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\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(c+1=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\dfrac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(VP=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}=\dfrac{2}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2}}\)
\(=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(\Rightarrow VT=VP\) (đpcm)
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
ta có \(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4a+4\sqrt{abc}\)
=> \(4a+4\sqrt{abc}=16-4b-4c\Leftrightarrow4a+4\sqrt{abc}+bc=16-4b-4c+bc\)
=> \(\left(2\sqrt{a}+\sqrt{bc}\right)^2=\left(4-b\right)\left(4-c\right)\Rightarrow a\left(4-b\right)\left(4-c\right)=a\left(2\sqrt{a}+\sqrt{bc}\right)^2\)
=> \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a}\left(2\sqrt{a}+\sqrt{bc}\right)=2a+\sqrt{abc}\)
tương tự như thế thay vào , thì A=8
Ta có:
\(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4c+4\sqrt{abc}\)
\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\Leftrightarrow4a+4\sqrt{abc}+bc=16-4b-4c+bc\)
\(\Rightarrow\left(2\sqrt{a}+\sqrt{bc}\right)^2=\left(4-b\right)\left(4-c\right)\Rightarrow a\left(4-b\right)\left(4-c\right)=a\left(2\sqrt{a}+\sqrt{bc}\right)^2\)
\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a}\left(2\sqrt{a}+\sqrt{bc}\right)=2a+\sqrt{abc}\)
Tương tự như thế thay vào, thì A = 8
Câu 1 :
a ) \(\sqrt{0,36.100}=\sqrt{36}=6\)
b ) \(\sqrt[3]{-0,008}=\sqrt[3]{\left(-0,2\right)^3}=-0,2\)
c ) \(\sqrt{12}+6\sqrt{3}+\sqrt{27}=2\sqrt{3}+6\sqrt{3}+3\sqrt{3}=11\sqrt{3}\)
Câu 2 :
a ) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}=a-\sqrt{ab}+b\)
bai nay t lam roi vao trang chu cua nick thangbnsh cua t keo xuong tim la thay
Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)
Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì \(\left(a-b=b-c\right)\)
\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\dfrac{\sqrt{a}-\sqrt{c}}{b-c}\left(\sqrt{a}+\sqrt{c}\right)\)
\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)