\(x-x^2+\dfrac{1}{4}\)

\(=-\left(x^2-x-\dfrac{1}{4}\right)\)

\(=-\left[\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{1}{4}-\dfrac{1}{4}\right]\)

\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\right]\)

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

Ta có :

\(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\)

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\) ≤ \(\dfrac{1}{2}< 0\)

đặt a = 2x+y+z ; b = 2y+z+x ; c = 2z+x+y => a+b+c = 4x+4y+4z 
=> a - (a+b+c)/4 = x => x = (3a-b-c)/4 ; tương tự y = (3b-c-a)/4 ; z = (3c-a-b)/4 
thay vào vế trái ta có 
P = (3a-b-c)/4a + (3b-c-a)/4b + (3c-a-b)/4c = 
= 9/4 - (b/4a + c/4a + c/4b + a/4b + a/4c + b/4c) 
= 9/4 - (1/4)(b/a+a/b + c/a+a/c + c/b+b/c) 
Côsi cho từng cặp ta có: b/a+a/b ≥ 2 ; c/a+a/c ≥ 2 ; c/b+b/c ≥ 2 
=> b/a+a/b + c/a+a/c + c/b+b/c ≥ 6 
=> -(1/4)(b/a+a/b +c/a+a/c + c/b+b/c) ≤ -6/4 thay vào P ta có: 
P ≤ 9/4 - 6/4 = 3/4 (đpcm) ; dấu "=" khi a = b = c hay x = y = z 
cách này tuy biến đổi dài nhưng dễ hiểu) 
------------ 
Cách khác: 
P = x/(2x+y+z) -1 + y/(2y+z+x) -1 + z/(2z+x+y) - 1 + 3 
= -(x+y+z)/(2x+y+z) -(x+y+z)/(2y+z+x) -(x+y+z)/(2z+x+y) + 3 
= -(x+y+z).[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] + 3 
- - - 
Côsi cho 3 số: 
2x+y+z + 2y+z+x + 2z+x+y ≥ 3.³√(2x+y+z)(2y+z+x)(2z+x+y) 
=> 4(x+y+z) ≥ 3.³√(2x+y+z)(2y+z+x)(2z+x+y) (1*) 
Côsi cho 3 số: 
1/(2x+y+z)+1/(2y+z+x)+1/(2z+x+y) ≥ 3³√1/(2x+y+z)(2y+z+x)(2z+x+y) (2*) 
Lấy (1*) *(2*) ta có: 
4(x+y+z)[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] ≥ 9 
=> -(x+y+z).[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] ≤ -9/4 
thay vào P ta có: 
P ≤ -9/4 + 3 = 3/4 (đpcm) ; dấu "=" khi x = y = z 

Bạn ơi vì sao lại nhân với 9/4 mình tưởng chỉ nhân với 3/4 thôi chứ nhỉ

\(\sqrt{ắdsadwaswa\hept{\begin{cases}\\\end{cases}}da\hept{\begin{cases}\\\\\end{cases}}wáda\orbr{\begin{cases}\\\end{cases}}ưa\hept{\begin{cases}\\\\\end{cases}}d\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\sigma ssssssa}\)

Toán lớp 1 đây à ?

a/

\(\Leftrightarrow\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+3x}+x^2-1\ge0\)

\(\Leftrightarrow\left(x^2-1\right)\left(\frac{x^2+1}{x^2+3x}+1\right)\ge0\)

\(\Leftrightarrow\left(x^2-1\right)\left(\frac{2x^2+3x+1}{x^2+3x}\right)\ge0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(2x+1\right)}{x\left(x+3\right)}\ge0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+1\right)\left(x+1\right)^2}{x\left(x+3\right)}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x< -3\\x=-1\\-\frac{1}{2}\le x< 0\\x\ge1\end{matrix}\right.\)

b/

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)\left(\frac{-2-2x}{x}\right)\le0\)

\(\Leftrightarrow\frac{-2.\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}{x}\le0\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)\left(x-2\right)\left(x+1\right)^2}{x}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x\le-2\\x=-1\\0< x\le1\\x\ge2\end{matrix}\right.\)

c/

\(\Leftrightarrow\left(\frac{4\left(x-1\right)-2x}{x\left(x-1\right)}\right)\left(\frac{x^2+1-2x}{x}\right)\le0\)

\(\Leftrightarrow\frac{\left(2x-4\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)

\(\Rightarrow1< x\le2\)

d/

ĐKXĐ: \(\left\{{}\begin{matrix}x^3-4x\ge0\\\frac{1+x}{x}-2\ge0\\x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-2\right)\left(x+2\right)\ge0\\\frac{1-x}{x}\ge0\\x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}-2\le x\le0\\x\ge2\end{matrix}\right.\\0< x\le1\\x\ne0\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ

Vậy BPT đã cho vô nghiệm

Ta có: \(f\left(-2\right)=4a-2b+c\)

 \(f\left(3\right)=9a+3b+c=13a+b+2c-4a+2b-c=-4a+2b-c\)

\(\Rightarrow f\left(-2\right).f\left(3\right)=\left(4a-2b+c\right)\left(-4a+2b-c\right)=-\left(4a-2b+c\right)^2\le0\) (đpcm)