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a, x = 79 => x + 1 = 80
Ta có:\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+...+80x+15\)
\(=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+\left(x+1\right)x+15\)
\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+...+x^2+x+15\)
\(=x+15=79+15=94\)
Còn lại tương tự
\(Q_{\left(x\right)}=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
\(A=2x^2+10x+8\)
\(=2x^2+2x+8x+8\)
\(=2x\left(x+1\right)+8\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+8\right)\)
\(=2\left(x+1\right)\left(x+4\right)\)
\(B=x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-5y\right)\)
\(C=5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(x+y\right)\left(5x+y\right)\)
\(D=x^3+8\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
\(E=x^4+64\)
\(=[\left(x^2\right)^2+2.x^2.8+64]-16x^2\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
\(F=x^6+27\)
\(=\left(x^2\right)^3+3^3\)
\(=\left(x^2+3\right)\left(x^4-3x^2+9\right)\)
\(=\left(x^2+3\right)\left[\left(x^4+6x^2+9\right)-9x^2\right]\)
\(=\left(x^2+3\right)\left[\left(x^2+3\right)^2-\left(3x\right)^2\right]\)
\(=\left(x^2+3\right)\left(x^2+3x+3\right)\left(x^2-3x+3\right)\)
\(G=x^4-2x^3+2x-1\)
\(=x^4-x^3-x^3+x^2-x^2+x+x-1\)
\(=x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\left(x^2-1\right)\)
\(=\left(x-1\right)^3\left(x+1\right)\)
\(H=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt \(x^2+5x+5=a\)
\(\Rightarrow H=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5\right)\left(x^2+5x+10\right)\)
A= \(2x^2\)+8x+2x+8
A= 2x(x+1)+8(x+1)
A= (x+1)(2x+8)
B=\(x^2\)-2xy-5xy+10\(y^2\)
B=x(x-2y)-5y(x-2y)
B=(x-2y)(x-5y)
C= 5\(x^2\)+5xy+xy+\(y^2\)
C= 5x(x+y)+y(x+y)
C= (x+y)(5x+y)
D= \(x^3\)+\(2^3\)
D= (x+2)(\(x^2\)-2x+4)
E= \(\left(x^2\right)^2\)+\(8^2\)
E=-\(\left[\left(x^2\right)^2-8^2\right]\)
E=-\(\left[\left(x^2+8\right)\left(x^2-8\right)\right]\)
F= \(\left(x^2\right)^3\)+\(3^3\)
F= (\(x^2\)+3)(\(x^4\)-3\(x^2\)+9)
G= \(\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
G= (x-1)(x+1)(\(x^2\)+1)-2x(x-1(x+1)
G= (x-1)(x+1)(\(x^2\)+1-2x)
G= (x-1)(x+1)\(\left(x-1\right)^2\)
G= \(\left(x-1\right)^3\)(x+1)
\(Tacó\): \(C=x^2+2xy+y^2+y^2-6y+15\)
\(=\left(x^2+2xy+y^2\right)+\left(y^2-6y+9\right)+6\)
\(=\left(x+y\right)^2+\left(y-3\right)^2+6\)
\(Mà\)\(\left(x+y\right)^2\ge0\)với mọi x,y
\(\left(y-3\right)^2\ge0\)với mọi y
\(\Rightarrow\left(x+y\right)^2+\left(y-3\right)^2+6>0\)
\(Hay\)\(x^2+2xy+y^2+y^2-6y+15>0\)\
:
Ta có C = (x2 + 2xy + y2) + (y2 - 6x + 9) + 6
= (x + y)2 + (y - 3)2 + 6 \(\ge6>0\)(đpcm)
C = x2 + 2xy + y2 + y2 - 6y + 15
C = ( x2 + 2xy + y2 ) + ( y2 - 6y + 9 ) + 6
C = ( x + y )2 + ( y - 3 )2 + 6 ≥ 6 > 0 ∀ x ( đpcm )
D = x2 + y2 + 6x + 10y + 30
D = ( x2 + 6x + 9 ) + ( y2 + 10y + 25 ) - 4
D = ( x + 3 )2 + ( y + 5 )2 - 4 ≥ -4 ( xem lại đề nhớ )
Lời giải:
a) Với \(x=79\)
\(P(x)=x^7-80x^6+80x^5-80x^4+...+80x+15\)
\(=(x^7-79x^6)-(x^6-79x^5)+(x^5-79x^4)-....-(x^2-79x)+x+15\)
\(=x^6(x-79)-x^5(x-79)+x^4(x-79)-...-x(x-79)+x+15\)
\(=(x^6-x^5+x^4-...-x)(x-79)+x+15\)
\(=(x^6-x^5+x^4-...-x)(79-79)+79+15=79+15=94\)
b) Hoàn toàn tương tự phần a.
\(Q(x)=(x^{14}-9x^{13})-(x^{13}-9x^{12})+(x^{12}-9x^{11})-...+(x^2-9x)-x+10\)
\(=x^{13}(x-9)-x^{12}(x-9)+x^{11}(x-9)-...+x(x-9)-x+10\)
\(=(x-9)(x^{13}-x^{12}+x^{11}-...+x)-x+10\)
\(=(9-9)(x^{13}-x^{12}+...+x)-9+10=0-9+10=1\)
c)
\(R(x)=(x^4-16x^3)-(x^3-16x^2)+(x^2-16x)-x+20\)
\(=x^3(x-16)-x^2(x-16)+x(x-16)-x+20\)
\(=(x-16)(x^3-x^2+x)-x+20\)
Với $x=16$ thì $Q(x)=(16-16)(x^3-x^2+x)-16+20=0-16+20=4$
d)
\(S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+x(x-12)-x+10\)
\(=x^9(x-12)-x^8(x-12)+x^7(x-12)-...+x(x-12)-x+10\)
\(=(x-12)(x^9-x^8+x^7-..+x)-x+10\)
\(=(12-12)(x^9-x^8+x^7-...+x)-12+10=-12+10=-2\)
3)
e)
b) Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3
= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1
= (x-3y)2 + (2x -1)2 + (y-1)2 +1
Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0
(2x -1)2 luôn lớn hơn hoặc bằng 0
(y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0
2.
Ta có hằng đẳng thức : \(\left(a-b\right)^2=a^2-2ab+b^2\left(1\right)\)
Lại có \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2-4ab=a^2+2ab-4ab+b^2\)
\(\Leftrightarrow\left(a+b\right)^2-4ab=a^2-2ab+b^2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left(a-b\right)^2=\left(a+b\right)^2-4ab\)( đpcm )
3.
Ta có hằng đẳng thức \(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy\)
Thay \(x+y=7\)và \(xy=-3\)vào ta được :
\(x^2+y^2=7^2-2\left(-3\right)\)
\(\Leftrightarrow x^2+y^2=49+6=55\)
Vậy ...
1.
a) Đặt \(A=x^2-6x+10\)
\(A=\left(x^2-6x+9\right)+1\)
\(A=\left(x-3\right)^2+1\)
Mà \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow A\ge1>0\)
Vậy ...
b) Đặt \(B=x^2-4x+7\)
\(B=\left(x^2-4x+4\right)+3\)
\(B=\left(x-2\right)^2+3\)
Mà \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow B\ge3\)
Vậy ...