A=[2(x^2-8x+22)-1]/(x^2-8x+22)

A=2-1/[(x-4)^2+6]

A nho nhat khi (x-4)^2=0=> x=4

min(A)=2-1/6

\(A=\dfrac{3x^2+9x+17}{3x^2+9x+7}=1+\dfrac{10}{3x^2+9x+7}=1+\dfrac{10}{3\left(x^2+2.x.\dfrac{9}{2}+\dfrac{81}{4}\right)-\dfrac{215}{4}}\\ =1+\dfrac{10}{3\left(x+\dfrac{9}{2}\right)^2-\dfrac{215}{4}}\le\dfrac{35}{43}\)

Câu khác giải TT

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)

=\(\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)

= \(\dfrac{3\cdot\left(x^2-2x+5\right)}{x^2-2x+5}+\dfrac{2}{x^2-2x+5}\)

= \(3+\dfrac{2}{x^2-2x+5}\)

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

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

vì (x-1)² ≥ 0 ∀ x

⇔ (x-1)² +4 ≥ 4

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

⇔\(3+\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{7}{2}\)

⇔ A \(\le\dfrac{7}{2}\)

⇔ Min A =\(\dfrac{7}{2}\)

khi x-1=0

⇔ x=1

vậy ....

Ta có:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)

\(B=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)

\(B=2-\dfrac{3}{x^2-8x+16+6}\)

\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)

\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)

d)\(D=\dfrac{x^6+512}{x^2+8}\)

\(D=\dfrac{x^6+8x^4-8x^4-64x^2+64x^2+512}{x^2+8}\)

\(D=\dfrac{x^4\left(x^2+8\right)-8x^2\left(x^2+8\right)+64\left(x^2+8\right)}{x^2+8}\)

\(D=\dfrac{\left(x^2+8\right)\left(x^4-8x^2+64\right)}{x^2+8}\)

\(D=x^4-8x^2+64\)

\(D=\left(x^2-4\right)^2+48\ge48\)

\(\Rightarrow MIND=48\Leftrightarrow x=\pm2\)

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

a:

ĐKXĐ: x<>-1/2

Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì

\(2x^3+x^2+2x+1+1⋮2x+1\)

=>\(2x+1\inƯ\left(1\right)\)

=>2x+1 thuộc {1;-1}

=>x thuộc {0;-1}

b:

ĐKXĐ: x<>1/3

 \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)

=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1

=>2 chia hết cho 3x-1

=>3x-1 thuộc {1;-1;2;-2}

=>x thuộc {2/3;0;1;-1/3}

mà x nguyên

nên x thuộc {0;1}

c: 

ĐKXĐ: x<>2

\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)

=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)

=>\(x+2⋮x-2\)

=>x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;0;6;-2}

1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)

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

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)

Để A=B thì \(\dfrac{3}{3x+1}-\dfrac{2}{3x-1}=\dfrac{x-5}{\left(3x-1\right)\left(3x+1\right)}\)

=>9x-3-6x-3=x-5

=>3x-6=x-5

=>2x=1

=>x=1/2

hai biểu thức A,B có cùng một giá trị

\(=>A=B\\ đk:\left\{{}\begin{matrix}x\ne\dfrac{1}{3}\\x\ne-\dfrac{1}{3}\end{matrix}\right.\\ =>\dfrac{3}{3x+1}+\dfrac{2}{1-3x}=\dfrac{x-5}{9x^2-1}\\ =>\dfrac{3}{3x+1}+\dfrac{-2}{3x-1}=\dfrac{x-5}{\left(3x-1\right)\left(3x+1\right)}\\ =>\dfrac{3\left(3x-1\right)-2\left(3x+1\right)}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{x-5}{\left(3x-1\right)\left(3x+1\right)}\\ =>9x-3-6x-2=x-5\\ =>3x-5=x-5\\ =>3x-x=-5+5\\ =>2x=0\\ =>x=0\left(t/m\right)\)