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\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
Ta có:\(\left|a\right|,\left|b\right|\) \(\leq\) \(1\)
\(\implies\) \(\left(1-a\right).\left(1-b\right)\) \(\geq\) \(0\)
\(\implies\) \(1-b-a+ab\)\(\geq\) \(0\)
\(\implies\) \(1+ab\) \(\geq\) \(a+b\)
\(\implies\) \(\left|1+ab\right|\) \(\geq\) \(\left|a+b\right|\) \(\left(đpcm\right)\)
a) \(\dfrac{a}{b}+\dfrac{-a}{b+1}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}+\dfrac{-a.b}{b\left(b+1\right)}=\dfrac{ab+a-ab}{b\left(b+1\right)}=\dfrac{a}{b\left(b+1\right)}\)
b) \(\dfrac{a}{b+1}+\dfrac{-a}{b}=\dfrac{ab}{b\left(b+1\right)}+\dfrac{-a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab-ab-a}{b\left(b+1\right)}=\dfrac{-a}{b\left(b+1\right)}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2\)
\(=\left(\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2=\left(\dfrac{b}{d}\right)^2\)(1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}\)
\(=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Lời giải:
Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$
$\Rightarrow x=at; y=bt; z=ct$. Ta có:
$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$
Mặt khác:
$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$
Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)
Ta có: VP = \(a\left(b^2-2bc+c^2\right)+b\left(c^2-2ac+a^2\right)+c\left(a^2-2ab+b^2\right)\)
= \(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2-6abc\)(1)
\(VT=\left(ab+b^2+ac+bc\right)\left(c+a\right)-8abc\)
\(=abc+b^2c+ac^2+bc^2+a^2b+b^2a+a^2c+abc-8abc\)
= \(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2-6abc\)(2)
Từ (1) ; (2) => VT = VP
Vậy đẳng thức luôn đúng.
2) 12723 < 12823= (27)23 = 2161
51318 >51218 = (29)18 = 2162
Vì 2161 < 2162 => 12723 < 2161 < 2162 < 51318
Vậy: 12723 < 51318
ai làm câu 1 tui tick đúng cho hehe^^