Bài 1: Tìm GTLN:
A=\(\dfrac{2016}{\left|2x+1\right|+2017}\)
B=\(\dfrac{\left(x^2-4\right)^2+2016}{-2017}\)
Câu 1 : Tìm GTLN
a) \(A=\dfrac{2003}{\left(x-2\right)^2+\left(x-y\right)^6+3}\)
b) \(B=3-\left(2x+\dfrac{1}{3}\right)^6\)
c) \(C=\dfrac{x^{2016}+2017}{x^{2016}+2015}\)
a: \(\left(x-2\right)^2+\left(x-y\right)^6+3\ge3\)
\(\Leftrightarrow A=\dfrac{2003}{\left(x-2\right)^2+\left(x-y\right)^6+3}\le\dfrac{2003}{3}\)
Dấu '=' xảy ra khi x=y=2
b: \(B=-\left(2x+\dfrac{1}{3}\right)^6+3\le3\forall x\)
Dấu '=' xảy ra khi x=-1/6
c: \(C=\dfrac{x^{2016}+2015+2}{x^{2016}+2015}=1+\dfrac{2}{x^{2016}+2015}\le\dfrac{2}{2015}+1=\dfrac{2017}{2015}\)
Dấu '=' xảy ra khi x=0
a, tính GT của đa thức \(f\left(x\right)=\left(x^4-3x+1\right)^{2016}\) tại \(x=9-\dfrac{1}{\sqrt{\dfrac{9}{4}-\sqrt{5}}}+\dfrac{1}{\sqrt{\dfrac{9}{4}+\sqrt{5}}}\)
b, so sánh \(\sqrt{2017^2-1}-\sqrt{2016^2-1}và\dfrac{2.2016}{\sqrt{2017^2-1}-\sqrt{2016^2-1}}\)
c, tính GTBT: \(sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)
d, biết \(\sqrt{5}\) là số hữu tỉ, hãy tìm các số nguyên a,b tm::
\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\)
a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
d.
\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)
Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì
\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)
Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)
Vậy \(\left(a;b\right)=\left(9;4\right)\)
Giải típ hộ mik nha !
\(\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{2016}{2017}=\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2016}{2017}=\dfrac{n+1-2}{2.\left(n+1\right)}=\dfrac{2016}{2017}=\dfrac{n-1}{2.\left(n+1\right)}=\dfrac{2016}{2017}=2017.\left(n-1\right)=2016.2\left(n+1\right)=...\)
Cho A=\(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right).\left(\dfrac{1}{2016}-1\right).\left(\dfrac{1}{2017}-1\right)\)
B=\(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right).\left(-1\dfrac{1}{2016}\right).\left(-1\dfrac{1}{2017}\right)\)
Tính M=A.B
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right)\left(\dfrac{1}{2016}-1\right)\left(\dfrac{1}{2017}-1\right)\\ A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)...\left(-\dfrac{2014}{2015}\right)\left(-\dfrac{2015}{2016}\right)\left(-\dfrac{2016}{2017}\right)\\ A=\dfrac{1.2.3.4...2014.2015.2016}{2.3.4...2015.2016.2017}=\dfrac{1}{2017}\)
\(B=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right)\left(-1\dfrac{1}{2016}\right)\left(-1\dfrac{1}{2017}\right)\\ B=\left(-\dfrac{3}{2}\right)\left(-\dfrac{4}{3}\right)\left(-\dfrac{5}{4}\right)...\left(-\dfrac{2016}{2015}\right)\left(-\dfrac{2017}{2016}\right)\left(-\dfrac{2018}{2017}\right)\\ B=\dfrac{3.4.5...2016.2017.2018}{2.3.4...2015.2016.2017}=\dfrac{2018}{2}=1009\)
\(M=A.B=\dfrac{1}{2017}.1009=\dfrac{1009}{2017}\)
Cho \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\) Hãy tính giá trị của biểu thức sau: \(A=f\left(\dfrac{1}{2017}\right)+f\left(\dfrac{2}{2017}\right)+...+f\left(\dfrac{2015}{2017}\right)+f\left(\dfrac{2016}{2017}\right)\)
Lời giải:
Ta thấy: \(f(x)=\frac{x^3}{1-3x+3x^2}\Rightarrow f(1-x)=\frac{(1-x)^3}{1-3(1-x)+3(1-x)^2}=\frac{(1-x)^3}{3x^2-3x+1}\)
\(\Rightarrow f(x)+f(1-x)=\frac{x^3}{1-3x+3x^2}+\frac{(1-x)^3}{3x^2-3x+1}=\frac{x^3+(1-x)^3}{3x^2-3x+1}=1\)
Do đó:
\(f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)=1\)
\(f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)=1\)
............
\(f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)=1\)
Cộng theo vế:
\(\Rightarrow A=f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+f\left(\frac{3}{2017}\right)+...f\left(\frac{2015}{2017}\right)+f\left(\frac{2016}{2017}\right)\)
\(=\underbrace{1+1+1...+1}_{1008}=1008\)
cho P=\(\left(\dfrac{2017}{x-1}-\dfrac{2016}{x+1}-\dfrac{2014+2016}{x^2-1}\right):\dfrac{x^2-4}{x^2-1}\)
a) tìm đk và rút gọn P
b)tìm x để P\(\ge0\)
c)tìm x nguyên để P nguyên
a: ĐKXĐ: x<>1; x<>2; x<>-2; x<>-1
\(P=\dfrac{2017x+2017-2016x+2016-2014x-2016}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2015x+2017}{x^2-4}\)
1) Tìm GTLN của các phân số:
a) \(\dfrac{2016}{\left|x\right|+2017}\) b) \(\dfrac{\left|x\right|+2016}{-2017}\)
2) Tìm GTNN của các phân số:
a) \(\dfrac{\left|x\right|+1945}{1975}\) b) \(\dfrac{-1}{\left|x\right|+1}\)
3) CMR: S=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+...+\(\dfrac{1}{100}\)>\(\dfrac{99}{100}\)
1a.Vì \(\left|x\right|\) là 1 số tự nhiên nên \(\left|x\right|+2017\ge2017\)(1)
Mà ta đã biết:\(\dfrac{a}{b}\ge\dfrac{a}{b+n}\)với n là một số tự nhiên.
Nên từ (1)suy ra\(\dfrac{2016}{\left|x\right|+2017}\le\dfrac{2016}{2017}\)
Vậy để \(\dfrac{2016}{\left|x\right|+2017}\)lớn nhất thì \(\dfrac{2016}{\left|x\right|+2017}=\dfrac{2016}{2017}\)
1b.Ta thấy:
\(\dfrac{\left|x\right|+2016}{-2017}=\dfrac{-\left(\left|x\right|+2016\right)}{2017}\)
Để \(\dfrac{-\left(\left|x\right|+2016\right)}{2017}\)lớn nhất thì \(-\left(\left|x\right|+2016\right)\)lớn nhất
Mà theo câu a,ta có:\(\left|x\right|\)+2016 là một số tự nhiên nên \(-\left(\left|x\right|+2016\right)\)mang dấu âm hay \(-\left(\left|x\right|+2016\right)\le0\)( chú ý \(-0=0\))
Vậy để \(-\left(\left|x\right|+2016\right)\)lớn nhất hay \(\dfrac{\left|x\right|+2016}{-2017}\)lớn nhất thì \(\left|x\right|+2016=0\)
\(\Rightarrow\)Để \(\dfrac{\left|x\right|+2016}{-2017}\)lớn nhất thì nó bằng \(\dfrac{0}{-2017}\)hay nó bằng 0
2)
a)Để \(\dfrac{\left|x\right|+1945}{1975}\)nhỏ nhất thì \(\left|x\right|+1945\) nhỏ nhất
Vì \(\left|x\right|\ge0\) nên \(\left|x\right|+1945\ge1945\)
\(\Rightarrow\)Để \(\left|x\right|+1945\) nhỏ nhất thì \(\left|x\right|+1945\) = 1945
\(\Rightarrow\)Để \(\dfrac{\left|x\right|+1945}{1975}\)bé nhất thì nó phải bằng \(\dfrac{1945}{1975}\)hay\(\dfrac{389}{395}\)
b)Để \(\dfrac{-1}{\left|x\right|+1}\)thì \(\left|x\right|+1\)bé nhất
Vì \(\left|x\right|\ge0\) nên \(\left|x\right|+1\ge1\)
\(\Rightarrow\)Để \(\left|x\right|+1\)bé nhất thì \(\left|x\right|+1\)\(=1\)
\(\Rightarrow\)GTNN của \(\dfrac{-1}{\left|x\right|+1}\)là \(\dfrac{-1}{1}\) hay -1
3)
Ta thấy:
S\(>\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{99.100}\)
Ta lại thấy:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{1}-\dfrac{1}{100}\)
=\(\dfrac{99}{100}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}>\dfrac{99}{100}\)(đpcm)
tính giá trị của biểu thức:
\(P=\left(2x^5+2x^4-x^3-1\right)^{2016}+\left(\sqrt{2x+2x-3x+3x+3}\right)^3+\dfrac{\left(2x^3+2x^2-x-3\right)^{2017}}{2x^4+2x^3-x^2-3^{2017}}\)
khi \(x=\sqrt{\dfrac{2-\sqrt{3}}{2}}\)
Câu 1:
a) \(\left|\left|3x-3\right|+2x+\left(-1\right)^{2016}\right|=3x+2017^0\)
b) Cho B = \(1+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+....+\dfrac{1}{x}.\left(1+2+3+...+4\right)=115\)(x thuộc số tự nhiên khác 0)