1.
a. a/a+1 và a+1/a+2 ; c. 18/21 và 19/20
b. 17/19 và 2017/2019
2.cho M = 2011 - 45/x+2
a. tính M biết x = 7
b. tính giá trị của x để M có giá trị nhỏ nhất
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\(a,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0.abc=0\)
Mà \(a+b+c=1=>\left(a+b+c\right)^2=1=>a^2+b^2+c^2+2ab+2bc+2ac=1\)
\(=>a^2+b^2+c^2+2\left(ab+bc+ac\right)=1=>a^2+b^2+c^2=1-0=1\) (vì ab+bc+ac=0)
\(b,S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3\)
\(=2014.\frac{1}{2014}-3=1-3=-2\)
Vậy.....................
B=(2+1)(22+1)(24+1)...(22016+1)+1
B=(2-1)(2+1)(22+1)...(22016+1)+1
B=(22-1)(22+1)...(22016+1)+1
B=(24-1)(24+1)...(22016+1)+1
...........................
B=(22016-1)(22016+1)+1
B=(22016)2-1+1=42016
a) \(\dfrac{1+\sqrt{a}}{1-\sqrt{a}}=\dfrac{a+2\sqrt{a}+1}{1-a}\)
b) \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}=\dfrac{-\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}=-\sqrt{a}\)
a) \(\dfrac{a}{3\sqrt{a}-1}=\dfrac{a\left(3\sqrt{a}+1\right)}{9a-1}\)
\(1,4a=5b\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{4}=\dfrac{b-a}{4-5}=\dfrac{27}{-1}=-27\\ \Leftrightarrow\left\{{}\begin{matrix}a=-135\\b=-108\end{matrix}\right.\\ 2,\dfrac{1}{3}x=\dfrac{1}{2}y=\dfrac{1}{5}z\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{5}=\dfrac{x+2y-z}{3+4-5}=\dfrac{8}{2}=4\\ \Leftrightarrow\left\{{}\begin{matrix}x=12\\y=8\\z=20\end{matrix}\right.\\ 3,\dfrac{1}{3}a=\dfrac{1}{2}b;\dfrac{1}{5}a=\dfrac{1}{7}c\\ \Leftrightarrow\dfrac{a}{15}=\dfrac{b}{10}=\dfrac{c}{21}=\dfrac{a+b+c}{15+10+21}=\dfrac{184}{46}=4\\ \Leftrightarrow\left\{{}\begin{matrix}a=60\\b=40\\c=84\end{matrix}\right.\)
(a + 1)(a + 2)(a + 3) - a(a + 1)(a + 2) = (a + 1)(a + 2)[(a + 3) - a] = 3(a + 1)(a + 2)
Ta có: (a+1).(a+2).(a+3) - a.(a+1).(a+2)=(a+1).(a+2).(a+3-a)=(a+1).(a+2).3
=>(a+1).(a+2).(a+3) - a.(a+1).(a+2) = 3.(a+1).(a+2)
a: Ta có: \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)
\(=\dfrac{6\sqrt{a}-6+10-2\sqrt{a}}{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)
\(=\dfrac{4\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\dfrac{1}{4\sqrt{a}}\)
\(=\dfrac{1}{\sqrt{a}}\)
a) \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{6\left(\sqrt{a}-1\right)+10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)
b) \(C=B.\left(a-\sqrt{a}+1\right)=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}-1+\dfrac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\dfrac{1}{\sqrt{a}}}-1=1\)(bất đẳng thức Cauchy cho 2 số dương)
mình cần gấp nhé
tớ ko bt