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(Mình giải theo cách lớp 8 nhé)
\(A=1^2-2^2+3^2-4^2+...+2015^2\)
\(=1+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2015^2-2014^2\right)\)
\(=1+\left(3-2\right)\left(3+2\right)+\left(5-4\right)\left(5+4\right)+...+\left(2015-2014\right)\left(2015+2014\right)\)
\(=1+\left(2+3\right)+\left(4+5\right)+...+\left(2014+2015\right)\)
\(=1+2+3+...+2015=B\)
\(\Leftrightarrow A=B\)
\(\Leftrightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Leftrightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x=2012\)
1,Thực hiện phép tính :
a, (x + 2)9 : (x + 2)6
=(x+2)9-6
=(x+2)3
b, (x - y) 4 : (x - 2)3
=(x-y)4-3
=x-y
c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)
=(x2+2x+4)5-1
=(x2+2x+4)4
d, 2(x2 + 1)3 : 1/3(x2 + 1)
=(2÷1/3).[(x2+1)3÷(x2+1)]
=6(x2+1)2
e, 5 (x - y)5 : 5/6 (x - y)2
=(5÷5/6).[(x-y)5÷(x-y)2]
=6(x-y))3
a: \(=\dfrac{5}{2}x-2x+\dfrac{7}{2}=\dfrac{1}{2}x+\dfrac{7}{2}\)
b: \(=\dfrac{-1}{4}x^4-3x^2+\dfrac{9}{4}x\)
c: \(=\dfrac{1}{5}x+\dfrac{1}{15}xy+\dfrac{7}{10}x^2\)
d: \(=-9x^3-1-12y+27xy\)
A=1^2-2^2+3^2-4^2+...+2015^2-2016^2+2017^2
A=2017^2-2016^2+2015^2-2014^2+...+5^2-4^2+3^2-2^2+1^2
A=(2017-2016)(2017+2016)+(2015-2014)(2015+2014)+...+(5-4)(5+4)+(3-2)(3+2)+1
A=2017+2016+2015+2014+...+5+4+3+2+1
A=(2017+1).2017:2
A=2018.2017:2
A=1009.2017=2035153
Hình như mình cũng làm như thế này thì phải mà sao vào thi violympic lại xai
Bài 3 :
Ta có : \(A=x^2+x+2012\)
=> \(A=x^2+x+\left(\frac{1}{2}\right)^2+\frac{8047}{4}\)
=> \(A=\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\)
- Ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\ge\frac{8047}{4}\forall x\)
- Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\)
<=> \(x=-\frac{1}{2}\)
Vậy MinA = \(\frac{8047}{4}\) <=> x = \(-\frac{1}{2}\) .
Bài 1 :
a, Ta có : \(\left(3x-2\right)\left(4+5x\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4+5x=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\5x=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{4}{5}\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = \(\frac{2}{3}\), x = \(-\frac{4}{5}\) .
b,- ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
=> \(x\ne\pm1\)
Ta có : \(\frac{x+1}{x-1}-\frac{4}{x+1}=\frac{3-x^2}{1-x^2}\)
=> \(\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}=\frac{x^2-3}{x^2-1}\)
=> \(\left(x+1\right)^2-4\left(x-1\right)=x^2-3\)
=> \(x^2+2x+1-4x+4=x^2-3\)
=> \(-2x=-3-5\)
=> \(x=4\left(TM\right)\)
Vậy phương trình có nghiệm là x = 4 .
c, Ta có : \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}-\frac{2-10x}{2014}\)
=> \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}+\frac{10x-2}{2014}\)
=> \(\frac{10x+3}{2009}+1+\frac{10x-1}{2013}+1=\frac{10x+1}{2011}+1+\frac{10x-2}{2014}+1\)
=> \(\frac{10x+3}{2009}+\frac{2009}{2009}+\frac{10x-1}{2013}+\frac{2013}{2013}=\frac{10x+1}{2011}+\frac{2011}{2011}+\frac{10x-2}{2014}+\frac{2014}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}=\frac{10x+2012}{2011}+\frac{10x+2012}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}-\frac{10x+2012}{2011}-\frac{10x+2012}{2014}=0\)
=> \(\left(10x+2012\right)\left(\frac{1}{2009}+\frac{1}{2013}-\frac{1}{2011}-\frac{1}{2014}\right)=0\)
=> \(10x+2012=0\)
=> \(x=-\frac{2012}{10}\)
Vậy phương trình có nghiệm là x = \(-\frac{2012}{10}\) .
Bài 3:
Giải:
Ta có : A = x2 + x + 2012
= x2 + 2.\(\frac{1}{2}\).x + \(\frac{1}{4}\) + \(\frac{8047}{4}\)
= (x + \(\frac{1}{2}\))2 + \(\frac{8047}{4}\) ≥ \(\frac{8047}{4}\)
⇒ Amin = \(\frac{8047}{4}\) ⇔ (x + \(\frac{1}{2}\))2 = 0 ⇔ x = \(-\frac{1}{2}\)
Vậy Amin = \(\frac{8047}{4}\) tại x = \(-\frac{1}{2}\)
Chúc bạn học tốt@@
Ta có : B = 202 - 192 + 182 - 172 + ..... + 22 - 12
=> B = (20 - 19)(20 + 19) + (18 - 17)(18 + 17) + ..... + (2 - 1)(2 + 1)
=> B = 39 + 35 + 31 + ..... + 3
Số số hạng của dãy trên là :
(39 - 3) : 4 + 1 = 10 (số)
Tổng B là :
(39 + 3) x 10 : 2 = 210
Vậy B = 210
Ta có : \(C=\left(15^4-1\right)\left(15^4+1\right)-3^8.5^8\)
\(\Rightarrow C=\left(15^4\right)^2-1-15^8\)
\(\Rightarrow C=15^8-1-15^8\)
=> C = -1
Vậy C = - 1
a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)
\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)
\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)
\(=-2x^2+2x+6\)
\(=-2\left(x^2-x-3\right)\)
b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)
\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
\(=4\left(x^2+5\right)\)
c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)
\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)
\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)
\(=-7x^2-20xy-17y^2+1\)
d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
\(=-3x^2\left(x^2-1\right)\)
\(=-3x^2\left(x-1\right)\left(x+1\right)\)
e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2\)
\(=\left(-2\right)^2=4\)
g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y+z\right)^2\)
\(=\left(x+2z\right)^2\)
h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)
\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)
\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)
\(=5x^2+2x^2+3x-1-3x-3\)
\(=7x^2-4\)
\(C=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2013-2014\right)\left(2013+2014\right)+2015^2\)
\(=2015^2-\left(1+2+3+4+...+2013+2014\right)\)
\(=2015^2-\dfrac{2015\cdot2014}{2}=2031120\)