a)\(\left(x^2-9\right)\left(x+2\right)=x+3\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+2\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(\left(x-3\right)\left(x+2\right)-1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-x-6-1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-x-7=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1\pm\sqrt{29}}{2}\end{cases}}\)

b)\(x^4-6x^2+4x=0\)

\(\Leftrightarrow x\left(x^3-6x+4\right)=0\)

\(\Leftrightarrow x\left[x^3+2x^2-2x-2x^2-4x+4\right]=0\)

\(\Leftrightarrow x\left[x\left(x^2+2x-2\right)-2\left(x^2+2x-2\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+2x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0;x=2\\x=\pm\sqrt{3}-1\end{cases}}\)

c)\(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)

Đặt \(a=\sqrt{x^2-3x+3}>0\Rightarrow a^2+3=x^2-3x+6\)

\(pt\Leftrightarrow a+\sqrt{a^2+3}=3\)\(\Leftrightarrow\sqrt{a^2+3}=3-a\)

\(\Leftrightarrow a^2+3=a^2-6a+9\)

\(\Leftrightarrow6a-6=0\Leftrightarrow6\left(a-1\right)=0\Rightarrow a=1\) (thỏa)

\(\sqrt{x^2-3x+3}=1\)\(\Rightarrow x^2-3x+3=1\)

\(\Rightarrow x^2-3x+2=0\Rightarrow\left(x-2\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\) (thỏa)

Em năm nay mới lên lớp sáu nên ko bít lm...

e cx thế

mới lên lp 6

Ta chứng minh :\(\sqrt{2\left(a^2+b^2\right)}\ge a+b\)

\(\sqrt{2\left(a^2+b^2\right)}\ge a+b\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

Đúng theo BĐT Cauchy-Schwarz 

Tương tự cho 2 BĐT còn lại cũng có:

\(\sqrt{2\left(b^2+c^2\right)}\ge b+c;\sqrt{2\left(a^2+c^2\right)}\ge a+c\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge a+b+b+c+c+a=2\left(a+b+c\right)=VP\)

Đẳng thức xảy ra khi \(a=b=c\)

\(A=\sqrt{9.7}-2\sqrt{25.7}+\sqrt{9.7.4}-\frac{1}{7}\sqrt{4.7}\)

\(=3\sqrt{7}-10\sqrt{7}+6\sqrt{7}-\frac{2}{7}\sqrt{7}\)

\(=\frac{-9}{7}\sqrt{7}\)

Nếu đúng tk nhé

a = \(\sqrt{63}-2\sqrt{175}+\sqrt{252}-\frac{1}{7}\sqrt{28}\)

  = \(\sqrt{\frac{4}{7}}\left(1,5-5+3-1\right)\)

 =  \(-1,5\sqrt{\frac{4}{7}}\)

Áp dụng BĐT AM-GM ta có:

\(P+3=a+b^2+1+c^3+1+1\)\(\ge a+2b+3c\)

Lại có \(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=6\) nên nhân theo vế rồi áp dụng BĐT Cauchy-Schwarz có:

\(6\left(P+3\right)=\left(a+2b+3c\right)\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right)\)

\(\ge\left(\sqrt{a\cdot\frac{1}{a}}+\sqrt{2b\cdot\frac{2}{b}}+\sqrt{3c\cdot\frac{3}{c}}\right)^2\)

\(=\left(1+2+3\right)^2=6^2=36\)

\(\Rightarrow6\left(P+3\right)\ge36\Rightarrow P+3\ge6\Rightarrow P\ge3\)

Đẳng thức xảy ra khi \(a=b=c=1\)