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\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
=> (a+b).\(\left(\dfrac{1}{b}+\dfrac{1}{b}\right)\ge\left(a+b\right).\dfrac{4}{a+b}=4\left(dpcm\right)\)
b)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+b+c}\)
=>\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{a+b+c}=9\left(dpcm\right)\)
b)\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}\) và \(\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
\(\Rightarrow1+\dfrac{a+b}{c}=1+\dfrac{b+c}{a}\)và \(1+\dfrac{b+c}{a}=1 +\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{c}{c}+\dfrac{a+b}{c}=\dfrac{a}{a}+\dfrac{b+c}{a}\)và \(\dfrac{a}{a}+\dfrac{b+c}{a}=\dfrac{b}{b}+\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}\)và \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{a+b+c}{a}=0\) \(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
và \(\dfrac{a+b+c}{a}-\dfrac{a+b+c}{b}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\)
+) Vì a,b,c đôi một khác 0
\(\Rightarrow a+b+c=0\)
\(\rightarrow a+b=\left(-c\right)\)
\(\rightarrow a+c=\left(-b\right)\)
\(\rightarrow b+c=\left(-a\right)\)
+) Ta có:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
Bài 1: Nhân chéo
Bài 2:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
\(\Rightarrowđpcm\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)
\(=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}\)
\(=\dfrac{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}\)
\(=\dfrac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow c=-c\)
\(\Rightarrow c+c=0\)
\(\Rightarrow2c=0\Rightarrow c=0\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)
\(=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
Bài 1:
a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)
=>2 căn x=6
=>căn x=3
=>x=9
b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)
=>x=1
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(=\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\)
\(=\dfrac{\left(a+b+b+c+c+a\right)-\left(c+a+b\right)}{a+b+c}\)
\(=\dfrac{2a+2b+2c-a-b-c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-c}{c}=1\\\dfrac{b+c-a}{a}=1\\\dfrac{c+a-b}{b}=1\end{matrix}\right.\)
\(PHUCDZ=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
\(PHUCDZ=\left(\dfrac{b+c-a}{a}+\dfrac{b}{a}\right)\left(\dfrac{c+a-b}{b}+\dfrac{c}{b}\right)\left(\dfrac{a+b-c}{c}+\dfrac{a}{c}\right)\)
\(PHUCDZ=\dfrac{b+c-a+b}{a}.\dfrac{c+a-b+c}{b}.\dfrac{a+b-c+a}{c}\)
\(PHUCDZ=\dfrac{2b+c-a}{a}.\dfrac{2c+a-b}{b}.\dfrac{2a+b-c}{c}\)
\(PHUCDZ=\dfrac{\left(2b+c-a\right)\left(2c+a-b\right)\left(2a+b-c\right)}{abc}\)
Ta có: \(a-b-c=0\)
\(\Rightarrow\left\{{}\begin{matrix}a=b+c\\b=a-c\\c=a-b\end{matrix}\right.\)
Ta thay: \(a=b+c;b=a-c;c=a-b\) vào biểu thức \(A\), ta đc:
\(A=\left(1-\dfrac{a-b}{a}\right)\left(1-\dfrac{b+c}{b}\right)\left(1+\dfrac{a-c}{c}\right)\)
\(=\left(1-\dfrac{a}{b}+\dfrac{b}{a}\right)\left(1-\dfrac{b}{b}-\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}-\dfrac{c}{c}\right)\)
\(\dfrac{b}{a}.\dfrac{-c}{b}.\dfrac{a}{c}=-1\)
Chúc bạn học tốt!
Bài 1a):
Ta có:
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\left(a+b\right).\dfrac{a+b}{ab}=\dfrac{a^2+2ab+b^2}{ab}=\dfrac{a^2+b^2}{ab}+2\)
Lại có: (a - b)2 = a2 - 2ab + b2 \(\ge\) 0
\(\Rightarrow\) a2 + b2 \(\ge\) 2ab
\(\Rightarrow\) \(\dfrac{a^2+b^2}{ab}\ge2\)
\(\Rightarrow\) \(\dfrac{a^2+b^2}{ab}+2\ge4\)
Vậy \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
Bài 2a):
Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\ge0\)
\(\Rightarrow a+b\ge2\sqrt{ab}\)
Vậy ta có đpcm