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a) \(25^2-15^2=\left(25-15\right)\left(25+15\right)=10.40=400\)
b) \(87^2+73^2-27^2-12^2\)
\(=\left(87^2-13^2\right)+\left(73^2-27^3\right)\)
\(=\left(87-13\right)\left(87+13\right)+\left(73-27\right)\left(73+27\right)\)
\(=74.100+100.46\)
\(=100\left(74+46\right)\)
\(=100.120=12000\)
a) 252-152 b) 872+732-272-132
=(25-15)(25+15) =(872-132)+(732-272)
=10*30 =[(87-13)(87+13)]+[(73-27)(73+27)]
=300 =7400+4600
=12000
Giải:
1). \(25^2-15^2\)
\(=\left(25-15\right)\left(25+15\right)\)
\(=10.40\)
\(=400\)
2). \(87^2+73^2-27^2-13^2\)
\(=\left(87^2-13^2\right)+\left(73^2-27^2\right)\)
\(=\left(87-13\right)\left(87+13\right)+\left(73-27\right)\left(73+27\right)\)
\(=74.100+46.100\)
\(=100\left(74+46\right)\)
\(=100.120\)
\(=12000\)
Chúc bạn học tốt!!!
Bài 1 :
Theo giả thiết đã ra ta có :
\(n^2\left(n+1\right)+2n\left(n+1\right)\)
\(=\left(n+1\right)\left(n^2+2n\right)=n\left(n+1\right)\left(n+2\right)\) .
\(n\left(n+1\right)\left(n+2\right)\) là tích của 3 số nguyên liên tiếp nên luôn chia hết cho 6 .
Vì vậy \(n^2\left(n+1\right)+2n\left(n+1\right)\) luôn chia hết cho 6 với mọi số nguyên n ( đpcm )
Bài 2 :
Câu a : \(25^2-15^2=\left(25-15\right)\left(25+15\right)=10.40=400\)
Câu b : \(87^2+73^2-27^2-13^2=\left(87^2-13^2\right)+\left(73^2-27^2\right)\)
\(=\left(87+13\right)\left(87-13\right)+\left(73+27\right)\left(73-27\right)=100.74+100.46\)
\(=100\left(74+46\right)=100.120=12000\)
Bài 3 :
Câu a :
\(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2-0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x=0\) và \(x=\dfrac{1}{2}\)
Câu b :
\(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
Tính nhanh :
a) 252 - 152 = (25 + 15)(25 - 15) = 40 . 10 = 400
b) 872 + 732 - 272 - 132 = (872 - 132) + (732 - 272)
= (87 + 13)(87 - 13) + (73 + 27)(73 - 27)
= 100 . 74 + 100 . 26 = 100 . (74 + 26) = 100 . 100 = 10000
Bài 1:
a)\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x\cdot2y=2\left(x+y\right)\)
b) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)
Bài 2:
a) \(25^2-15^2=\left(25-15\right)\left(25+15\right)=10\cdot40=400\)
b) \(87^2+73^2-27^2-13^2=\left(87^2-27^2\right)+\left(73^2-13^2\right)\\ =\left(87-27\right)\left(87+27\right)+\left(73-13\right)\left(73+13\right)\)
\(=60\cdot114+60\cdot86=60\cdot\left(114+86\right)=60\cdot200=12000\)
Bài 2:
a) \(x^3-0,25\cdot x=0\)
\(\Leftrightarrow x^2\left(x-0,25\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-0,25=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=0,25\end{array}\right.\)
b) \(x^2-10=-25\)
\(\Leftrightarrow x^2=-15\) (vô nghiệm0
c) \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
d) \(8x^3+12x^2+6x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^3=0\)
\(\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
a, 25^2 - 15^2 = ( 25 - 15 )( 25 + 15) = 10 . 40 = 400
b, 87^2 + 73^2 - 27^2 - 13^2
= 87^2 - 27^2 + 73^2 - 13^2
= ( 87 - 27)( 87 + 27) + (73 - 13 )(73+ 13)
= 60 . 114 + 60 . 86
= 60( 114 + 86)
= 60 .200
= 12000
c, x^3 + 27 + 9 x^2 + 27x
= x^3 + 27x + 9x^2 + 27
=(x + 3)^3
thay x =97 ta có
= (97 + 3)^3
= 100^3
=1000000
d, 1,6^2 + 4.0,8.3,4 + 3,4^2 ( nè 3,4^2 chứ không phải 3,42)
= 1,6^2 + 2.2.0,8.3,4 + 3,4^2
=1,6^2 + 2.1,6.3,4 + 3,4^2
= (1,6 + 3,4)^2
= 5^2
= 25
e, x = 11 => 12 =x + 1 thay vào ta có
x^4 - ( x+ 1)x^3 + (x+1)x^2 -(x+1)x + 11
= x^4 - x^4 - x^13 + x^3 + x^2 - x^2 - x + 11
= -x + 11
= -11 + 11
= 0
ĐÚng ch o tui nha
872+732-272-132
=(872-132)+(732-272)
=(87+13)(87-13)+(73+27)(73-27)
=100.74+100.46
=100.(74+46)
=100.120
=12000
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y-x+y\right)\left(x+y+x-y\right)\)
\(=2y.2x\)
\(=4xy\)
b) (3x+1)2-(x+1)2
=(3x+1+x+1)[(3x+1)-(x+1)]
=(4x+2)2x
=2(2x+1)2x
=4x(2x+1)
c) x3+y3+z3-3xyz
=(x+y)3-3xy(x+y)+z3-3xyz
=[(x+y)3+z3]-3xy(x+y+z)
=(x+y+z)[(x+y)2-z(x+y)+z2]-3xy(x+y+z)
=(x+y+z)(x2+2xy+y2-xz-zy+z2-3xy)
=(x+y+z)(x2+y2+z2-xy-yz-zx)
a) \(25^2-15^2=\left(25-15\right)\left(25+15\right)\)
= 400
b) \(87^2+73^2-27^2-13^2\)
\(\Leftrightarrow\left(87^2-13^2\right)\)+\(\left(73^2-27^2\right)\)
\(\Leftrightarrow\left(87+13\right)\left(87-13\right)+\left(73+27\right)\left(73-27\right)\)
\(\Leftrightarrow7400+4600=12000\)