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a) \(x^2-2xy-4z^2+y^2\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)-\left(2z\right)^2\)
\(\Leftrightarrow\left(x-y\right)^2-\left(2z\right)^2\)
\(\Leftrightarrow\left[\left(x-y\right)+2z\right]\left[\left(x-y\right)-2z\right]\)
\(\Leftrightarrow\left(x-y+2z\right)\left(x-y-2z\right)\)
Tại x=6, y=-4, z=45
\(\left[6-\left(-4\right)+2.45\right]\left[6-\left(-4\right)-2.45\right]=100.\left(-80\right)=-8000\)
b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(\Leftrightarrow3\left(x^2+7x-3x-21\right)+\left(x^2-4x+4\right)+48\)
\(\Leftrightarrow3x^2+21x-9x-63+x^2-4x+4+48\)
\(\Leftrightarrow4x^2+8x-11\)
Tại x=0,5 ta có:
\(4.\left(0,5\right)^2+8.0,5-11=-6\)
a)Đặt \(A=x^2-2xy-4z^2+y^2\)
\(=\left(x^2-2xy+y^2\right)-\left(2z\right)^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y-2z\right)\left(x-y+2z\right)\)
Thay \(x=6;y=-4;z=45\) vào A, ta có:
\(A=\left[6-\left(-4\right)-2\cdot45\right]\left[6-\left(-4\right)+2\cdot45\right]\)
\(=100\cdot\left(-80\right)\)
\(=-8000\)
Vậy \(A=-8000\)
b) Đặt \(B=3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=3\left(x^2+7x-3x-21\right)+x^2-4x+4+48\)
\(=3x^2+12x-63+x^2-4x+52\)
\(=4x^2+8x-11\)
Thay \(x=0,5\) vào B, ta có:
\(B=4\cdot\left(0,5\right)^2+8\cdot0,5-11\)
\(=1\cdot4-11\)
\(=-6\)
Vậy \(B=-6\)
a)\(P=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x=0\)
Vậy g/t P không phụ thuộc vào biến.
b)\(Q=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6\left(x^2-1\right)=-6x^2-2+6x^2-6=-8\)
Vậy g/t Q không phụ thuộc vào biến.
b) Ta có: \(Q=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)
\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)
\(=-2\left(x^2-2x+1+x^2-1+x^2+2x+1\right)+6\left(x^2-1\right)\)
\(=-2\left(3x^2+1\right)+6\left(x^2-1\right)\)
\(=-6x^2-2+6x^2-6\)
=-8
\(A=x^2-2xy-4z^2+y^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y+2z\right)\left(x-y-2z\right)\)
\(=\left(6+4+45\right)\left(6+4-45\right)\)
\(=-1925\)
`@` `\text {Ans}`
`\downarrow`
`P(x)+Q(x)-R(x)`
`= 5x^2 + 5x - 4 +2x^2 - 3x + 1 - (4x^2 - x + 3)`
`= 5x^2 + 5x - 4 + 2x^2 - 3x + 1 - 4x^2 + x - 3`
`= (5x^2 + 2x^2 - 4x^2) + (5x - 3x + x) + (-4 + 1 - 3)`
`= 3x^2 + 3x - 6`
Thay `x=-1/2`
`3*(-1/2)^2 + 3*(-1/2) - 6`
`= 3*1/4 - 3/2 - 6`
`= 3/4 - 3/2 - 6`
`= -3/4 - 6 = -27/4`
Vậy, khi `x=-1/2` thì GTr của đa thức là `-27/4`
P(x)+Q(x)-R(x)
=5x^2+5x-4+2x^2-3x+1-4x^2+x-3
=2x^2+3x-6(1)
Khi x=-1/2 thì (1) sẽ là 2*1/4+3*(-1/2)-6=1/2-3/2-6=-7
1, \(A=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(A=5x^3-15x+7x^2-5x^3-7x^2\)
\(A=\left(5x^3-5x^3\right)+\left(7x^2-7x^2\right)-15x\)
\(A=-15x\)
Thay \(x=-5\) vào A ta được:
\(-15\cdot-5=75\)
Vậy: ....
2. \(B=x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(B=x^3-3x+7x^2-5x^3-7x^2\)
\(B=\left(x^3-5x^3\right)+\left(7x^2-7x^2\right)-3x\)
\(B=-4x^3-3x\)
Thay \(x=10,y=-1\) vào B ta được:
\(-4\cdot10^3-3\cdot10=-4\cdot1000-3\cdot10=-4000-30=-4030\)
Vậy: ....
1: A=4x^2+12x+9-4x^2+4x-1-6x=10x+8
Khi x=201 thì A=10*201+8=2018
2: B=4x^2+20x+25-4x^2+12=20x+37
Khi x=1/20 thì B=1+37=38
1, \(A=\left(2x+3\right)^2-\left(2x-1\right)^2-6x\)
\(A=\left[\left(2x+3\right)+\left(2x-1\right)\right]\left[\left(2x+3\right)-\left(2x-1\right)\right]-6x\)
\(A=\left(2x+3+2x-1\right)\left(2x+3-2x+1\right)-6x\)
\(A=4\left(4x+2\right)-6x\)
\(A=16x+8-6x\)
\(A=10x+8\)
Thay \(x=201\) vào A ta có:
\(A=10\cdot201+8=2010+8=2018\)
Vậy: ....
2, \(B=\left(2x+5\right)^2-4\left(x+3\right)\left(x-3\right)\)
\(B=\left(2x+5\right)^2-4\left(x^2-9\right)\)
\(B=4x^2+20x+25-4x^2+36\)
\(B=20x+61\)
Thay \(x=\dfrac{1}{20}\) vào B ta có:
\(B=20\cdot\dfrac{1}{20}+61=1+61=62\)
Vậy: ...
Đặt \(g\left(x\right)=f\left(x\right)-x-1\Rightarrow g\left(2\right)=g\left(3\right)=g\left(4\right)=0\)
\(\Rightarrow g\left(x\right)\) có 3 nghiệm 2;3;4
\(\Rightarrow g\left(x\right)=a\left(x-2\right)\left(x-3\right)\left(x-4\right)\)
\(\Rightarrow f\left(x\right)=g\left(x\right)+x+1=a\left(x-2\right)\left(x-3\right)\left(x-4\right)+x+1\)
\(f\left(5\right)=10\Rightarrow a\left(5-2\right)\left(5-3\right)\left(5-4\right)+5+1=10\)
\(\Rightarrow a=\dfrac{2}{3}\)
\(\Rightarrow f\left(x\right)=\dfrac{2}{3}\left(x-2\right)\left(x-3\right)\left(x-4\right)+x+1\)
\(\Rightarrow f\left(6\right)=\dfrac{2}{3}.4.3.2+6+1=...\)