a: Xét ΔBAC có 

D là trung điểm của AB

M là trung điểm của AC

Do đó: DM là đường trung bình của ΔABC

Suy ra: DM//BC và \(DM=\dfrac{BC}{2}=3.5\left(cm\right)\)

Bài 4: 

a: \(A=\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)

\(=2x^2+3x-10x-15-2x^2+6x+x+7\)

=-8

Bài 5: 

a: \(x\left(x-1\right)-x^2+4x=-3\)

\(\Leftrightarrow x^2-x-x^2+4x=-3\)

hay x=-1

i: \(x^2-9x+8=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

11) 3x(x-5)²-(x+2)³+2(x-1)³-(2x+1)(4x²-2x+1)=3x(x²-10x+25)-(x³+6x²+12x+8)+2(x³-3x²+3x-1)-(8x³+1)=3x³-30x²+75x-x³-6x²-12x-8+2x³-6x²+6x-2-8x³-1=-4x³-42x²+63x-11

\(M=35\left(6^2+1\right)\left(6^4+1\right)\left(6^8+1\right)\left(6^{12}+1\right)-6^{24}\)

\(=\left(6^2-1\right)\left(6^2+1\right)\left(6^4+1\right)\left(6^8+1\right)\left(6^{12}+1\right)-6^{24}\)

\(=\left(6^4-1\right)\left(6^4+1\right)\left(6^8+1\right)\left(6^{12}+1\right)-6^{24}\)

\(=6^{24}-1-6^{24}=-1\)

\(M=\left(6^2-1\right)\left(6^2+1\right)\left(6^4+1\right)\left(6^8+1\right)\left(6^{12}+1\right)-6^{24}\)

\(=\left(6^4-1\right)\left(6^4+1\right)\left(6^8+1\right)\left(6^{12}+1\right)-6^{24}\)

\(=\left(6^8-1\right)\left(6^8+1\right)\left(6^{12}+1\right)-6^{24}\)

\(=\left(6^{16}-1\right)\left(6^{12}+1\right)-6^{24}\)

\(=6^{28}+6^{16}-6^{12}-1-6^{24}\)

c: \(\dfrac{2}{x-1}=\dfrac{2x+2}{\left(x-1\right)\left(x+1\right)}\)

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

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

a. Áp dụng BĐT Bunhiacopxky:

$\text{VT}^2\leq [(a+2)+(b+2)+(c+2)](1+1+1)=3(a+b+c+6)=3.12=36$

$\Rightarrow \text{VT}\leq 6$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

b.

Áp dụng BĐT Cô-si:

$\text{VT}\leq \frac{a+3b+8+8}{12}+\frac{b+3c+8+8}{12}+\frac{c+3a+8+8}{12}=\frac{a+b+c}{3}+4=\frac{6}{3}+4=6$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=2$

c. Áp dụng BĐT Cô-si:

\(\text{VT}\leq \frac{9a+a(b+7)}{6}+\frac{9b+b(c+7)}{6}+\frac{9c+c(a+7)}{6}=\frac{16(a+b+c)}{6}+\frac{ab+bc+ac}{6}=16+\frac{ab+bc+ac}{6}\)

Cũng áp dụng BĐT Cô-si:

$ab+bc+ac\leq \frac{(a+b+c)^2}{3}=\frac{6^2}{3}=12$

$\Rightarrow \text{VT}\leq 16+\frac{12}{6}=18$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$
d.

Áp dụng BĐT Cô-si:
\(\text{VT}\leq \frac{4a+(b+6)}{4}+\frac{4b+(c+6)}{4}+\frac{4c+(a+6)}{4}=\frac{5(a+b+c)+18}{4}=\frac{30+18}{4}=12\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

e.

Áp dụng BĐT Cô-si:
\(\text{VT}\leq \frac{a+6+8+8}{12}+\frac{b+6+8+8}{12}+\frac{c+6+8+8}{12}=\frac{a+b+c+66}{12}=\frac{6+66}{12}=\frac{72}{12}=6\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

đọc đề mà quạo lun ă:>

dễ mà tự lm ik:333