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\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{c+b}{c}\right)\left(\frac{a+c}{a}\right)\)
Mà a+b+c = 0 nên a + c = -b
a + b = -c
b + c = -a
\(A=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
1.
Ta có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\Leftrightarrow ab+ad< ad+bc\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\) (1)
Lại có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow bc>ad\Leftrightarrow bc+cd>ad+cd\Leftrightarrow c\left(b+d\right)>d\left(a+c\right)\Leftrightarrow\frac{c}{d}>\frac{a+c}{b+d}\) (2)
Từ (1) và (2) suy ra \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
2.
Ta có: a(b + n) = ab + an (1)
b(a + n) = ab + bn (2)
Trường hợp 1: nếu a < b mà n > 0 thì an < bn (3)
Từ (1),(2),(3) suy ra a(b + n) < b(a + n) => \(\frac{a}{n}< \frac{a+n}{b+n}\)
Trường hợp 2: nếu a > b mà n > 0 thì an > bn (4)
Từ (1),(2),(4) suy ra a(b + n) > b(a + n) => \(\frac{a}{b}>\frac{a+n}{b+n}\)
Trường hợp 3: nếu a = b thì \(\frac{a}{b}=\frac{a+n}{b+n}=1\)
Bài 2 : Theo ví dụ trên ta có : \(\frac{a}{b}< \frac{c}{d}\)=> ad < bc
Suy ra :
\(\Leftrightarrow ad+ab< bc+ba\Leftrightarrow a(b+d)< b(a+c)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\)
Mặt khác : ad < bc => ad + cd < bc + cd
\(\Leftrightarrow d(a+c)< (b+d)c\Leftrightarrow\frac{a+c}{b+d}< \frac{c}{d}\)
Vậy : ....
b, Theo câu a ta lần lượt có :
\(-\frac{1}{3}< -\frac{1}{4}\Rightarrow-\frac{1}{3}< -\frac{2}{7}< -\frac{1}{4}\)
\(-\frac{1}{3}< -\frac{2}{7}\Rightarrow-\frac{1}{3}< -\frac{3}{10}< -\frac{2}{7}\)
\(-\frac{1}{3}< -\frac{3}{10}\Rightarrow-\frac{1}{3}< -\frac{4}{13}< -\frac{3}{10}\)
Vậy : \(-\frac{1}{3}< -\frac{4}{13}< -\frac{3}{10}< -\frac{2}{7}< -\frac{1}{4}\)
a, ( x - 3 ) . ( x - 4 ) = 0
=> x - 3 = 0 hoặc x - 4 = 0
Nếu x - 3 = 0 => x = 3
Nếu x - 4 = 0 => x = 4
b, (\(\frac{1}{2}\)x - 4 ) . ( x - \(\frac{1}{4}\)) = 0
=>( \(\frac{1}{2}\)x - 4 ) = 0 Hoặc ( x - \(\frac{1}{4}\)) = 0
Nếu ( \(\frac{1}{2}\)x - 4 ) = 0 => x = \(\frac{8}{1}\)
Nếu ( x - \(\frac{1}{4}\)) = 0 => x = \(\frac{1}{4}\)
c, (\(\frac{1}{3}\)- x ) . ( \(\frac{1}{2}\)+ 1 : x ) = 0
=> ( \(\frac{1}{3}\)- x ) = 0 Hoặc ( \(\frac{1}{2}\)+ 1 : x ) = 0
Nếu (\(\frac{1}{3}\)- x ) = 0 => x = \(\frac{1}{3}\)
Nếu ( \(\frac{1}{2}\)+ 1 : x ) = 0 => x = \(\frac{-2}{1}\)
d, ( x + 3 ) . ( x - 4 ) + 2.(x + 3 ) = 0
=> (X + 3 ) = 0 Hoặc ( x - 4 ) = 0 Hoặc 2. ( x + 3 ) = 0
Nếu x + 3 = 0 => x = 0
Nếu ( x - 4 ) = 0 => x = 4
Nếu 2.(x + 3) = 0 => x = 3
# Cụ MAIZ
a. ( x - 3 ) ( x - 4 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
b. \(\left(\frac{1}{2}x-4\right)\left(x-\frac{1}{4}\right)=0\)
<=> \(\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
Bài làm :
\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
\(b\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=0+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1\div x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-0\\1\div x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
\(d\text{)}...\Leftrightarrow\left(x+3\right)\left(x-4+2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Bài làm :
\(a,\left(x-3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
\(b,\left(\frac{1}{2}x-4\right)\left(x-\frac{1}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
\(c,\left(\frac{1}{3}-x\right).\left(\frac{1}{2}+1:x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1:x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
\(d,\left(x+3\right)\left(x-4\right)+2\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-4+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Học tốt nhé
Bài làm :
\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
\(b\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=0+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1\div x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-0\\1\div x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
\(d\text{)}...\Leftrightarrow\left(x+3\right)\left(x-4+2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
a)
Với A=0
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
với A<0
\(\Rightarrow x\left(x-4\right)< 0\)
\(th1\orbr{\begin{cases}x< 0\\x-4>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 0\\x>4\end{cases}\Leftrightarrow4< x< 0\left(vl\right)}\)
\(th2\orbr{\begin{cases}x>0\\x-4< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>0\\x< 4\end{cases}\Leftrightarrow0< x< 4\left(tm\right)}\)
\(\Leftrightarrow0< x< 4\Leftrightarrow x\in\left\{1;2;3\right\}\)
Với A>0
\(\Rightarrow x\left(x-4\right)>0\)
\(th1\orbr{\begin{cases}x>0\\x-4>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>0\\x>4\end{cases}}\Leftrightarrow x>4\)
\(th2\orbr{\begin{cases}x< 0\\x-4< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 0\\x< 4\end{cases}}\Leftrightarrow x< 0\)
b)
Với B=0
\(\Rightarrow\frac{x-3}{x}=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\Rightarrow x=3\\x=0\left(l\right)\end{cases}}\)
vậy x=3 thì B = 0
Với B < 0
\(\Rightarrow\frac{x-3}{x}< 0\)
\(th1\orbr{\begin{cases}x-3>0\\x< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>3\\x< 0\end{cases}\Leftrightarrow3< x< 0\left(vl\right)}\)
\(th2\orbr{\begin{cases}x-3< 0\\x>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 3\\x>0\end{cases}\Leftrightarrow0< x< 3\left(tm\right)\Leftrightarrow x\in\left\{1;2\right\}}\)
Với B > 0
\(th1\orbr{\begin{cases}x-3>0\\x>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>3\\x>0\end{cases}\Leftrightarrow x>3}\)
\(th2\orbr{\begin{cases}x-3< 0\\x< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 0\end{cases}\Leftrightarrow x< 0}\)
Với N=0
=> a.b=0
=> \(\hept{\begin{cases}a=0\\\forall b\end{cases}}\)hoặc \(\hept{\begin{cases}b=0\\\forall a\end{cases}}\)
Với N>0
=> \(\orbr{\begin{cases}\hept{\begin{cases}a>0\\b>0\end{cases}}\\\hept{\begin{cases}a< 0\\b< 0\end{cases}}\end{cases}}\)