Vi anh hung duoc tren man=> anh that=> k<0

\(-\dfrac{d'}{d}=-3\Leftrightarrow d'=3d\)

\(\dfrac{1}{f}=\dfrac{1}{d}+\dfrac{1}{d'}\Leftrightarrow\dfrac{1}{15}=\dfrac{1}{d}+\dfrac{1}{3d}\Leftrightarrow d=...\)

Tks u

2b)

Áp dụng BĐT bunhiacopxki có:

\(\left(1+1\right)\left(x^4+y^4\right)\ge\left(x^2+y^2\right)^2\)

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

\(\Rightarrow2\left(x^4+y^4\right)\ge\dfrac{\left(x+y\right)^4}{4}\Leftrightarrow x^4+y^4\ge\dfrac{1}{8}.\left(x+y\right)^4\)

Dấu "=" xảy ra khi x=y

3)

Áp dụng bđt Holder có:

\(\left(x^3+y^3+z^3\right)\left(1+1+1\right)\left(1+1+1\right)\ge\left(x+y+z\right)^3\)

\(\Leftrightarrow x^3+y^3+z^3\ge\dfrac{1}{9}\left(x+y+z\right)^3\)

Dấu "=" xảy ra khi x=y=z

3)(Nếu không dùng Holder)

Với x,y,z >0, ta có bđt sau:\(2x^3+2y^3+2z^3\ge xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\) (1)

Thật vậy (1)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)+\left(y+z\right)\left(y^2-yz+z^2\right)-yz\left(y+z\right)+\left(z+x\right)\left(z^2-zx+x^2\right)-zx\left(x+z\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^2+\left(y+z\right)\left(y-z\right)^2+\left(z+x\right)\left(z-x\right)^2\ge0\) (lđ)

Áp dụng AM-GM có:

\(x^3+y^3+z^3\ge3xyz\)

\(\Leftrightarrow\dfrac{2\left(x^3+y^3+z^3\right)}{3}\ge2xyz\) (2)

Từ (1) và (2), cộng vế với vế \(\Rightarrow\dfrac{8}{3}\left(x^3+y^3+z^3\right)\ge xy\left(x+y\right)+yz\left(x+z\right)+xz\left(x+z\right)+2xyz\)

\(\Leftrightarrow\dfrac{8}{3}\left(x^3+y^3+z^3\right)\ge\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(\Leftrightarrow9\left(x^3+y^3+z^3\right)\ge x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)^3\)

\(\Rightarrow x^3+y^3+z^3\ge\dfrac{1}{9}\left(x+y+z\right)^3\) (đpcm)

Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>\(b=ck;a=bk=ck\cdot k=ck^2\)

\(\frac{a}{c}=\frac{ck^2}{c}=k^2\)

\(\frac{2a^2+5b^2}{2b^2+5c^2}=\frac{2\cdot\left(ck^2\right)^2+5\cdot\left(ck\right)^2}{2\cdot\left(ck\right)^2+5\cdot c^2}=\frac{c^2k^2\left(2k^2+5\right)}{c^2\left(2k^2+5\right)}=k^2\)

Do đó: \(\frac{a}{c}=\frac{2a^2+5b^2}{2b^2+5c^2}\) (ĐPCM)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

(3x)^2=3^2.x^2=9x^2

Pt/c: Thân cao x thân lừn

F1: 100% đậu thân lùn

=> Thân lùn trội hoàn toàn so với thân cao

Quy ước: A: thân lùn; a : thân cao

P: AA    x      aa

G A                a

F1: Aa(100% thân lùn)

F1xF1 : Aa    x      Aa

G          A, a            A, a

F2: 1AA: 2Aa : 1aa

TLKH: 3 thân lùn: 1 thân cao