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1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
a) 2x2 + 4x + xy + 2y
= (2x2 + xy) + (4x + 2y)
= x(2x + y) + 2(2x + y)
= (x + 2)(2x + y)
b) x2 + xy - 7x - 7y
= x(x + y) - 7(x + y)
= (x - y)(x + y)
\(e,-5x+x^2-14\)
\(=x^2+2x-7x-14\)
\(=x\left(x+2\right)-7\left(x+2\right)\)
\(=\left(x+2\right)\left(x-7\right)\)
\(f,x^3+8+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+8x+4\right)\)
\(g,15x^2-7xy-2y^2\)
\(=15x^2+3xy-10xy-2y^2\)
\(=3\left(5x+y\right)-2y\left(5x+y\right)\)
\(=\left(5x+y\right)\left(3-2y\right)\)
\(h,3x^2-16x+5\)
\(=3x^2-x-15x+5\)
\(=x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x+5\right)\)
\(a,x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)\)
\(=x\left(x+y\right)^2\)
\(b,4x^2-9y^2+4x-6y\)
\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)
\(=\left(2x+1\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(c,-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
\(d,x^2+4x-12\)
\(=x^2-2x+6x-12\)
\(=x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(x-2\right)\left(x+6\right)\)
Lời giải:
a)
\(S=12(x^3+y^3)+16x^2y^2+34xy\)
\(=12[(x+y)^3-3xy(x+y)]+16x^2y^2+34xy\)
\(=12(1-3xy)+16x^2y^2+34xy=12+16x^2y^2-2xy\)
\(=(4xy-\frac{1}{4})^2+\frac{191}{16}\geq \frac{191}{16}\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix} x+y=1\\ xy=\frac{1}{16}\end{matrix}\right.\Leftrightarrow (x,y)=(\frac{2+\sqrt{3}}{4}, \frac{2-\sqrt{3}}{4})\)
Vậy \(S_{\min}=\frac{191}{16}\) khi \(\Leftrightarrow (x,y)=(\frac{2+\sqrt{3}}{4}, \frac{2-\sqrt{3}}{4})\) và có hoán vị.
b)
\(A=5(x^3+y^3)+12xy+4x^2y^2\)
\(=5[(x+y)^3-3xy(x+y)]+12xy+4x^2y^2\)
\(=5(1-3xy)+12xy+4x^2y^2\)
\(=5+4x^2y^2-3xy\)
Áp dụng BĐT Cô-si: $1=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}$
$A=4x^2y^2-3xy+5=xy(4xy-1)-\frac{1}{2}(4xy-1)+4,5=(xy-\frac{1}{2})(4xy-1)+4,5$
Vì $xy\leq \frac{1}{4}\Rightarrow 4xy-1\leq 0; xy-\frac{1}{2}< 0\Rightarrow (xy-\frac{1}{2})(4xy-1)\geq 0$
$\Rightarrow A=(xy-\frac{1}{2})(4xy-1)+4,5\geq 4,5$
Vậy $A_{\min}=4,5$ khi $x=y=\frac{1}{2}$
1, \(\left(2x^4-5x^2y^2+3xy^3\right)\left(5x^3+x^2y-y^3\right)\)
\(=10x^7-25x^5y^2+15x^4y^3+2x^6y-5x^4y^3+5x^2y^5+3xy^6\)
2, a, \(4-2x+5x^2-4x^2\&5x-3+x^2\)
Sắp xếp: \(4-2x+5x^2-4x^2=5x^2-4x^2-2x+4=x^2-2x+4\)
\(5x-3+x^2=x^2+5x-3\)
- \(\left(x^2-2x+4\right)\left(x^2+5x-3\right)=x^4+3x^3-9x^2-14x-12\)
b, Làm tương tự câu a
1 ) \(\left(2x^4-5x^2y^2+3xy^3\right)\left(5x^3+x^2y-y^3\right)\)
\(=2x^4\left(5x^3+x^2y-y^3\right)-5x^2y^2\left(5x^3+x^2y-y^3\right)+3xy^3\left(5x^3+x^2y-y^3\right)\)\(=10x^7+2x^6y-2x^4y-25x^5y^2-5x^4y^3+5x^2y^5+15x^4y^3+3x^3y^4-3xy^6\)2 ) a ) \(4-2x+5x^2-4x^2=x^2-2x+4\)
\(5x-3+x^2=x^2+5x-3\)
\(\left(x^2-2x+4\right)\left(x^2+5x-3\right)\)
\(=x^4-2x^3+4x^2+5x^3-10x^2+20x-3x^2+6x-12\)
\(=x^4+3x^3-9x^2+26x-12\)
b ) \(10-x^4+3x-4x^2=-x^4-4x^2+3x+10\)
\(2x+x^3-1=x^3+2x-1\)
\(\left(-x^4-4x^2+3x+10\right)\left(x^3+2x-1\right)\)
\(=-x^4\left(x^3+2x-1\right)-4x^2\left(x^3+2x-1\right)+3x\left(x^3+2x-1\right)+10\left(x^3+2x-1\right)\)\(=-x^7-2x^5+x^4-4x^5-8x^3+4x^2+3x^4+6x^2-3x+10x^3+20x-10\)\(=-x^7-\left(2x^5+4x^5\right)+\left(3x^4+x^4\right)+\left(10x^3-8x^3\right)+\left(4x^2+6x^2\right)+\left(20x-3x\right)-10\)\(=-x^7-6x^5+4x^4+2x^3+10x^2+17x-10\)
Bài nhiều quá... nhìn mik nổi gai ốc lun...oh my god sao mà nhiều vậy nè .
Mik định giải giúp bạn nhưng bây h mik hoảng quá ... nhiều vậy chắc mik chết mất... ToT ... >.< =)))
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)