Áp dụng BĐT Bunhiacopxki,ta có:

\(a^4+b^4\) \(\geq\) \(\dfrac{\left(a^2+b^2\right)^2}{2}\) \(\geq\) \(\dfrac{\left(\dfrac{1}{2}\left(a+b\right)^2\right)^2}{2}\) = \(\dfrac{\dfrac{1}{4}\left(a+b\right)^4}{2}\) = \(\dfrac{\left(a+b\right)^4}{8}\)

Dấu = xảy ra khi a=b

Ta có: B=\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Leftrightarrow\) 2B= \(2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3-1\right).\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

= \(\left(3^{16}-1\right)\left(3^{16}+1\right)\)

= \(3^{32}-1\)

\(\Rightarrow\) B= \(\dfrac{3^{32}-1}{2}\)

Mà ta có A= \(3^{32}-1\)

\(\Rightarrow\) A=2B

Ta có: P= \(x^2+xy+y^2-2x-3y+2010\)

\(\Leftrightarrow\) 4P= \(4\left(x^2+xy+y^2-2x-3y+2010\right)\)

= \(4x^2+4xy+4y^2-8x-12y+8040\)

= \(\left(4x^2+y^2+4+4xy-8x-8y\right)+3y^2-8y+8036\)

= \(\left(2x+y-2\right)^2+3y^2-8y+\dfrac{16}{3}-\dfrac{16}{3}+8036\)

= \(\left(2x+y-2\right)^2+3\left(y^2-\dfrac{8}{3}y+\dfrac{16}{9}\right)+\dfrac{24092}{3}\)

= \(\left(2x+y-2\right)^2+3\left(y-\dfrac{4}{3}\right)^2+\dfrac{24092}{3}\) \(\geq\) \(\dfrac{24092}{3}\)

\(\Rightarrow\) 4P \(\geq\) \(\dfrac{24092}{3}\) \(\Rightarrow\) P \(\geq\) \(\dfrac{6023}{3}\)

Dấu = xảy ra khi \(\begin{cases} (2x+y-2)^{2}=0\\ (y-\dfrac{4}{3})^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x+y-2=0\\ y-\dfrac{4}{3}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x=-(y-2)\\ y=\dfrac{4}{3} \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x=-(\dfrac{4}{3}-2)\\ y=\dfrac{4}{3} \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} x=\dfrac{1}{3}\\ y=\dfrac{4}{3} \end{cases} \)

Từ đó suy ra Min P= \(\dfrac{6023}{3}\) khi \(\begin{cases} x=\dfrac{1}{3}\\ y=\dfrac{4}{3} \end{cases} \)

Chúc bạn học tốt.

\(0,8=\frac{8}{10}=\frac{8.10}{10.10}=\frac{80}{100}=0,80\)

Ta có: 3n+7 chia hết cho n+1

=>3n+3+4 chia hết cho n+1

=>3.(n+1)+4 chia hết cho n+1

=>4 chia hết cho n+1

=>n+1=Ư(4)=(-1,-2,-4,1,2,4)

=>n=(0,-1,-3,2,3,5)

Ta có:

\(\left(\left(5a-3b\right)+8c\right).\left(\left(5a-3b\right)-8c\right)=\left(3a-5b\right)^2\)

\(\Leftrightarrow\) \(\left(5a-3b\right)^2-\left(8c\right)^2=\left(3a-5b\right)^2\)

Ta có: \(a^2-b^2=4c^2\) \(\Leftrightarrow\) \(16\left(a^2-b^2\right)=64c^2=\left(8c\right)^2\)

\(\Rightarrow\) VT= \(\left(5a-3b\right)^2-16\left(a^2-b^2\right)\)

= \(25a^2-30ab+9b^2-16a^2+16b^2\)

= \(9a^2-30ab+25b^2\)

= \(\left(3a-5b\right)^2\)

\(\Rightarrow\) đpcm

a) Ta có:

\(a^2+b^2+c^2=ab+bc+ca\)

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

\(\Leftrightarrow\) \(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

\(\Leftrightarrow\) \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

\(\Leftrightarrow\) \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\) (1)

Ta có: (a-b)² \(\geq\) 0; (b-c)² \(\geq\) 0; (a-c)² \(\geq\) 0 (2)

(1)(2) \(\Rightarrow\) \(\begin{cases} (a-b)^{2}=0\\ (b-c)^{2}=0\\ (a-c)^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a-b=0\\ b-c=0\\ a-c=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a=b\\ b=c\\ a=c \end{cases} \) \(\Leftrightarrow\) a=b=c

b) Ta có: \(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab+2ac+2bc=3a^2+3b^2+3c^2\)

\(\Leftrightarrow\) \(3a^2+3b^2+3c^2-a^2-b^2-c^2-2ac-2bc-2ab=0\)

\(\Leftrightarrow\) \(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

\(\Leftrightarrow\) \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

\(\Leftrightarrow\) \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

Ta có: (a-b)² \(\geq\) 0; (b-c)² \(\geq\) 0; (a-c)² \(\geq\) 0 (2)

(1)(2) \(\Rightarrow\) \(\begin{cases} (a-b)^{2}=0\\ (b-c)^{2}=0\\ (a-c)^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a-b=0\\ b-c=0\\ a-c=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a=b\\ b=c\\ a=c \end{cases} \) \(\Leftrightarrow\) a=b=c

c. Ta có: \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)

\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab+2ac+2bc=3ab+3bc+3ac\)

\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac=0\)

\(\Leftrightarrow\) \(a^2+b^2+c^2-ab-bc-ac=0\)

\(\Leftrightarrow\) \(2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\) \(\left(a^2-2bc+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

\(\Leftrightarrow\) \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

Ta có: (a-b)² \(\geq\) 0; (b-c)² \(\geq\) 0; (a-c)² \(\geq\) 0 (2)

(1)(2) \(\Rightarrow\) \(\begin{cases} (a-b)^{2}=0\\ (b-c)^{2}=0\\ (a-c)^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a-b=0\\ b-c=0\\ a-c=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a=b\\ b=c\\ a=c \end{cases} \) \(\Leftrightarrow\) a=b=c

Chúc bạn học tốt

a) Áp dụng công thức là xong hết có gì khó đâu

Ta có số số hạng của dãy số: \(\dfrac{\left(100-1\right)}{1}+1=99+1=100\)

Tổng của dãy số: \(\left(100+1\right).100:2\) = 5050

b) Ta có: \(\dfrac{5}{4}-\dfrac{1}{100}.25=\dfrac{5}{4}-\dfrac{25}{100}=\dfrac{5}{4}-\dfrac{1}{4}=\dfrac{4}{4}=1\)