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a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
b: \(B=\left(2x-4\right)^2+2\cdot\left(2x-4\right)\left(x+1\right)+\left(x+1\right)^2\)
=(2x-4+x+1)^2
=(3x-3)^2
Khi x=-1/2 thì B=(-3/2-3)^2=(-9/2)^2=81/4
c: \(C=x^2\left(5-4\right)+y^2\left(4-6\right)+z^2\left(6+4\right)\)
=x^2-2y^2+10z^2
=6^2-2*5^2+10*4^2
=146
d: x=9 thì x+1=10
\(D=x^{2017}-x^{2016}\left(x+1\right)+x^{2015}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-\left(x+1\right)\)
=x^2017-x^2017+x^2016+...-x^3-x^2+x^2+x-x-1
=-1
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
a) Thay \(x=0,25y\) vào M ta có:
\(M=26\cdot\left(0,25y\right)^2+y\left(2\cdot0,25y+y\right)-10\cdot0,25y\cdot\left(0,25y+y\right)\)
\(M=1,625y^2+y\cdot1,5y-2,5y\cdot1,25y\)
\(M=1,625y^2+1,5y^2-3,125y^2\)
\(M=0\)
b) Thay \(x+6y=9\Rightarrow x=9-6y\) vào N ta có:
\(N=50y^2+\left(9-6y\right)\left(9-6y-2y\right)+14y\left(9-6y-y\right)\)
\(N=50y^2+\left(9-6y\right)\left(9-8y\right)+14\left(9-7y\right)\)
\(N=50y^2+81-72y-54y+48y^2+126-98y\)
\(N=2y^2-224y+207\)
\(a,M=26x^2+y\left(2x+y\right)-10x\left(x+y\right)\\ =26x^2+2xy+y^2-10x^2-10xy\\ =16x^2-8xy+y^2\\ =16\left(x^2-\dfrac{1}{2}xy+\dfrac{1}{16}y^2\right)\\ =16\left(x^2-2.x.y.\dfrac{1}{4}+\dfrac{1}{16}y^2\right)=16\left(x-\dfrac{1}{4}y\right)^2\\ Vì:x=0,25y\Rightarrow y=4x\\ Vậy:M=16\left(x-\dfrac{1}{4}y\right)^2=16\left(x-x\right)^2=16.0^2=0\\ Vậy:tại.x=0,25y.thìM=0\)
a, \(A=x^3-30x^2-31x+1\)
\(=x^3-31x^2+x^2-31x+1\)
\(=x^2\left(x-31\right)+x\left(x-31\right)+1\)
\(=\left(x^2+x\right)\left(x-31\right)+1\)
Thay x = 31 \(\Rightarrow A=1\)
Vậy A = 1 khi x = 31
b, tách ra làm tương tự phần a
x=9
=>x+1=10
\(A=x^{10}-10x^9+10x^8-...+10x^2-10x+1\)
\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+1\)
\(=x^{10}-x^{10}-x^9+x^8+...+x^3+x^2-x^2-x+1\)
=-x+1
=-9+1=-8
a: A=2/3x^2y+4x^2y=14/3x^2y
=14/3*9*7=294
b: B=xy^2(1/2+1/3+1/6)=xy^2=3/4*1/4=3/16
c: C=x^3y^3(2+10-20)=-8x^3y^3
=-8*1^3(-1)^3=8
d: D=xy^2(2018+16-2016)
=18xy^2
=18(-2)*1/9=-4
\(A=x^{14}-10x^{13}+10x^2-10x^{11}\)\(+...+10x^{12}-10x+10\)
Thay x = 9 vào biểu thức A
\(\Rightarrow A=9^{14}-\left(9+1\right).9^{13}+\left(9+1\right).9^{12}\)\(-...+9+1\)
\(\Rightarrow A=9^{14}-9^{14}-9^{13}+9^{12}+...-9+9+1\)
\(\Rightarrow A=1\)
P/s tham khảo thêm trên google
Ta có 10=9+1=x+1(Vì x=9)
=>B= x14-(x+1)x13+(x+1)x12-(x+1)x11+.........-(x+1)x+10
=>B= x14-x14-x13+x13+x12-x12-x11+.....-x2-x+10
=>B=-x+10
Thay x=9, ta có
B=-9+10=1
\(B=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+...+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
Bài 1:
a: \(M=x^2-10x+3\)
\(=x^2-10x+25-22\)
\(=\left(x^2-10x+25\right)-22\)
\(=\left(x-5\right)^2-22>=-22\forall x\)
Dấu '=' xảy ra khi x-5=0
=>x=5
b: \(N=x^2-x+2\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x-1/2=0
=>x=1/2
c: \(P=3x^2-12x\)
\(=3\left(x^2-4x\right)\)
\(=3\left(x^2-4x+4-4\right)\)
\(=3\left(x-2\right)^2-12>=-12\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
`a)100x^2-20x+1`
`=(10x-1)^2`
Thay `x=1/10`
`=>100x^2-20x+1=(1-1)^2=0`
`b)49x^2-42x+10`
`=49*4/49-42*2/7+10`
`=4-12+10=2`
`c)25x^2+40x+16y^2`
`=(5x+4y)^2=(2+3)^2=25`
phan h 10=9+1
x=9=>10=x+1
thqy 10=x+1 vào A
ta có A=x^14 - (x+1)x^13+(x+1)x^12-(x+1)x^11+...+(x+1)x^2-(x+1)x+10
=x^14-x^14-x^13+x^13+x^12-x^12-x^11+...+x^3+x^2-x^2_x+10
=x+10
mà x=9
=>A=19