^2-3xy\right)\)
g. \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y^2\right)-4z^2\right]\)
\(=5\left(x-y+z\right)\left(x-y-z\right)\)
h. \(x^3-x+3x^2y+3xy^3+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
i. \(x^2+7x-8\)
\(=x^2-x+8x-8\)
\(=x\left(x-1\right)+8\left(x-1\right)\)
\(=\left(x+8\right)\left(x-1\right)\)
2: \(=3\left(x-2y\right)+y\left(x-2y\right)=\left(x-2y\right)\left(y+3\right)\)
3: \(=x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
4: \(x^2+2x+1-16y^2\)
\(=\left(x+1\right)^2-16y^2\)
\(=\left(x+1+4y\right)\left(x+1-4y\right)\)
5: \(x^2-y^2+5x+5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+5\right)\)
6: \(=25-\left(x^2-2xy+y^2\right)\)
\(=25-\left(x-y\right)^2\)
\(=\left(5-x+y\right)\left(5+x-y\right)\)
Bài 1:
\(x^2-8x+y^2+6y+25=0\)
\(\Leftrightarrow\)\(\left(x^2-8x+16\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\)\(\left(x-4\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x-4=0\\y+3=0\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x=4\\y=-3\end{cases}}\)
Vậy...
Bài 2:
Phương trình có nghiệm duy nhất là x = -2/3 nên ta có:
\(\left(4+a\right).\frac{-2}{3}=a-2\)
\(\Leftrightarrow\)\(-\frac{8}{3}-\frac{2}{3}a=a-2\)
\(\Leftrightarrow\)\(a+\frac{2}{3}a=2-\frac{8}{3}\)
\(\Leftrightarrow\)\(\frac{5}{3}a=-\frac{2}{3}\)
\(\Leftrightarrow\)\(a=-\frac{2}{5}\)
Bài 3:
\(A=a^4-2a^3+3a^2-4a+5\)
\(=a^3\left(a-1\right)-a^2\left(a-1\right)+2a\left(a-1\right)-2\left(a-1\right)+3\)
\(=\left(a-1\right)\left(a^3-a^2+2a-2\right)+3\)
\(=\left(a-1\right)\left[a^2\left(a-1\right)+2\left(a-1\right)\right]+3\)
\(=\left(a-1\right)^2\left(a^2+2\right)+3\ge3\)
\(\text{Vậy Min A=3. Dấu "=" xảy ra khi và chỉ khi }a-1=0\Leftrightarrow a=1\)
Bài 4:
\(xy-3x+2y=13\)
\(\Leftrightarrow x\left(y-3\right)+2\left(y-3\right)=7\)
\(\Leftrightarrow\left(x+2\right)\left(y-3\right)=7=1.7=7.1=-1.-7=-7.-1\)
| x+2 | -7 | -1 | 1 | 7 |
| y-3 | -1 | -7 | 7 | 1 |
| x | -9 | -3 | -1 | 5 |
| y | 2 | -4 | 10 | 4 |
Vậy...
Bài 5:
\(xy-x-3y=2\)
\(\Leftrightarrow x\left(y-1\right)-3\left(y-1\right)=5\)
\(\Leftrightarrow\left(x-3\right)\left(y-1\right)=5=1.5=5.1=-1.-5=-5.-1\)
| x-3 | -5 | -1 | 1 | 5 |
| y-1 | -1 | -5 | 5 | 1 |
| x | -2 | 2 | 4 | 8 |
| y | 0 | -4 | 6 | 2 |
Vậy....
a, x4 + 2x3 +x2 = x4 +x3 +x3 +x2 =(x4+x3 )+(x3 +x2 ) =x3(x +1 ) + x2 (x+1 ) =(x+1)(x3+x2)
a) x4 + 2x3 + x2
= x2(x2 + 2x + 1)
= x2(x + 1)2
= [x(x + 1)]2
= (x2 + x)2
b) 5x3 - 10xy + 5y2 - 20z2
= 5(x3 - 2xy + y2 - 4z2)
c) x2y - xy2 + x3 - y3
= xy(x - y) + (x - y)(x2 + xy + y2)
= (x - y)(x2 + 2xy + y2)
= (x - y)(x + y)2
d) x2 - xy + 4x - 2y + 4
= (x2 + 4x + 4) - (xy + 2y)
= (x + 2)2 - y(x + 2)
= (x + 2)(x + 2 - y)
d) x2 - x - 6
= x2 - 3x + 2x - 6
= x(x - 3) + 2(x - 3)
= (x + 2)(x - 3)
f) 3x2 - 5x - 8
= 3x2 + 3x - 8x - 8
= 3x(x + 1) - 8(x + 1)
= (3x - 8)(x + 1)
g) x3 + 3x2 + 6x + 4
= (x3 + 3x2 + 3x + 1) + (3x + 3)
= (x + 1)3 + 3(x + 1)
= (x + 1)[(x + 1)2 + 3]
h) 3x3 - 5x2 - 6x + 8
= 3x3 - 3x2 - 2x2 - 6x + 8
= 3x3 - 3x2 - 2x2 + 2x - 8x + 8
= 3x2(x - 1) - 2x(x - 1) - 8(x - 1)
= (3x2 - 2x - 8)(x - 1)
a) \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(5x^2-10xy+5y^2-20z^2\) (đã sửa đề)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
c) \(x^2y-xy^2+x^3-y^3\)
\(=xy\left(x-y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\)
\(=\left(x+2\right)^2-y\left(x+2\right)\)
\(=\left(x+2\right)\left(x-y+2\right)\)
e) \(x^2-x-6=\left(x+2\right)\left(x-3\right)\)
f) \(3x^2-5x-8\)
\(=\left(3x^2+3x\right)-\left(8x+8\right)\)
\(=3x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-8\right)\)