tk nhé

Ông An cao 180 cm, vòng bụng 108 cm.

Ông Chung cao 160 cm, vòng bụng 70 cm.

Với mọi x;y dương ta có:

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

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

Áp dụng:

\(P\ge\sqrt{\dfrac{3}{4}\left(a+b\right)^2}+\sqrt{\dfrac{3}{4}\left(b+c\right)^2}+\sqrt{\dfrac{3}{4}\left(c+a\right)^2}\)

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

\(P\ge\sqrt{3}\left(a+b+c\right)=\sqrt{3}\)

\(P_{min}=\sqrt{3}\) khi \(a=b=c=\dfrac{1}{3}\)

a: |2,5|+|7,5|=2,5+7,5=10

b: \(1,2\cdot\left|-3\right|+6,4=1,2\cdot3+6,4=3,6+6,4=10\)

c: \(\left|-\dfrac{7}{2}\right|+\left|\dfrac{15}{2}\right|=\dfrac{7}{2}+\dfrac{15}{2}=\dfrac{22}{2}=11\)

\(\left(3x-2^4\right)\cdot7^{13}=2\cdot7^{14}\)

=>\(3x-16=2\cdot\dfrac{7^{14}}{7^{13}}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\dfrac{30}{3}=10\)

\(\left(3x-2^4\right).7^{13}=2.7^{14}\\ \Rightarrow\left(3x-16\right).7^{13}=2.7.7^{13}\\ \Rightarrow3x-16=2.7\\ \Rightarrow3x-16=14\\ \Rightarrow3x=30\\ \Rightarrow x=30:3\\ \Rightarrow x=10\)

Vậy: \(x=10\)

a: \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}\)

=>\(1-\dfrac{AB'}{AB}=1-\dfrac{AC'}{AC}\)

=>\(\dfrac{BB'}{AB}=\dfrac{CC'}{AC}\)

=>\(\dfrac{BB'}{CC'}=\dfrac{AB}{AC}\)

mà \(\dfrac{AB}{AC}=\dfrac{AB'}{AC'}\)

nên \(\dfrac{AB'}{AC'}=\dfrac{BB'}{CC'}\)

=>\(\dfrac{AB'}{BB'}=\dfrac{AC'}{CC'}\)

b: \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}\)

=>\(1-\dfrac{AB'}{AB}=1-\dfrac{AC'}{AC}\)

=>\(\dfrac{BB'}{AB}=\dfrac{CC'}{AC}\)

Đa thức này thu gọn rồi bạn

Biểu thức này ko thu gọn được nữa em (đồng thời cũng ko phân tích được thành nhân tử)

\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55=7^3.7.11.5=5.77.7^3\)

Do 77 chia hết 77 \(\Rightarrow5.77.7^3⋮77\)

Vậy \(\left(7^6+7^5-7^4\right)⋮77\)

a.

\(\Leftrightarrow\sqrt{2x^2+5x+3}=-x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x-3\ge0\\2x^2+5x+3=\left(-x-3\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-3\\2x^2+5x+3=x^2+6x+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-3\\x^2-x-6=0\end{matrix}\right.\)

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

Vậy pt vô nghiệm

b.

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

\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\2x^2+x+3=\left(1-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\2x^2+x+3=1-2x+x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^2+3x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x=-1\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy pt vô nghiệm

\(\Leftrightarrow\left(x+5\right)^2-9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+5-9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)

\(\left(x+5\right)^2-9x-45x=0\\ < =>\left(x^2+10x+25\right)-54x=0\\ < =>x^2+10x+25-54x=0\\ < =>x^2-44x+25=0\\ < =>\left(x^2-44x+484\right)-459=0\\ < =>\left(x-22\right)^2-459=0\\ < =>\left(x-22\right)^2=459\\ < =>\left[{}\begin{matrix}x-22=\sqrt{459}\\x-22=-\sqrt{459}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=22+\sqrt{459}\\x=22-\sqrt{459}\end{matrix}\right.\)

a: Xét ΔMNP có \(\dfrac{MH}{MN}=\dfrac{MK}{MP}\)

nên HK//PN

Xét tứ giác NHKP có HK//NP

nên NHKP là hình thang

Hình thang NHKP có \(\widehat{HNP}=\widehat{KPN}\)(ΔMNP cân tại M)

nên NHKP là hình thang cân