\(k\left(x\right)=\dfrac{5x^2-22x+25}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5x^2-20x+20-x+2-x+2+1}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{\left(5x^2-20x+20\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x^2-4x+4\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2-\left(x-2\right)-\left(x-2\right)+1}{\left(x-2\right)^2}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}+\dfrac{1}{\left(x-2\right)^2}\)

\(\Leftrightarrow k\left(x\right)=5-\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{\left(x-2\right)^2}\)

Đặt \(y=\dfrac{1}{x-2}\)

\(\Rightarrow k\left(x\right)=5-y-y+y^2=y^2-2y+1+4=\left(y-1\right)^2+4\ge4\)

Vậy GTNN của \(k\left(x\right)=4\) khi \(y=1\Rightarrow\dfrac{1}{x-2}=1\Leftrightarrow x=3\)

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

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

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

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

Đặt \(y=\dfrac{1}{x-1}\)

\(\Rightarrow h\left(x\right)=1+y+y^2\)

\(\Rightarrow h\left(x\right)=y^2+y+1\)

\(\Rightarrow h\left(x\right)=y^2+2.y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

=> GTNN của \(h\left(x\right)=\dfrac{3}{4}\) khi \(y+\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{-1}{2}\)

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

\(\Leftrightarrow x=-1\)

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

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

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

=>\(y'=1+\dfrac{-4\cdot2\cdot\left(x-2\right)'\left(x-2\right)}{\left(x-2\right)^4}\)

=>\(y'=1-\dfrac{8}{\left(x-2\right)^3}\)

Đặt y'=0

=>\(\dfrac{8}{\left(x-2\right)^3}=1\)

=>\(\left(x-2\right)^3=8\)

=>x-2=2

=>x=4

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

\(f\left(4\right)=4+\dfrac{4}{\left(4-2\right)^2}=4+1=5\)

\(f\left(0\right)=0+\dfrac{4}{\left(0-2\right)^2}=0+\dfrac{4}{4}=1\)

\(f\left(5\right)=5+\dfrac{4}{\left(5-2\right)^2}=5+\dfrac{4}{9}=\dfrac{49}{9}\)

Vì f(0)<f(4)<f(5)

nên \(f\left(x\right)_{max\left[0;5\right]\backslash\left\{2\right\}}=f\left(5\right)=\dfrac{49}{9}\) và \(f\left(x\right)_{min\left[0;5\right]\backslash\left\{2\right\}}=1\)

2: \(y=cos^22x-sinx\cdot cosx+4\)

\(=1-sin^22x-\dfrac{1}{2}\cdot sin2x+4\)

\(=-sin^22x-\dfrac{1}{2}\cdot sin2x+5\)

\(=-\left(sin^22x+\dfrac{1}{2}\cdot sin2x-5\right)\)

\(=-\left(sin^22x+2\cdot sin2x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{81}{16}\right)\)

\(=-\left(sin2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}\)

\(-1< =sin2x< =1\)

=>\(-\dfrac{3}{4}< =sin2x+\dfrac{1}{4}< =\dfrac{5}{4}\)

=>\(0< =\left(sin2x+\dfrac{1}{4}\right)^2< =\dfrac{25}{16}\)

=>\(0>=-\left(sin2x+\dfrac{1}{4}\right)^2>=-\dfrac{25}{16}\)

=>\(\dfrac{81}{16}>=-sin\left(2x+\dfrac{1}{4}\right)^2+\dfrac{81}{16}>=-\dfrac{25}{16}+\dfrac{81}{16}=\dfrac{7}{2}\)

=>\(\dfrac{81}{16}>=y>=\dfrac{7}{2}\) 

\(y_{min}=\dfrac{7}{2}\) khi \(sin2x+\dfrac{1}{4}=\dfrac{5}{4}\)

=>\(sin2x=1\)

=>\(2x=\dfrac{\Omega}{2}+k2\Omega\)

=>\(x=\dfrac{\Omega}{4}+k\Omega\)

\(y_{max}=\dfrac{81}{16}\) khi sin 2x=-1

=>\(2x=-\dfrac{\Omega}{2}+k2\Omega\)

=>\(x=-\dfrac{\Omega}{4}+k\Omega\)

Câu a:
\(A=x^2-4x+1=(x^2-4x+4)-3\)

\(=(x-2)^2-3\geq 0-3=-3\)

Dấu "=" xảy ra khi $(x-2)^2=0$ hay $x=2$

Vậy GTNN của $A$ là $-3$ khi $x=2$

Câu b:

\(B=5-8x-x^2=21-(x^2+8x+16)\)

\(=21-(x+4)^2\leq 21-0=21\)

Dấu "=" xảy ra khi $(x+4)^2=0$ hay $x=-4$

Vậy GTLN của $B$ là $21$ khi $x=-4$

Câu c:

\(C=5x-x^2=-(x^2-5x)=\frac{25}{4}-(x^2-5x+\frac{5^2}{2^2})\)

\(=\frac{25}{4}-(x-\frac{5}{2})^2\leq \frac{25}{4}-0=\frac{25}{4}\)

Dấu "=" xảy ra khi \((x-\frac{5}{2})^2=0\Leftrightarrow x=\frac{5}{2}\)

Vậy GTLN của $C$ là $\frac{25}{4}$ khi $x=\frac{5}{2}$

Câu d:

\(D=(x-1)(x+3)(x+2)(x+6)=[(x-1)(x+6)][(x+3)(x+2)]\)

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

\(=(x^2+5x)^2-6^2=(x^2+5x)^2-36\geq 0-36=-36\)

Dấu "=" xảy ra khi \((x^2+5x)^2=0\Leftrightarrow \left[\begin{matrix} x=0\\ x=-5\end{matrix}\right.\)

Vậy GTNN của $D$ là $-36$ khi $x=0$ hoặc $x=-5$

Câu e:
\(E=\frac{1}{x^2+5x+4}=\frac{1}{x^2+5x+\frac{25}{4}+\frac{31}{4}}=\frac{1}{(x+\frac{5}{2})^2+\frac{31}{4}}\)

Vì \((x+\frac{5}{2})^2\geq 0, \forall x\Rightarrow (x+\frac{5}{2})^2+\frac{31}{4}\geq \frac{31}{4}\)

\(\Rightarrow E\leq \frac{1}{\frac{31}{4}}=\frac{4}{31}\)

Vậy GTLN của $E$ là \(\frac{4}{31}\) tại \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Câu f:

\(F=\frac{2x^2+4x+10}{x^2+2x+3}=\frac{2(x^2+2x+3)+4}{x^2+2x+3}=2+\frac{4}{x^2+2x+3}\)

\(=2+\frac{4}{(x^2+2x+1)+2}=2+\frac{4}{(x+1)^2+2}\)

Vì \((x+1)^2\geq 0\Rightarrow \frac{4}{(x+1)^2+2}\leq \frac{4}{2}=2\)

\(\Rightarrow F\leq 2+2=4\)

Vậy GTLN của $F$ là $4$ khi $x=-1$

Bài 2 :

a )

\(\left(4x-3\right)\left(4x+3\right)-15\left(x-1\right)\left(x+1\right)-\left(x+6\right)-3x=1\)

\(\Leftrightarrow16x^2-9-15x^2+15-x-6-3x=1\)

\(\Leftrightarrow x^2-4x-1=0\)

\(\Delta=16+4=20>0\)

\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{4+\sqrt{20}}{2}=2+\sqrt{5}\\\dfrac{4-\sqrt{20}}{2}=2-\sqrt{5}\end{matrix}\right.\)

Vậy \(x=2-\sqrt{5}\) hoặc \(x=2+\sqrt{5}\)

b )

\(\left(5x+1\right)\left(5x-1\right)-25\left(x+3\right)\left(x-1\right)=4\)

\(\Leftrightarrow25x^2-1-25x^2-50x+75=4\)

\(\Leftrightarrow-50x+70=0\)

\(\Leftrightarrow x=\dfrac{70}{50}\)

Vậy \(x=\dfrac{70}{50}\)

1) A=x²-4x+4-3=(x-2)²-3

(x-2)²≥0 (Với mọi x)

=> (x-2)²-3 ≥ -3 (V...)

=> Min A=-3

Làm tương tự với những câu khác nha

2) a)( 4x-3)(4x+3)-15(x-1)(x+1)-x-6-3x=1

<=> 16x²-9-15x²+15-4x-6=1

<=> x²-4x=1 <=> x²-4x-1=0 <=> (x-2)²=5 =>x-2=\(\sqrt{5}\) => x=\(\sqrt{5}+2\)

làm tương tự nha

a,

\(\sqrt{1-4x+4x^2}=5\\ \sqrt{\left(2x-1\right)^2}=5\\ \left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\\ \left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

b,

\(\sqrt{4-5x}=12\\ 4-5x=144\\ x=-28\)

c,

\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\\ \sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\\ 2\left(1-\sqrt{x-1}\right)=0\\ 1-\sqrt{x-1}=0\\ \sqrt{x-1}=1\\ x-1=1\\ x=2\)