Violympic toán 9

a) \(\sqrt{\dfrac{7}{27}}\) \(=\sqrt{\dfrac{21}{81}}\) \(=\dfrac{\sqrt{21}}{\sqrt{81}}\)  \(=\dfrac{\sqrt{21}}{9}\)

b) \(\sqrt{\dfrac{5}{11}}=\sqrt{\dfrac{55}{121}}\) \(=\dfrac{\sqrt{55}}{\sqrt{121}}\) \(=\dfrac{\sqrt{55}}{11}\)

Có \(x^2+x-1=\left(\dfrac{-1-\sqrt{5}}{2}\right)^2+\dfrac{-1-\sqrt{5}}{2}-1\)

\(=\dfrac{6+2\sqrt{5}}{4}+\dfrac{-3-\sqrt{5}}{2}\)\(=\dfrac{3+\sqrt{5}}{2}+\dfrac{-3-\sqrt{5}}{2}\)

\(=0\)

\(\Rightarrow x^2=1-x\) và \(-x^2-x=-1\)

Có \(M=\left(4x^5+4x^4-5x^3+2x-2\right)^{2018}+2019\)

\(=\left[4x^3\left(x^2+x-1\right)-x^3-2\left(1-x\right)\right]^{2018}+2019\)

\(=\left[-x.x^2-2x^2\right]^{2018}+2019\)

\(=\left[-x\left(1-x\right)-2x^2\right]^{2018}+2019\)

\(=\left(-x^2-x\right)^{2018}+2019\)

\(=\left(-1\right)^{2018}+2019\)

\(=2020\)

Đặt \(4x^5+4x^4-5x^3+2x-2=A\)

Xét pt \(t^2+t-1=0\)

\(\Delta=b^2-4ac=1+4=5\Rightarrow\left[{}\begin{matrix}t=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{5}}{2}\\t=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)

\(\Rightarrow x\) là nghiệm của pt \(t^2+t-1=0\Rightarrow x^2+x-1=0\)

\(A=4x^5+4x^4-4x^3-x^3-x^2+x+x^2+x-1-1\)

\(=4x^3\left(x^2+x-1\right)-x\left(x^2+x-1\right)+x^2+x-1-1=-1\)

\(\Rightarrow M=\left(-1\right)^{2018}+2019=2020\)

Đk: \(x\ge3\)

PT \(\Leftrightarrow25.\dfrac{\sqrt{x-3}}{5}-7.\sqrt{\dfrac{4\left(x-3\right)}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9\left(x^2-9\right)}{81}}=0\)

\(\Leftrightarrow5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\dfrac{1}{3}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\dfrac{1}{3}=\sqrt{x+3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=\dfrac{-26}{9}\left(L\right)\end{matrix}\right.\)

Vậy..

\(\sqrt{18x+9}-\sqrt{8x+4}+\dfrac{1}{3}\sqrt{2x+1}=4\) (đk:\(x\ge-\dfrac{1}{2}\))

\(\Leftrightarrow\sqrt{9\left(2x+1\right)}-\sqrt{4\left(2x+1\right)}+\dfrac{1}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\dfrac{1}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\sqrt{2x+1}=3\)

\(\Leftrightarrow x=4\)(TM)

Vạy...

b) Có \(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\left(\dfrac{5}{2}\right)^2.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)

\(6\sqrt{\dfrac{1}{37}}=\sqrt{\dfrac{6^2}{37}}=\sqrt{\dfrac{36}{37}}\)

Có \(\dfrac{25}{24}>1>\dfrac{36}{37}\)

\(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)

a)\(A=\sqrt{12}+3\sqrt{27}-5\sqrt{48}\)

\(=\sqrt{2^2.3}+3\sqrt{3^2.3}-5\sqrt{4^2.3}\)

\(=2\sqrt{3}+9\sqrt{3}-20\sqrt{3}\)\(=-9\sqrt{3}\)

b)\(B=3\sqrt{a^2+3}-3\sqrt{16a^2+48}+4\sqrt{25a^2+75}\)

\(=3\sqrt{a^2+3}-3\sqrt{16\left(a^2+3\right)}+4\sqrt{25\left(a^2+3\right)}\)

\(=3\sqrt{a^2+3}-12\sqrt{a^2+3}+20\sqrt{a^2+3}\)\(=11\sqrt{a^2+3}\)

Đề bài sai, bạn kiểm tra lại điều kiện \(a^2+b^2+c^2=1\)

Đề bài sai, bạn có thể thử kiểm tra với \(a=1.0001\) và \(b=0.9999\)