So sánh(không dùng bảng số hay máy tính cầm tay)
a)\(\dfrac{1}{7}\sqrt{51}\) với \(\dfrac{1}{9}\sqrt{150}\)
b)\(\sqrt{2017}-\sqrt{2016}\) với \(\sqrt{2016}-\sqrt{2015}\)
so sánh (ko dùng bảng số hay máy tính cầm tay):
a) \(\frac{1}{7}\sqrt{51}với\frac{1}{9}\sqrt{150}\)
b) \(\sqrt{2017}-\sqrt{2016}với\sqrt{2016}-\sqrt{2015}\)
So sánh (không dùng bảng số hay máy tính cầm tay ) :
a. \(\frac{1}{7}\).\(\sqrt{51}\)với \(\frac{1}{9}\).\(\sqrt{150}\)
b. \(\sqrt{2017}\)-\(\sqrt{2016}\)với \(\sqrt{2016}\)-\(\sqrt{2015}\)
a)1/7\(\sqrt{51}\)=\(\sqrt{\frac{51}{49}}\);1/9\(\sqrt{150}=\sqrt{\frac{150}{81}}=\sqrt{\frac{50}{27}}\)
\(\frac{51}{49}=1+\frac{1}{49}+\frac{1}{49}\);\(\frac{50}{27}=1+\frac{23}{27}>1+\frac{23}{36}>\)\(1+\frac{2}{36}=1+\frac{1}{36}+\frac{1}{36}\)
1/49<1/36 nên 51/49<50/27 =>1/7\(\sqrt{51}\)<1/9\(\sqrt{150}\)
b) \(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}\)+\(\sqrt{2015}\)
=>\(\frac{1}{\sqrt{2017}+\sqrt{2016}}< \)\(\frac{1}{\sqrt{2016}+\sqrt{ }2015}\) <=> \(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}\)-\(\sqrt{2015}\)
so sanh
\(\dfrac{1}{7}\sqrt{51}với\dfrac{1}{9}\sqrt{150}\)
\(\sqrt{2017}-\sqrt{2016}với\sqrt{2016}-\sqrt{2015}\)
\(\dfrac{1}{7}\sqrt{51}với\dfrac{1}{9}\sqrt{150}\)
<=> \(\dfrac{\sqrt{51}}{7}với\dfrac{\sqrt{150}}{9}\)
<=> \(9\sqrt{51}với7\sqrt{150}\)
<=> \(\sqrt{4131}với\sqrt{7350}\)
=> \(\sqrt{4131}< \sqrt{7350}\)
=> \(\dfrac{1}{7}\sqrt{51}< \dfrac{1}{9}\sqrt{150}\)
So sánh: \(\frac{2016}{\sqrt{2015}}+\frac{2015}{\sqrt{2016}}\text{ và }\sqrt{2015}+\sqrt{2016}\)(không dùng máy tính cầm tay)
So sánh ( ko dung bảng số, máy tính )
a) \(\dfrac{1}{7}\sqrt{51}\) vs \(\dfrac{1}{9}\sqrt{150}\)
b) \(\sqrt{2017}-\sqrt{2016}\) vs \(\sqrt{2016}-\sqrt{2015}\)
Lời giải:
a)
Ta có: \(\frac{1}{7}\sqrt{51}< \frac{1}{7}\sqrt{64}=\frac{8}{7}\)
\(\frac{1}{9}\sqrt{150}> \frac{1}{9}\sqrt{144}=\frac{12}{9}=\frac{4}{3}=\frac{8}{6}> \frac{8}{7}\)
Do đó: \(\frac{1}{7}\sqrt{51}< \frac{1}{9}\sqrt{150}\)
b)
\(\sqrt{2017}-\sqrt{2016}=\frac{2017-2016}{\sqrt{2017}+\sqrt{2016}}=\frac{1}{\sqrt{2017}+\sqrt{2016}}< \frac{1}{\sqrt{2016}+\sqrt{2015}}\)
\(\sqrt{2016}-\sqrt{2015}=\frac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\frac{1}{\sqrt{2016}+\sqrt{2015}}\)
Do đó:
\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
Không dùng máy tính, hãy so sánh \(\sqrt{2017}-\sqrt{2016}\) và \(\sqrt{2016}-\sqrt{2015}\)
\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
2017>2015
=>căn 2017>căn 2015
=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)
=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
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 \(\left(a;b\right)=\left(9;4\right)\)
1.So sánh A = \(\sqrt{2014}+\sqrt{2015}+\sqrt{2016}\) và B = \(\sqrt{2011}+\sqrt{2013}+\sqrt{2021}\) mà không dùng máy tính và bảng số.
2.Giải phương trình : \(\sqrt{\left(x-2015\right)^{14}}+\sqrt{\left(x-2016\right)^{10}}=1\)
Không dùng máy tính, hãy so sánh: \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}v\text{à}\sqrt{2016}+\sqrt{2017}\)
\(\frac{2016}{\sqrt{2016}}=\sqrt{2016}\)
\(\frac{2017}{\sqrt{2017}}=\sqrt{2017}\)
=> Bằng nhau
\(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}=\left(\frac{2016}{\sqrt{2017}}-\sqrt{2017}\right)+\left(\frac{2017}{\sqrt{2016}}-\sqrt{2016}\right)\)
\(=\frac{2016-2017}{\sqrt{2017}}+\frac{2017-2016}{\sqrt{2016}}=\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)
vì \(2016< 2017\Rightarrow\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)
\(\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}>0\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2016}+\sqrt{2017}\)