tìm n \(\in\) N để \(n^2+n+6\) là số chính phương
tìm n \(\in\) N để \(n^2+n+6\) là số chính phương
Đặt \(n^2+n+6=a^2\)
\(\Leftrightarrow4n^2+4n+24=4a^2\)
\(\Leftrightarrow4n^2+4n+1+23=4a^2\)
\(\Leftrightarrow\left(2n+1\right)^2+23=4a^2\)
\(\Leftrightarrow4a^2-\left(2n+1\right)^2=23\)
\(\Leftrightarrow\left(2a-2n-1\right)\left(2a+2n+1\right)=23\)
\(\forall n\in N\)thì \(2a+2n+1>2a-2n-1>0\)
\(\Rightarrow\left\{{}\begin{matrix}2a+2n+1=23\\2a-2n-1=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=6\\n=5\end{matrix}\right.\)
Vậy n = 5
1,rút gọ các phân thức sau
a,\(\frac{2a^2+b^5}{3a^2b^2}\)
b\(\frac{x^2+y^2-4+2xy}{x^2-y^2+4+4x}\)
2, rút gọn
A=\(\frac{a^2+ax+ab+bx}{a^2+ã-ab-bx}\)
1, b) \(\frac{x^2+y^2-4+2xy}{x^2-y^2+4+4x}\) = \(\frac{\left(x^2+2xy+y^2\right)-4}{\left(x^2+4x+4\right)-y^2}\) =\(\frac{\left(x+y\right)^2-2^2}{\left(x+2\right)^2-y^2}\)= \(\frac{\left(x+y+2\right)\left(x+y-2\right)}{\left(x+2+y\right)\left(x+2-y\right)}\) = \(\frac{x+y-2}{x+2-y}\)
2, A= \(\frac{a^2+ax+ab+bx}{a^2+ax-ab-bx}\) = \(\frac{\left(a^2+ax\right)+\left(ab+bx\right)}{\left(a^2+ax\right)-\left(ab+bx\right)}\) = \(\frac{a\left(a+x\right)+b\left(a+x\right)}{a\left(a+x\right)-b\left(a+x\right)}\)= \(\frac{\left(a+x\right)\left(a+b\right)}{\left(a+x\right)\left(a-b\right)}\)= \(\frac{a+b}{a-b}\)
Tính nhanh
Ta có :
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+....+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+....+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)
\(=\frac{6}{x\left(x+6\right)}\)
thực hiện phép tính
a)(5x-3)(5x+3); b) (2x^2)-3x(4x^2-5x+6); c) 16x^3y^4:4x^2y^2
a: \(\left(5x-3\right)\left(5x+3\right)=25x^2-9\)
b: \(=2x^2-12x^3+15x^2-18x=-12x^3+17x^2-18x\)
c: \(\dfrac{16x^3y^4}{4x^2y^2}=4xy^2\)
Cho f(x)= x^10+ax^3+b
g(x)=x^2-1
Tìm a,b để f(x) : g(x) dư 2x+1
Kí hiệu thương trong phép chia f(x) cho g(x) là q(x) thì ta có:
f(x)=q(x).(x2-1)+(2x+1)
=q(x).(x-1).(x+1)+(2x+1)
Có f(1)=110+a.13+b=2.1+1
=>a+b+1=3=>a+b=2 (1)
Có f(-1)=(-1)10+a.(-1)3+b=2.(-1)+1
=>-a+b+1=-1=>-a+b=-2 (2)
Cộng từng vế (1) và (2):
a+b-a+b=0<=>2b=0<=>b=0<=>a=2
Vậy a=2;b=0
Cho a^3 + b^3 + c^3 = 3abc . Tính số trị biểu thức : N=bc/a^2+ca/b^2+ab/c^2.
Áp dụng hằng đẳng thức
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
Do \(a^3+b^3+c^3=3abc\) nên \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0.\)
Do đó : \(\left[\begin{array}{nghiempt}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{array}\right.\)
Nếu \(a+b+c=0\) thì do \(a,b,c\ne0\),ta có :\(P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)
Nếu \(a^2+b^2+c^2-ab-bc-ac=0\) thì ta suy ra\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Điều này chỉ xảy ra khi \(a-b=0;b-c=0;a-c=0\Leftrightarrow a=b=c.\)
Khi đó \(P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\).
Vậy \(P=-1\) hoặc \(P=8.\)
cho \(\frac{a}{b}=\frac{10}{3}\)
tính M=\(\frac{16a^2-40ab}{8a^2-24ab}\)
\(\frac{a}{b}=\frac{10}{3}\)
\(\Rightarrow a=\frac{10}{3}b\)
\(M=\frac{16a^2-40ab}{8a^2-24ab}=\frac{8a\left(2a-5b\right)}{8a\left(a-3b\right)}=\frac{\frac{20}{3}b-5b}{\frac{10}{3}b-3b}=\frac{\frac{5}{3}b}{\frac{1}{3}b}=5\)
cho 4a2 + b2 =5ab với 2a > b > 0
tính M = \(\frac{ab}{4a^2-b^2}\)
\(4a^2+b^2=5ab\)
\(4a^2-5ab+b^2=0\)
\(4a^2-4ab-ab+b^2=0\)
\(4a\left(a-b\right)-b\left(a-b\right)=0\)
\(\left(a-b\right)\left(4a-b\right)=0\)
\(\left[\begin{array}{nghiempt}a-b=0\\4a-b=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=b\\4a=b\end{array}\right.\)
mà \(2a>b>0\)
\(\Rightarrow a=b\)
Thay a = b vào M, ta có:
\(M=\frac{b\times b}{4b^2-b^2}\)
\(=\frac{b^2}{3b^2}\)
\(=\frac{1}{3}\)
Vậy . . .
Cho \(M=\frac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Tính giá trị của M tại \(x=2014^{2015}-20142015;y=20142015-2015^{2014};z=2015^{2014}-2014^{2015}\)
Ta có:
\(M=\frac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{xyz-x^3+xyz-y^3+xyz-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{3xyz-x^3-y^3-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(-M=\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Xét đẳng thức phụ:
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=\left[\left(a +b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-ab\right]=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-abc-ac\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
Thay vào -M ta có:
\(-M=\frac{\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{1}{2}\left(x+y+z\right)\Rightarrow M=-\frac{1}{2}\left(x+y+z\right)\)
Giờ thay: \(x=2014^{2015}-20142015;y=20142015-2015^{2014};z=2015^{2014}-2014^{2015}\)
Ta có:
\(M=-\frac{1}{2}\left(2014^{2015}-20142015+20142015-2015^{2014}+2015^{2014}-2014^{2015}\right)=0\)
Bài 5:Cho x+y+z=0 chứng minh : \(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}+3=0\)
HELP ME....MAI MÌNH NỘP RỒI
mình cảm ơn
\(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}+3=0\)
\(\Leftrightarrow\frac{y+z}{x}+1+\frac{x+z}{y}+1+\frac{x+y}{z}+1=0\)
\(\Leftrightarrow\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\), luôn đúng
=> đpcm