\(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\)

vì \(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\le0,\forall x\inℝ\)

\(\Rightarrow B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\le3\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{4}{9}x-\dfrac{2}{15}=0\Rightarrow\dfrac{4}{9}x=\dfrac{2}{15}\Rightarrow x=\dfrac{9}{15}\)

Vậy \(GTLN\left(B\right)=3\left(tạix=\dfrac{9}{15}\right)\)

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

vì \(\left(2x+\dfrac{1}{3}\right)^4\ge0,\forall x\inℝ\)

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

Dấu "=" xảy ra khi và chỉ khi

\(2x+\dfrac{1}{3}=0\Rightarrow2x=-\dfrac{1}{3}\Rightarrow x=-\dfrac{1}{6}\)

\(\Rightarrow GTNN\left(A\right)=-1\left(tạix=-\dfrac{1}{6}\right)\)

a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

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

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

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

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

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

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

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

\(A=\dfrac{1}{2}+\left|2x-1\right|\ge\dfrac{1}{2}\forall x\)

\(minA=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{\left|x\right|+2007}{2008}\ge\dfrac{0+2007}{2008}=\dfrac{2007}{2008}\)

\(minB=\dfrac{2007}{2008}\Leftrightarrow x=0\)

a: ta có: \(\dfrac{\left(x+2\right)^2}{2}+\dfrac{\left(2x+1\right)^2}{4}+\dfrac{\left(2x-1\right)^2}{8}-\left(x+1\right)^2=0\)

\(\Leftrightarrow4\left(x^2+4x+4\right)+2\left(4x^2+4x+1\right)+4x^2-4x+1-8\left(x+1\right)^2=0\)

\(\Leftrightarrow4x^2+16x+16+8x^2+8x+2+4x^2-4x+1-8\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow16x^2+20x+19-8x^2-16x-8=0\)

\(\Leftrightarrow8x^2+4x+11=0\)

\(\text{Δ}=4^2-4\cdot8\cdot11=-336< 0\)

Vì Δ<0 nên phương trình vô nghiệm

b.

PT \(\Leftrightarrow \frac{x^2+2x+1}{2}-\frac{4x^2-4x+1}{3}+\frac{4x^2+4x+1}{4}-\frac{x^2-10x+25}{6}=0\)

\(\Leftrightarrow \left(\frac{x^2+2x+1}{2}+\frac{4x^2+4x+1}{4}\right)-\left(\frac{4x^2-4x+1}{3}+\frac{x^2-10x+25}{6}\right)=0\)

\(\Leftrightarrow \frac{6x^2+8x+3}{4}-\frac{9x^2-18x+27}{6}=0\)

\(\Leftrightarrow \frac{3(6x^2+8x+3)-2(9x^2-18x+27)}{12}=0\)

$\Leftrightarrow 5x-\frac{15}{4}=0$

$\Leftrightarrow x=\frac{3}{4}$

c.

PT $\Leftrightarrow (x^3+9x^2+27x+27)-(3x^3+12x^2)+(x^3+6x^2+12x+8)=(-x^3+3x^2-3x+1)-8$

$\Leftrightarrow 42x+42=0$

$\Leftrightarrow x=-1$

a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)

\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)

\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)

\(A=\left|2021-x\right|+\dfrac{1}{2}\left|4040-2x\right|\)

\(A=\left|2021-x\right|+\left|2020-x\right|\)

\(A=\left|2021-x\right|+\left|x-2020\right|\ge\left|2021-x+x-2020\right|=1\)

\(A_{min}=1\) khi \(2020\le x\le2021\)

C=|2x-3/5|+4/3>=4/3

Dấu = xảy ra khi x=3/10

D=|x-3|+|-x-2|>=|x-3-x-2|=5

Dấu = xảy ra khi -2<=x<=3

ĐKXĐ: \(x\notin\left\{-1;-\dfrac{1}{2}\right\}\)

a) Ta có: \(P=\left(\dfrac{2x}{x^3+x^2+x+1}+\dfrac{1}{x+1}\right):\left(1+\dfrac{x}{x+1}\right)\)

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

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

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

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

b) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ

nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(P=\dfrac{x^2+2x+1}{\left(2x+1\right)\left(x^2+1\right)}\), ta được:

\(P=\left[\left(\dfrac{1}{4}\right)^2+2\cdot\dfrac{1}{4}+1\right]:\left[\left(2\cdot\dfrac{1}{4}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)

\(=\left(\dfrac{1}{16}+\dfrac{1}{2}+1\right):\left[\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)

\(=\dfrac{25}{16}:\dfrac{51}{32}=\dfrac{25}{16}\cdot\dfrac{32}{51}=\dfrac{50}{51}\)

Vậy: Khi \(x=\dfrac{1}{4}\) thì \(P=\dfrac{50}{51}\)

a: ĐKXĐ: x<>2; x<>0

b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)

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

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

\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7