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\(a,A=\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\\ A=\left(5+7-36\right)\left(5+7+36\right)=-24\cdot48=-1152\\ b,B=\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\\ B=\left(2+2\right)\left(2-2-1\right)=4\cdot\left(-1\right)=-4\)
a: \(N=\left(2x-3y\right)\left(2x+3y\right)=\left(2x\right)^2-\left(3y\right)^2\)
\(=4x^2-9y^2\)
Thay x=1/2 và y=1/3 vào N, ta được:
\(N=4\cdot\left(\dfrac{1}{2}\right)^2-9\left(\dfrac{1}{3}\right)^2\)
\(=4\cdot\dfrac{1}{4}-9\cdot\dfrac{1}{9}\)
=1-1
=0
b: \(N=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)
\(=\left(2x\right)^3-y^3=8x^3-y^3\)
Khi x=1 và y=3 thì \(N=8\cdot1^3-3^3=8-27=-19\)
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349
1/
a, \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
b, \(4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+2xy+y^2\right)\left(2x^2-2xy+y^2\right)\)
c, \(x^4+324=x^4+36x^2+324-36x^2=\left(x^2+18\right)^2-\left(6x\right)^2=\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)
2/
a, \(x^2+\frac{1}{3}x+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2=\left(\frac{35}{6}+\frac{1}{6}\right)^2=6^2=36\)
b, \(x^2-y^2+2y-1=x^2-\left(y-1\right)^2=\left(x+y-1\right)\left(x-y+1\right)=\left(100+1-1\right)\left(100-1+1\right)=100.100=10000\)
a: \(N=\left(5x\right)^3-\left(2y\right)^3=1^3-1^3=0\)
b: \(Q=x^3+27y^3=\dfrac{1}{8}+\dfrac{27}{8}=\dfrac{28}{8}=\dfrac{7}{2}\)
Bài giải:
a) x2 + x+ tại x = 49,75
Ta có: x2 + x+ = x2 + 2 . x . + =
Với x = 49,75: = (49,75 + 0,25)2 = 502 = 2500
b) x2 – y2 – 2y – 1 tại x = 93 và y = 6
Ta có: x2 – y2 – 2y – 1 = x2 – (y2 + 2y + 1)
= x2 - (y + 1)2 = (x - y - 1)(x + y + 1)
Với x = 93, y = 6: (93 - 6 - 1)(93 + 6 + 1) = 86 . 100 = 8600
a) \(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\) tại \(x = 49,75\)
Ta có : \(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\) \(=\left(x^2+2.x.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right)\)
\(=\left(x+\dfrac{1}{4}\right)^2\)
Khi \(x = 49,75\) ,ta có :
\(\left(49,75+\dfrac{1}{4}\right)^2\) \(=\left(\dfrac{200}{4}\right)^2\)
\(= 50^2\)
\(= 2500\)
b) \(x^2 - y^2 - 2y - 1\) tại \(x = 93\) và \(y = 6\)
Ta có : \(x^2 - y^2 - 2y - 1 = x^2 - (y^2 + 2y +1)\)
\(= x^2 - (y + 1)^2\)
\(= (x- y - 1) ( x+ y +1)\)
Khi \(x = 93\) và \(y = 6\) , ta có :
\((93 - 6 - 1) ( 93 + 6 + 1)\) \(= 86 . 100\)
\(= 8600\)
\(a,A=y^2-\dfrac{1}{2}y+\dfrac{1}{16}\)
\(=y^2-2.y.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\)
\(=\left(y-\dfrac{1}{4}\right)^2\)
Với \(y=100,25\), ta được:
\(A=\left(100,25-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{401}{4}-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{400}{4}\right)^2=100^2=10000\)
\(------\)
\(b,B=4x^2-9y^2-6y-1\)
\(=\left(2x\right)^2-\left[\left(3y\right)^2+2.3y.1+1\right]\)
\(=\left(2x\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
Với \(x=23;y=1\), ta được:
\(B=\left(2.23-3.1-1\right)\left(2.23+3.1+1\right)\)
\(=\left(46-4\right)\left(46+4\right)\)
\(=42.50=2100\)