Lời giải:

a) 

Áp dụng BĐT Bunhiacopxky:

\((y-2x)^2\leq (16y^2+36x^2)(\frac{1}{16}+\frac{1}{9})=9.\frac{25}{144}\)

\(\Rightarrow \frac{-5}{4}\leq y-2x\leq \frac{5}{4}\Rightarrow \frac{15}{4}\leq y-2x+5\leq \frac{25}{4}\)

Vậy $A_{\min}=\frac{15}{4}$ và $A_{\max}=\frac{25}{4}$

b) 

Áp dụng BĐT Bunhiacopxky:

\((2x-y)^2\leq (\frac{x^2}{4}+\frac{y^2}{9})(16+9)=25\)

\(\Rightarrow -5\leq 2x-y\leq 5\Leftrightarrow -7\leq 2x-y-2\leq 3\)

Vậy $B_{min}=-7; B_{\max}=3$

\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)

GIÚP MÌNH VỚI MÌNH ĐANG GẤP ! XIN CẢM ƠN !!!

1. a. \(A=8a-8a^2+3=-8\left(a-\frac{1}{2}\right)^2+5\)

Vì \(\left(a-\frac{1}{2}\right)^2\ge0\forall a\)\(\Rightarrow-8\left(a-\frac{1}{2}\right)^2+5\le5\)

Dấu "=" xảy ra \(\Leftrightarrow-8\left(a-\frac{1}{2}\right)^2=0\Leftrightarrow a-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{2}\)

Vậy Amax = 5 <=> a = 1/2

b. \(B=b-\frac{9b^2}{25}=-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\)

Vì \(\left(b-\frac{25}{18}\right)^2\ge0\forall b\)\(\Rightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\le\frac{25}{36}\)

Dấu "=" xảy ra \(\Leftrightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2=0\Leftrightarrow b-\frac{25}{18}=0\Leftrightarrow b=\frac{25}{18}\)

Vậy Bmax = 25/36 <=> b = 25/18

a,\(A=8a-8a^2+3\)

       \(=-8\left(a^2-a\right)+3\)

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

       \(=-8\left[\left(a-\frac{1}{2}\right)^2-\frac{1}{4}\right]+3\)

       \(=-8\left(a-\frac{1}{2}\right)^2+2+3\)

       \(=-8\left(a-\frac{1}{2}\right)^2+5\le5\forall a\) 

Dấu"=" xảy ra khi \(\left(a-\frac{1}{2}\right)^2=0\Rightarrow a=\frac{1}{2}\)

Vậy \(Max_A=5\)khi\(a=\frac{1}{2}\)

bài 2:

b,\(D=d^2+10e^2-6de-10e+26\)

\(=d^2-23de+\left(3e\right)^2+e^2-2.5e+5^2+1\)

\(=\left(d-3e\right)^2+\left(e-5\right)^2+1\ge1\forall d,e\)

Dấu"=" xảy ra khi\(\orbr{\begin{cases}\left(d-3e\right)^2=0\\\left(e-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}d=15\\e=5\end{cases}}}\)

vậy \(D_{min}=1\)khi \(d=15;e=5\)

c,:\(E=4x^4+12x^2+11\)

\(=\left(2x^2\right)^2+2.2x^2.3+3^2+2\)

\(=\left(2x^2+3\right)^2+2\ge2\forall x\)

còn 1 đoạn nx bạn tự lm tiếp,lm giống như D

2. a. \(C=\frac{1}{16}c^2-9c+10=\frac{1}{16}\left(x-72\right)^2-314\)

Vì \(\left(x-72\right)^2\ge0\forall x\)\(\Rightarrow\frac{1}{16}\left(x-72\right)^2-314\ge-314\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{1}{16}\left(x-72\right)^2=0\Leftrightarrow x-72=0\Leftrightarrow x=72\)

Vậy Cmin = - 314 <=> x = 72

b. \(D=d^2+10e^2-6de-10e+26=\left(d^2-6de+9e^2\right)+\left(e^2-10e+25\right)+1\)

\(=\left(d-3e\right)^2+\left(e-5\right)^2+1\)

Vì \(\left(d-3e\right)^2\ge0;\left(e-5\right)^2\ge0\forall d;e\)\(\Rightarrow\left(d-3e\right)^2+\left(e-5\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left(d-3e\right)^2=0\\\left(e-5\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}d-3e=0\\e=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}d=15\\e=5\end{cases}}\)

Vậy Dmin = 1 <=> d = 15 ; e = 5

c. \(E=4x^4+12x^2+11=\left(2x^2+3\right)^2+2\)

Vì \(\left(2x^2+3\right)^2\ge0\forall x\)\(\Rightarrow\left(2x^2+3\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2x^2+3\right)^2=0\Leftrightarrow2x^2+3=0\Leftrightarrow x^2=-\frac{3}{2}\left(vo-ly\right)\)

Không thể xảy ra dấu "=" trong th này

Vậy để Emin thì \(\left(2x^2+3\right)^2_{min}=\left(3^2\right)=9\Leftrightarrow2x^2=0\Leftrightarrow x=0\)

Vậy Emin = 9 + 2 = 11 <=> x = 0

a, Ta có: \(\left(x+1\right)^2\ge0\)

\(\Rightarrow2\left(x+1\right)^2\ge0\)

\(\Rightarrow2\left(x+1\right)^2-17\ge-17\)

\(\Rightarrow B\ge-17\)
Dấu "=" xảy ra <=> (x+1)² = 0 <=> x = -1

Vậy GTNN của B là -17 khi x = -1

b, Ta có: \(\left(x-2\right)^2\ge0\)

\(\Rightarrow25-\left(x-2\right)^2\ge25\)

\(\Rightarrow B\ge25\)

Dấu "=" xảy ra <=> (x-2)² = 0 <=> x = 2

Vậy GTLN của B là 25 khi x = 2

thanh you very much

kết bạn không

\(P=\sqrt{\left(x-3\right)^2+4^2}+\sqrt{\left(y-3\right)^2+4^2}+\sqrt{\left(z-3\right)^2+4^2}\)

\(P\ge\sqrt{\left(x-3+y-3+z-3\right)^2+\left(4+4+4\right)^2}=6\sqrt{5}\)

\(P_{min}=6\sqrt{5}\) khi \(x=y=z=1\)

Mặt khác với mọi \(x\in\left[0;3\right]\) ta có:

\(\sqrt{x^2-6x+25}\le\dfrac{15-x}{3}\)

Thật vậy, BĐT tương đương: \(9\left(x^2-6x+25\right)\le\left(15-x\right)^2\)

\(\Leftrightarrow8x\left(3-x\right)\ge0\) luôn đúng

Tương tự: ...

\(\Rightarrow P\le\dfrac{45-\left(x+y+z\right)}{3}=14\)

\(P_{max}=14\) khi \(\left(x;y;z\right)=\left(0;0;3\right)\) và hoán vị

tach 14-x = 10-4-x roi sau do chac ban cung phai tu biet lam