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ta có : \(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)
\(=20^2-19^2+18^2-17^2+...+2^2-1^2\)
\(=\left(20^2-1^2\right)-\left(19^2-2^2\right)+\left(18^2-3^2\right)-...-\left(11^2-10^2\right)\)
\(=21.\left(20-1\right)-21\left(19-2\right)+21\left(18-3\right)-...-21\left(11-10\right)\)
\(=21.19-21.17+21.15-...-21.1\)
\(=21\left(19-17+15-13+...+3-1\right)\)
\(=21\left(2+2+...+2\right)=21.2.5=210\)
Ta có:\(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)
\(=20^2+18^2+16^2+...+4^2+2^2-19^2-17^2-15^2-...-3^2-1^2\)
\(=(20^2-19^2)+(18^2-17^2)+...+(4^2-3^2)+(2^2-1^2)\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)
\(=20+19+18+17+...+4+3+2+1\)
\(=\dfrac{\left(20+1\right).20}{2}=\dfrac{21.20}{2}=210\)
Bài 1: Rút gọn
a) Ta có: \(A=\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2\)
\(=2x+15\)
b) Ta có: \(B=\left(2x-1\right)^2+2\left(2x-1\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(2x-1+x+1\right)^2\)
\(=\left(3x\right)^2=9x^2\)
Bài 2: Tính nhanh
a) Ta có: \(A=138^2+124\cdot138+62^2\)
\(=138^2+2\cdot138\cdot62+62^2\)
\(=\left(138+62\right)^2\)
\(=200^2=40000\)
b) Ta có: \(B=\left(100^2+98^2+...+2^2\right)-\left(99^2+97^2+...+3^2+1^2\right)\)
\(=100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+..+2+1\)
\(=5050\)
Bài 3: Chứng minh rằng các biểu thức sau luôn nhận giá trị dương với mọi giá trị của biến
a) Ta có: \(x^2-5x+10\)
\(=x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}+\frac{75}{4}\)
\(=\left(x-\frac{5}{2}\right)^2+\frac{75}{4}\)
Ta có: \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{5}{2}\right)^2+\frac{75}{4}\ge\frac{75}{4}\forall x\)
hay \(x^2-5x+10>0\forall x\)(đpcm)
b) Ta có: \(\left(x-1\right)\left(x-2\right)+5\)
\(=x^2-3x+2+5\)
\(=x^2-3x+7\)
\(=x^2-2\cdot x\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{19}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\)
Ta có: \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)
hay \(\left(x-1\right)\left(x-2\right)+5>0\forall x\)(đpcm)
#)Giải :
Bài 2 :
A = 1 + 5 + 52 + 53 + ... + 549 + 550
=> 5A = 5 + 52 + 53 + ...+ 550 + 551
=> 5A - A = 4A = ( 5 + 52 + 53 + ... + 550 + 551 ) - ( 1 + 5 + 52 + 53 + ... + 549 + 550 )
=> 4A = 551 - 1
=> A = 551 - 1 / 4
#)Giải :
Bài 1 :
a) ( x - 1/2 )2 + ( y + 1/2 )2 = 0
Ta có : ( x - 1/2 )2 ≥ 0 ; ( y + 1/2 )2 ≤ 0
=> ( x - 1/2 )2 = 0 ; ( y + 1/2)2 = 0
=> ( x - 1/2 )2 = 0 => x - 1/2 = 0 => x = 1/2
=> ( y + 1/2 )2 = 0 => y + 1/2 = 0 => y = -1/2
Vậy x = 1/2 ; y = -1/2
P/s : Maybe right ...
A = x 2x2 - 4 và 24và2 tại x = 1.856; y = -0,988
B = ( x 4 - y 4 )(x4-và4) : ( x 2 + y 2 )(x2+và2) tại x = 2003 ; y= 2004
A= chắc sai đề
B=( x 4 - y 4 )(x4-và4) : ( x 2 + y 2 )
=(x^2+y^2).(x^2-y^2)/(x^2+y^2)
=x^2-y^2
=(x-y)(x+y)
thay số =(2003-2004)(2003+2004)=-4007
#)Giải :
B = 2100 - 299 + 298 - 297 + ... + 22 - 2
=>2B = 2101 - 2100 + 299 - 298 + ... + 23 - 22
=>2B + B = ( 2101 - 2100 + 299 - 298 + ... + 23 - 22 ) + ( 2100 - 299 + 298 - 297 + ... + 22 - 2 )
=>3B = 2201 - 2
=>B = 2201 - 2 / 3
\(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow2B+B=2^{101}-2^2\)
\(\Rightarrow3B=2^{101}-2^2\)
\(\Rightarrow B=\frac{2^{101}-2^2}{3}\)
A = 1002 - 992 + 982 - 972 + . . . + 22 - 12
= (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + . . . (2 - 1)(2 + 1)
= 199 + 195 + . . . + 3
= 5050
B = 3(22 + 1)(24 + 1) . . . (264 + 1) + 1
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1)(264 + 1) + 1
= (24 - 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (28 - 1)(28 + 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (216 - 1)(216 + 1)(232 + 1)(264 + 1) + 1
= (232 - 1)(232 + 1)(264 + 1) + 1
= (264 - 1)(264 + 1) + 1
= 2128 - 1 + 1
= 2128
Các bạn ơi, mình làm đc rồi nên các bạn ko cần giúp mình nữa đâu mà giúp các bạn khác đi nha!!
Chúc các bạn thành công, may mắn và vui vẻ!!!
\(\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+2}\)
\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+abc}\)
\(=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{b+1+bc}+\frac{2c}{c\left(a+ab+2\right)}\)
\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{2}{a+2+ab}\)
\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{bc}{b+bc+1}\)
\(=\frac{b+bc+1}{b+bc+1}=1\)
Theo bài ra , ta có :
\(M=\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+2}\)
\(\Leftrightarrow\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2bc}{b\left(ac+2c+2\right)}\)(Vì abc = 2 )
\(\Leftrightarrow\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2bc}{abc+2bc+2b}\)
\(\Leftrightarrow\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2bc}{2+2bc+2b}\)( Vì abc = 2 )
\(\Leftrightarrow\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2bc}{2\left(1+bc+b\right)}\)
\(\Leftrightarrow\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)
\(\Leftrightarrow\frac{1+b+bc}{b+1+bc}=1\)
Vậy M=1
Chúc bạn học tốt =))
Phan Cả Phát xin hết !!!
ta có : \(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-1^2\right)-\left(99^2-2^2\right)+\left(98^2-3^2\right)-...+\left(52^2-49^2\right)-\left(51^2-50^2\right)\)
\(=101\left(100-1\right)-101\left(99-2\right)+101\left(98-3\right)-...+101\left(52-49\right)-101\left(51-50\right)\)
\(=101.99-101.97+101.95-...+101.3-101.1\)
\(=101\left(99-97+95-93+...+3-1\right)\)
\(=101.\left(2+2+2+...+2\right)=101.2.25=5050\)
Mình có cách khác nha !
\(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(97-97\right)+...\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\dfrac{100.101}{2}=5050\)