a)đkxđ: \(x+1\ne0\Leftrightarrow x\ne-1\)

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

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

Đặt \(\frac{1}{x+1}=a\)\(\Rightarrow B=3a^2-3a+1=3\left(a^2-a+\frac{1}{3}\right)=3\left(a^2-2a.\frac{1}{2}+\frac{1}{4}+\frac{1}{12}\right)=3\left(a-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(\left(a-\frac{1}{2}\right)^2\ge0\Leftrightarrow B\ge\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=\frac{1}{2}\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}\Leftrightarrow x+1=2\Leftrightarrow x=1\left(nhận\right)\)

Vậy GTNN của B là \(\frac{1}{4}\)khi \(x=1\)

b) đkxđ \(x-1\ne0\Leftrightarrow x\ne1\)\(E=\frac{3x^2-8x+6}{x^2-2x+1}=\frac{3\left(x^2-2x+1\right)-2x+3}{x^2-2x+1}=3-\frac{2x-3}{\left(x-1\right)^2}=3-\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}\)

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

Đặt \(\frac{1}{x-1}=b\)\(\Rightarrow E=b^2-2b+3=b^2-2b+1+2=\left(b-1\right)^2+2\)

Vì \(\left(b-1\right)^2\ge0\Leftrightarrow B\ge2\)

Dấu "=" xảy ra khi \(b-1=0\Leftrightarrow b=1\Leftrightarrow\frac{1}{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\left(nhận\right)\)

Vậy GTNN của B là 2 khi x = 2

Gọi \(A=\frac{x^2+x+1}{x^2-x+1}\)ta có :

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

\(=\frac{1}{3}+\frac{\frac{2}{3}\left(x-1\right)^2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\frac{1}{3}\forall x\) có GTNN là \(\frac{1}{3}\) tại \(x=1\)

Sr nhìn lộn

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

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

Bài 2 : 

 Ta có : \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=\frac{3}{4}+\left(x-\frac{1}{2}\right)^2\ge\frac{3}{4}\forall x\in R\)

Vậy A_min = \(\frac{3}{4}\) dấu "=" chỉ sảy ra khi x = \(\frac{1}{2}\)

Cảm ơn bạn nhiều nha

Còn câu b bạn suy nghĩ được chưa

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

Vậy Min = \(\frac{1}{3}\Leftrightarrow x=-1\)

Ta có : \(\frac{x^2+x+1}{x^2-x+1}=\frac{-2\left(x^2-2x+1\right)+3\left(x^2-x+1\right)}{x^2-x+1}=\frac{-2\left(x-1\right)^2}{x^2-x+1}+3\le3\)

Vậy Max = 3  <=> x = 1

diều kiện x >= 0

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

= \(\frac{x+2-x+\sqrt{x}-1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)

=\(\frac{\sqrt{x}+1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)=\(\frac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)

P=8/9

<=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

<=> \(3\sqrt{x}=2x-2\sqrt{x}+1\)

<=> \(2x-5\sqrt{x}+2=0\)

<=> \(\left[\begin{array}{nghiempt}x=4\\x=\frac{1}{4}\end{array}\right.\)

vậy x=4 hoặc x=1/4 thì p=8/9

a) \(P=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\left(ĐK:x\ge0;x\ne-1\right)\)

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

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

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

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

b) Để P=8/9

\(\Leftrightarrow\)\(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)

\(\Leftrightarrow24x-24\sqrt{x}+24-36\sqrt{x}=0\)

\(\Leftrightarrow24x-60\sqrt{x}+24=0\)

\(\Leftrightarrow12\left(2x-5\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\left(2x-\sqrt{x}\right)-\left(4\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{array}\right.\)

1. Câu hỏi của Trần Dương An - Toán lớp 7 - Học toán với OnlineMath