Where is "y"? Do vậy mình sẽ sửa đề nhé! Vả lại bài này

Tìm tìm GTLN \(P=\sqrt{x-2}+\sqrt{y-3}\) biết  x + y = 6

ĐK: \(\hept{\begin{cases}\sqrt{x-2}\ne\sqrt{2}\\\sqrt{y-3}\ne\sqrt{2}\end{cases}\Rightarrow}\hept{\begin{cases}x\ne4\\y\ne5\end{cases}}\)

Ta có: \(P=\sqrt{x-2}+\sqrt{y-3}\)

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

\(P^2=x-2+y-3=\left(x+y\right)-\left(2+3\right)\)

Thay x + y = 6 vào,ta có: \(P^2=6-5=1\Leftrightarrow\hept{\begin{cases}P=1\\P=-1\end{cases}}\)

Mà đề bài là tìm GTLN nên P = 1

Dấu "=" xảy ra \(\Leftrightarrow x+y=6\)

Vậy \(P_{max}=1\Leftrightarrow x+y=6\)

Woa dung la tu duy cua mot huyen thoai OLM that khac biet.

Woa dung la tu duy cua mot huyen thoai 0 L M that khac biet.

Bài 2 : 

Tìm min : Bình phương 

Tìm max : Dùng B.C.S ( bunhiacopxki )

Bài 3 : Dùng B.C.S

KP9

nói thế thì đừng làm cho nhanh bạn ạ

Người ta cũng có chút tôn trọng lẫn nhau nhé đừng có vì dăm ba cái tích 

toàn 1 lũ hãm điểm

Áp dụng bunhiacopxki ta có

\(A^2\)\(\le\)(1+1)(x-2+y-3)=2(x+y-5)=2(vì x+y=6)\(\Rightarrow\)A\(\le\)\(\sqrt{2}\)

dấu = xảy ra\(\Leftrightarrow\)x=\(\frac{23}{8}\).y=\(\frac{25}{8}\)vì x\(\ge\)2......            y\(\ge\)3

Toán lớp 9 nha

Bạn ghi rõ GTLN là gì đi

_@Yumi, GTLN là giá trị lớn nhất đó

Bài 1: \(x+y+z+11=2\sqrt{x}+4\sqrt{y-1}+6\sqrt{z-2}\)

ĐKXĐ:\(x\ge0;y\ge1;z\ge2\)

\(\Leftrightarrow x-2\sqrt{x}+1+\left(y-1\right)-2\cdot\sqrt{y-1}\cdot2+4+\left(z-2\right)-2\cdot\sqrt{z-2}\cdot3+9=0\)\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-2\right)^2+\left(\sqrt{z-2}-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{y-1}=2\\\sqrt{z-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=5\\z=11\end{matrix}\right.\)

Bài 2: 

Q=|x+2|+|x-2|>=|x+2+2-x|=4

Dấu = xảy ra khi (x+2)(x-2)<=0

=>-2<=x<=2

ĐKXĐ: x>=0; x<>1/4

Ta có: \(A=\frac{\sqrt{x}+1}{2\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{x+6\sqrt{x}+2}{2x+5\sqrt{x}-3}\)

\(=\frac{\sqrt{x}+1}{2\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{x+6\sqrt{x}+2}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+\sqrt{x}\left(2\sqrt{x}-1\right)-x-6\sqrt{x}-2}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+4\sqrt{x}+3+2x-\sqrt{x}-x-6\sqrt{x}-2}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2x-3\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+3}\)

Ta có: P=A*B

\(=\frac{\sqrt{x}-1}{\sqrt{x}+3}\cdot\frac{\sqrt{x}+3}{x+8}=\frac{\sqrt{x}-1}{x+8}\)

=>\(\frac{1}{P}=\frac{x+8}{\sqrt{x}-1}=\frac{x-1+9}{\sqrt{x}-1}=\sqrt{x}+1+\frac{9}{\sqrt{x}-1}=\sqrt{x}-1+\frac{9}{\sqrt{x}-1}+2\ge2\cdot\sqrt{\left(\sqrt{x}-1\right)\cdot\frac{9}{\sqrt{x}-1}}+2=2\cdot3+2=8\forall x\) thỏa mãn ĐKXĐ

=>\(P\le\frac18\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}-1\right)^2=9;\sqrt{x}-1>0\)

=>\(\sqrt{x}-1=3\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)

\(\dfrac{M}{N}=\left(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) (ĐKXĐ: \(x\ge0;x\ne4;x\ne9\))

\(=\left[\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)\(=\left[\dfrac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

\(=\left[\dfrac{2\sqrt{x}-9-x+9+x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

\(=\dfrac{2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

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

\(=\dfrac{2}{\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{M}{N}+1=\dfrac{2}{\sqrt{x}+2}+1\)

Ta thấy: \(\sqrt{x}\ge0\forall x\)

\(\Rightarrow\sqrt{x}+2\ge2\forall x\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\forall x\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}+1\le2\forall x\)

\(\Rightarrow Max_P=2\Leftrightarrow\dfrac{2}{\sqrt{x}+2}+1=2\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=1\)

\(\Leftrightarrow\sqrt{x}+2=2\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

#Urushi☕

Bạn tự rút gọn nha .

c) Ta có : \(P\text{=}\dfrac{M}{N}+1\text{=}\dfrac{2}{\sqrt{x}+2}+1\)

Để P có giá trị lớn nhất.

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}cóGTLN\)

\(\Leftrightarrow\sqrt{x}+2cóGTNN\)

Mà : \(\sqrt{x}+2\ge2\)

\(\Rightarrow\) Để : \(\left(\sqrt{x}+2\right)_{min}\) \(\Leftrightarrow\sqrt{x}\text{=}0\Leftrightarrow x\text{=}0\)

Vậy............