a. ĐKXĐ : x>1.

b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)

c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:

\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)

Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).

Ta có: \(\left|x-\dfrac{2}{3}\right|\ge0\forall x\)

\(\Leftrightarrow\left|x-\dfrac{2}{3}\right|-1\ge-1\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{2}{3}\)

GTLN :

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

Vì \(\frac{x^2}{x^2+x+1}=\frac{x^2}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\ge0\forall x\) nên \(A=1-\frac{x^2}{x^2+x+1}\le1\forall x\) có GTLN là 1

GTNN : 

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

\(=-\frac{1}{3}+\frac{\frac{1}{3}\left(x+2\right)^2}{x^2+x+1}=-\frac{1}{3}+\frac{\left(x+2\right)^2}{3\left(x^2+x+1\right)}\ge-\frac{1}{3}\) có GTNN là \(-\frac{1}{3}\)

a: \(A=\left|x+1\right|+5\ge5\forall x\)

Dấu '=' xảy ra khi x=-1

b: \(B=\dfrac{x^2+3+12}{x^2+3}=1+\dfrac{12}{x^2+3}\le\dfrac{12}{3}+1=4+1=5\)

Dấu '=' xảy ra khi x=0

a)ĐKXĐ:x>0

P=\(\left(\frac{3}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\left(vớix>0\right)\)

=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]:\frac{1}{\sqrt{x}+1}\)

=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\frac{1}{\sqrt{x}+1}\)

= \(\left[\frac{3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\frac{1}{\sqrt{x}+1}\)

=\(\frac{4-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{1}\)

=\(\frac{4-\sqrt{x}}{\sqrt{x}-1}\)

b)Để P=\(\frac{5}{4}\left(vớix>0\right)\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}-\frac{5}{4}=0\)

\(\Leftrightarrow\frac{4\left(4-\sqrt{x}\right)}{4\left(\sqrt{x}-1\right)}-\frac{5\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}=0\)

\(\Rightarrow16-4\sqrt{x}-5\sqrt{x}+5=0\)

\(\Leftrightarrow21-9\sqrt{x}=0\)

\(\Leftrightarrow-9\sqrt{x}=-21\)

\(\Leftrightarrow\sqrt{x}=\frac{7}{3}\)

\(\Leftrightarrow x=\frac{21}{9}\)

Vậy:Để P=\(\frac{5}{4}\)thì x=\(\frac{21}{9}\)

c)Còn phần c thì mik chịu

...

Nhỏ nhất:

D có giá trị nhỏ nhất khi: (x + 5)² = 0 và (2y - 6)² = 0

(x + 5)² = 0

(x + 5)² = 0²

=> x + 5 = 0

         x   = 0 - 5

         x   = -5

(2y - 6)² = 0

(2y - 6)² = 0²

=> 2y - 6 = 0

        2y   = 0 + 6

         2y  = 6

            y = 6 : 2

            y = 3

Ta có: D = 0 + 0  + 1 = 1

Lớn nhất:(không có giá trị lớn nhất)

GIÚP MÌNH VỚI

LÀM ƠN

Đề là như thế này à bạn

Tìm GTNN của \(\frac{x^2+2}{\sqrt{x^2}+1}\)