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a, Đặt \(A=x^2-y^2\)
Thay x = 87, y = 13 có:
\(A=87^2-13^2=\left(87+13\right)\left(87-13\right)\)
\(=100.74=7400\)
Vậy A= 7400 khi x = 87, y = 13
b, Đặt \(B=x^3-3x^2=x^3-3x^2+3x-1-3x+1\)
\(=\left(x-1\right)^3-\left(3x-1\right)\)
Thay x = 101 có:
\(B=100^3-302=999698\)
Vậy \(B=999698\) khi x = 101
c, Đặt \(C=x^3+9x^2+27x+27=\left(x+3\right)^3\)
Thay x = 97 \(\Rightarrow C=100^3=1000000\)
Vậy C = 1000000 khi x = 97
a, \(x^2-y^2=\left(x+y\right)\left(x-y\right)\)
Tại x = 87 ; y= 13
Ta có:
\(\left(87+13\right)\left(87-13\right)=100.74=7400\)
\(b,x^3-3x^2+3x-1-3x+1=\left(x-1\right)^3-\left(3x-1\right)\)Tại x = 101
Ta có :
\(\left(101-1\right)^3-\left(3x-1\right)=1000000-303+1=999698\)\(c,x^3+9x^2+27x+27=\left(x+3\right)^3\)
Tại x = 97
Ta có:
\(\left(97+3\right)^3=1000000\)
Bài 1:
Giải:
Đặt \(a=5k+4\)
Ta có: \(a^2=\left(5k+4\right)^2=25k^2+40k+16\)
\(=25k^2+40k+15+1\)
\(=5\left(5k^2+8k+3\right)+1\)
\(\Rightarrow a^2\) chia 5 dư 1
Vậy...
Bài 2:
a, Thay x = 87, y = 13 vào A có:
\(A=87^2-13^2=\left(87+13\right)\left(87-13\right)\)
\(=100.74=7400\)
Vậy A = 7400
b, \(B=x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
Thay x = 101 \(\Rightarrow B=100^3=1000000\)
Vậy B = 1000000
c, \(C=x^3+9x^2+27x+27=\left(x+3\right)^3\)
Thay x = 79 \(\Rightarrow C=82^3\)
Vậy \(C=82^3\)
1.
Theo đề bài ta có:
\(a=5x+4\left(x\in N\right)\)
\(a^2=\left(5x+4\right)^2=25x^2+40x+16=25x^2+40x+15+1=5\cdot\left(5x^2+8x+3\right)+1\)
\(a^2-1=5\cdot\left(5x^2+8x+3\right)⋮5\)
\(\Rightarrow a^2\) chia 5 dư 1
2.
a,
\(x^2-y^2\\ =\left(x+y\right)\left(x-y\right)\\ =\left(87+13\right)\left(87-13\right)\\ =100\cdot74\\ =7400\)
b,
\(x^3-3x^2+3x-1\\ =\left(x-1\right)^3\\ =\left(101-1\right)^3\\ =100^3\\ =1000000\)
c,
\(x^3+9x^2+27x+27\\ =\left(x+3\right)^3\\ =\left(97+3\right)^3\\ =100^3\\ =1000000\)
C1
a) -7x(3x-2)=-21x^2+14x
b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2
C2
a) (x-5)(x+5)
b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0
\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)
Vậy S={-5;2/3}
C3:
a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3
b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)
\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)
2.
a) . -x3 + 3x2 - 3x + 1
=13-3.12x+3.1.x2-x3
=(1-x)3
b)8- 12x + 6x2 - x3
=23-3.22.x+3.2.x2-x3
=(2-x)3
3.
a) x3 + 12x2 + 48x + 64 tại x = 6
=x3+3.x2.4+3x4+432
=(x+4)3thay x=6 ta được :
(6+4)3=103=1000
b) x3 - 6x2 + 12x - 8 tại x= 22
=x3-3.x2.2+3.x.22 -23
=(x-2)3 thay x=22 ta đc:
=(22-2)3=203=8000
\(A=\left(3x-y\right)^2-\left(3x+y\right)^2=\left(3x-y+3x+y\right)\left(3x-y-3x-y\right)\)
\(=6x.\left(-2y\right)=6.\frac{1}{2}.\left(-2.\frac{1}{3}\right)=2.\left(-1\right)=-2\)
\(B=\left(2x+3y\right)^2+\left(2x-3y\right)^2\)
\(=\left(2.\frac{1}{2}+3.\frac{1}{3}\right)^2+\left(2.\frac{1}{2}-3.\frac{1}{3}\right)^2\)
\(=\left(1+1\right)^2+\left(1-1\right)^2\)
\(=4+0=4\)
1.
a) \(\left(-2x^3\right)\)\(\left(x^2+5x-\frac{1}{2}\right)\) = \(-2x^5\)\(-10x^4\) \(+x^3\)
b) (\(6x^3-7x^2\)\(-x+2\))\(:\left(2x+1\right)\)=\(3x^2-5x+2\)
2.
a) 9x(3x-y) + 3y (y-3x)=9x(3x-y)-3y(3x-y)
= (9x-3y)(3x-y)
= 3(3x-y)(3x-y)
= 3(3x-y)^2
b) \(x^3-3x^2\)\(-9x+27\)= \(\left(x^3-3x^2\right)\)\(-\left(9x-27\right)\)
= \(x^2\left(x-3\right)\)\(-9\left(x-3\right)\)
= \(\left(x^2-9\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)^2\)
Bài 1 ) a ) \(\left(-2x^3\right)\left(x^2+5x-\frac{1}{2}\right)\)
\(=-2x^5-10x^4+x^3\)
b ) \(\left(6x^3-7x^2+x+2\right):\left(2x+1\right)\)
\(=3x^2-5x+2\)
2 ) a ) \(9x\left(3x-y\right)+3y\left(y-3x\right)\)
\(=9x\left(3x-y\right)-3y\left(3x-y\right)\)
\(=\left(3x-y\right)\left(9x-3y\right)\)
\(=3\left(3x-y\right)\left(x-y\right)\)
b ) \(x^3-3x^2-9x+27\)
\(=\left(x^3-3x^2\right)-\left(9x-27\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x^2-9\right)\left(x-3\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x-3\right)\)
\(b.x^4+4x^2-5=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(c.x^3-19x-30=x^3-25x+6x-30\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
a) Thay x=87 và y=13 vào biểu thức ta có:
x2-y2=(x-y)(x+y)=(87-13)(87+13)=74.100=7400
b) Thay x=101 và biểu thức ta có:
x3-3x2+3x-1=(x-1)3=(101-1)3=1003=1000000
c) Thay x=97 vào biểu thức ta có:
x3+9x2+27x+27=(x+3)3=(97+3)3=1003=1000000
a)x3-3x2+3x-1=(x-1)3
Thay x=101 vào ta được 1003=1 000 000
b)x2-y2=(x-y)(x+y)
Thay x=87;y=13 vào được
74*100=7400