a)

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

\(\Leftrightarrow3x^2-x+3x-1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)

b)

\(x^2-5x+6=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Áp dụng BĐT \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\) và BĐT \(xy\le\dfrac{1}{4}\left(x+y\right)^2\), ta có:

\(\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\)

\(\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2\)\(=\dfrac{1}{2}\left(1+\dfrac{x+y}{xy}\right)^2\)

\(\ge\dfrac{1}{2}\left(1+\dfrac{1}{\dfrac{1}{4}\left(x+y\right)^2}\right)^2=\dfrac{25}{2}\left(x+y=1\right)\)

Dấu "=" xảy ra khi x = y = 0,5

Áp dụng BĐT Cauchy Schwarz, BĐT AM - GM và BĐT \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\), ta có:

\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(a+c\right)}+\dfrac{1}{c^3\left(a+b\right)}\)

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

\(\ge\dfrac{3^2}{3\sqrt[3]{a^3b^3c^3\left(b+c\right)\left(a+c\right)\left(a+b\right)}}\)

\(\ge\dfrac{3^2}{3\sqrt[3]{8abc}}=\dfrac{3}{2}\left(abc=1\right)\)

Dấu "=" xảy ra khi a = b = c = 1

Câu 2:

Sửa lại đề: Cho đa thức bậc ba f(x) sao cho f(x) chia x² + 1 dư 2x + 3 và chia x² + 2x dư - 3x + 2. Tính f(2013)

Đặt f(x) = ax³ + bx² + cx + d, ta có:

(ax³ + bx² + cx + d) : (x² + 1) dư (c - a)x + (d - b)

(ax³ + bx² + cx + d) : (x² + 2x) dư \(\left[c-2\left(b-2a\right)\right]x+d=\left(c-2b+4a\right)x+d\)

Theo đê bài \(\Rightarrow\left\{{}\begin{matrix}\left(c-a\right)x+\left(d-b\right)=2x+3\\\left(c-2b+4a\right)x+d=-3x+2\end{matrix}\right.\)

Áp dụng định lí Bezout, ta có:

\(\left\{{}\begin{matrix}c-a=2\\d-b=3\\c-2b+4a=-3\\d=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{7}{5}\\b=-1\\c=\dfrac{3}{5}\\d=2\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)=-\dfrac{7}{5}x^3-x^2+\dfrac{3}{5}x+2\)

Đặt \(\sqrt{x}=a\left(a>0;a\ne1\right)\), ta có:

\(A=\dfrac{a^4+a^2+1}{a^2+a+1}=\dfrac{a^4-a+a^2+a+1}{a^2+a+1}\)

\(=\dfrac{a\left(a^3-1\right)}{a^2+a+1}+\dfrac{a^2+a+1}{a^2+a+1}\)

\(=\dfrac{a\left(a-1\right)\left(a^2+a+1\right)}{a^2+a+1}+1\)

\(=a\left(a-1\right)+1=a^2-a+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(x=\dfrac{1}{4}\)

Để a chia hết cho 2 thì a \(\in\){0,2,4,6,8}
Để a chia cho 5 dư 3 thì a = 8
Vậy a = 8

gà 22

chó 14

Coi x=20k, y=20h

xy=20k.20h=400.h.k =420...

Tên: Thiên Băng

Lớp: 8

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