Question

a) Giải hệ phương trình: \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

b)Giải phương trình: \(\sqrt[3]{2+x}+\sqrt[3]{5-x}=1\)

c) Cho a, b, c > 0 thỏa mãn biểu thức a+b+c=1

Chứng minh rằng: \(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\le2\)

Mn giúp mình với ạ, mk cần gấp lắm a~. Cảm ơn mn nhiều

Answer 1

a) \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\) (ĐKXĐ: \(x\ne-1;y\ne-4\))

Đặt \(\dfrac{x}{x+1}=a;\dfrac{1}{y+4}=b\left(a\ne0;b\ne0\right)\)

Hệ phương trình đã cho trở thành

\(\left\{{}\begin{matrix}3a-2b=4\left(1\right)\\2a-5b=9\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow2a=9+5b\Leftrightarrow a=\dfrac{9+5b}{2}\)

Thay \(a=\dfrac{9+5b}{2}\) vào \(\left(1\right)\), ta có:

\(\dfrac{3\left(9+5b\right)}{2}-2b=4\)

\(\Leftrightarrow27+15b-4b=8\)

\(\Leftrightarrow11b=-19\Leftrightarrow b=\dfrac{-19}{11}\)

Thay \(b=\dfrac{-19}{11}\) vào \(\left(2\right)\), ta có:

\(2a-5\cdot\dfrac{-19}{11}=9\)

\(\Leftrightarrow a=\dfrac{2}{11}\)

Với \(a=\dfrac{2}{11}\Rightarrow\dfrac{x}{x+1}=\dfrac{2}{11}\)

\(\Leftrightarrow11x=2x+2\Leftrightarrow x=\dfrac{2}{9}\)

Với \(b=\dfrac{-19}{11}\Rightarrow\dfrac{1}{y+4}=\dfrac{-19}{11}\)

\(\Leftrightarrow-19y-76=11\)

\(\Leftrightarrow y=\dfrac{-90}{19}\)

Answer 2

b,Ta có:

\(PT\Leftrightarrow7+3.\sqrt[3]{2+x}.\sqrt[3]{5-x}\left(\sqrt[3]{2+x}+\sqrt[3]{5-x}\right)=1\)

Thay \(\sqrt[3]{2+x}+\sqrt[3]{5-x}=1\) vào PT

\(\Rightarrow\) \(3.\sqrt[3]{2+x}.\sqrt[3]{5-x}=-6\)

\(\Leftrightarrow\sqrt[3]{2+x}.\sqrt[3]{5-x}=-2\)

\(\Leftrightarrow\left(2+x\right)\left(5-x\right)=-8\)

\(\Leftrightarrow x^2-3x-18=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)

Thử lại thấy x= - 3, x=6 thỏa mãn

Vậy x= -3, x = 6

Answer 3

b, Đặt \(\sqrt[3]{2+x}=a;\sqrt[3]{5-x}=b\)
Theo đề bài: a+b=1\(\Leftrightarrow a=1-b\)
Ta có \(a^3+b^3=2+x+5-x=7\)(1)
Thay a=1-b vào pt(1) ta được:
\(a^3+b^3=\left(1-b\right)^3+b^3=7\)
\(\Leftrightarrow1-3b+3b^2-b^3+b^3=7\)
\(\Leftrightarrow3b^2-3b-6=0\Leftrightarrow b^2-b-2=0\)
\(\Leftrightarrow\left(b+1\right)\left(b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=-1\\b=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{5-x}=-1\\\sqrt[3]{5-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5-x=-1\\5-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-3\end{matrix}\right.\)
Vậy nghiệm của hệ phương trình x=\(\left\{-3;6\right\}\)

Answer 4

\(P=\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\)

\(\Rightarrow\dfrac{2}{3}P=\sqrt{\left(a+bc\right).\dfrac{4}{9}}+\sqrt{\left(b+ac\right).\dfrac{4}{9}}+\sqrt{\left(c+ab\right).\dfrac{4}{9}}\)

Áp dụng BĐT AM-GM, ta có:

\(\sqrt{\left(a+bc\right).\dfrac{4}{9}}\le\dfrac{a+bc+\dfrac{4}{9}}{2}\)

Tương tự cho 2 cái kia

\(\Rightarrow\dfrac{2}{3}P\le\dfrac{a+b+c+ab+bc+ca+\dfrac{4}{9}.3}{2}\)

Mặt khác ta có:

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow ab+bc+ca\le\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}P\le\dfrac{1+\dfrac{1}{3}+\dfrac{4}{3}}{2}=\dfrac{4}{3}\)

\(\Rightarrow P\le2\) (đpcm)

\(''=''\Leftrightarrow a=b=c=\dfrac{1}{3}\)