\(x^4+3x^2=0\)

Có \(x^4\ge0;\forall x\); \(3x^2\ge0;\forall x\)

=> VT\(\ge0;\forall x\)

Dấu = xảy ra <=> x=0 

Ý C

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

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

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

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

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

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

\(=2020\)

Có các phần tử của A là bội của 6

Các phần tử của B là bội của 15

Các phần tử của C là bội của 30

mà [6;15]=30

=> Những phần tử vừa chia hết cho 6; vừa chia hết cho 15 thì sẽ chia hết cho 30

Hay \(C=A\cap B\) 

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

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

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

a)Có \(b^2+c^2-a^2=cosA.2bc\)

\(S=\dfrac{1}{2}bc.sinA\)\(\Rightarrow4S=2bc.sinA\)

\(\Rightarrow\dfrac{b^2+c^2-a^2}{4S}=\dfrac{cosA.2bc}{2bc.sinA}=cotA\) (dpcm)

b) CM tương tự câu a \(\Rightarrow\dfrac{a^2+c^2-b^2}{4S}=\dfrac{cosB.2ac}{2ac.sinB}=cotB\); \(\dfrac{a^2+b^2-c^2}{4S}=\dfrac{cosC.2ab}{2ab.sinC}=cotC\)

Cộng vế với vế \(\Rightarrow cotA+cotB+cotC=\dfrac{b^2+c^2-a^2}{4S}+\dfrac{a^2+c^2-b^2}{4S}+\dfrac{a^2+b^2-c^2}{4S}\)\(=\dfrac{a^2+b^2+c^2}{4S}\) (dpcm)

c) Gọi m_a;m_b;m_c là độ dài các đường trung tuyến kẻ từ đỉnh A;B;C của tam giác ABC 

Có \(GA^2+GB^2+GC^2=\dfrac{4}{9}\left(m_a^2+m_b^2+m_b^2\right)\)\(=\dfrac{4}{9}\left[\dfrac{2\left(b^2+c^2\right)-a^2}{4}+\dfrac{2\left(a^2+c^2\right)-b^2}{4}+\dfrac{2\left(b^2+c^2\right)-a^2}{4}\right]\)

\(=\dfrac{4}{9}.\dfrac{3\left(a^2+b^2+c^2\right)}{4}=\dfrac{a^2+b^2+c^2}{3}\) (đpcm)

d) Có \(a\left(b.cosC-c.cosB\right)=ab.cosC-ac.cosB\)

\(=\dfrac{a^2+b^2-c^2}{2}-\dfrac{a^2+c^2-b^2}{2}\)

\(=b^2-c^2\) (dpcm)