Bài 1:

a: \(A=x^2+2x+4\)

\(=x^2+2x+1+3\)

\(=\left(x+1\right)^2+3>=3\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

Vậy: \(A_{min}=3\) khi x=-1

b: \(B=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1>=1\forall x\)

Dấu '=' xảy ra khi x-10=0

=>x=10

Vậy: \(B_{min}=1\) khi x=10

c: \(C=x^2-2x+y^2+4y+8\)

\(=x^2-2x+1+y^2+4y+4+3\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+3>=3\forall x\)

Dấu '=' xảy ra khi x-1=0 và y+2=0

=>x=1 và y=-2

Vậy: \(C_{min}=3\) khi (x,y)=(1;-2)

Bài 2:

a: \(A=5-8x-x^2\)

\(=-\left(x^2+8x\right)+5\)

\(=-\left(x^2+8x+16-16\right)+5\)

\(=-\left(x+4\right)^2+16+5=-\left(x+4\right)^2+21< =21\forall x\)

Dấu '=' xảy ra khi x+4=0

=>x=-4

b: \(B=x-x^2\)

\(=-\left(x^2-x\right)\)

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

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

Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)

=>\(x=\dfrac{1}{2}\)

c: \(C=4x-x^2+3\)

\(=-x^2+4x-4+7\)

\(=-\left(x^2-4x+4\right)+7\)

\(=-\left(x-2\right)^2+7< =7\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

d: \(D=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x^2-6x+9+2\right)\)

\(=-\left(x-3\right)^2-2< =-2\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

\(A=\frac{2x^2-6x+5}{x^2-2x+1}=\frac{x^2-4x+4+x^2-2x+1}{x^2-2x+1}\)

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

Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}}\)\(\Rightarrow\frac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge0\)\(\Rightarrow\frac{\left(x-2\right)^2}{\left(x-1\right)^2}+1\ge1\)

\(\Rightarrow A\ge1\).Nên GTNN của \(A=1\) đạt được khi \(x=2\)

dòng thứ 2 ko hiểu

a)  -3/2

cần cách làm k cần kết quả

a) ta có : \(2x^2+2x-1=2\left(x^2+x-\frac{1}{2}\right)\))  = \(2\left(x^2+2.\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}-\frac{1}{2}\right)\)

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

   Vì \(\left(x+\frac{1}{2}\right)^2>=0\) => \(2\left(x+\frac{1}{2}\right)^2-\frac{3}{2}>=0\) => \(2.\left(x+\frac{1}{2}\right)^2>=-\frac{3}{2}\)

   Vậy GTNN của A là \(\frac{-3}{2}\). dấu " = " xảy ra khi và chỉ khi x =\(\frac{-1}{2}\)

\(A=\frac{3.\left(x^2-2x+5\right)+2}{x^2-2x+5}=3+\frac{2}{x^2-2x+1+4}=3+\frac{2}{\left(x-1\right)^2+4}\ge3+\frac{1}{2}=\frac{7}{2}\)

Dấu = xảy ra khi x-1=0

=> x=1

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

\(A=\frac{2x^2-4x+10+x^2-2x+7}{x^2-2x+5}\)

\(A=\frac{2\left(x^2-2x+5\right)+x^2-2x+5+2}{x^2-2x+5}\)

\(A=\frac{2\left(x^2-2x+5\right)}{x^2-2x+5}+\frac{x^2-2x+5}{x^2-2x+5}+\frac{2}{x^2-2x+5}\)

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

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

Vì \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow A\le3+\frac{2}{4}=\frac{7}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

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

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

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

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

\(B=4-\frac{\left(2x+1\right)^2}{x^2+1}\le4\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow2x+1=0\Leftrightarrow x=\frac{-1}{2}\)

Vậy....

1. Cho bt P= (1/√x+2 + 1/√x-2 ) . √x-2/√x với x>0, x khác 4

a) rút gọn P

b) tìm x để P>1/3

c) tìm các giá trị thực của x để Q=9/2P có giá trị nguyên

2. Cho 2 biểu thức

A= 1-√x / 1+√ x và B= ( 15-√x/ x-25 + 2/√x+5) : √x+1/√ x-5 với x lớn hơn hoặc bằng 0, x khác 25

a) tính giá trị của A khi x= 6-2√5

b) rút gọn B

c) tìm a để pt A-B=a có nghiệm

chúc bạn học tốt

Bài 1 :

\(a,P=\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}=\left[\frac{x}{\left(x+6\right)\left(x-6\right)}-\frac{x-6}{x\left(x+6\right)}\right]:\frac{2x-6}{x\left(x+6\right)}\)

\(=\frac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}.\frac{x\left(x+6\right)}{2x-6}=\frac{6\left(2x-6\right)}{x\left(x+6\right)\left(x-6\right)}.\frac{x\left(x+6\right)}{2x-6}\)

\(=\frac{6}{x-6}\)

\(b,\)Với \(x\ne-6;x\ne6;x\ne0;x\ne3\)  Thì

\(P=1\Rightarrow\frac{6}{X-6}=1\Rightarrow6=x-6\Rightarrow x=12\)(Thỏa mãn \(ĐKXĐ\))

\(c,\)Ta có :

\(P< 0\Rightarrow\frac{6}{X-6}< 0\Rightarrow X-6< 0\Rightarrow X< 6\)

Do :  \(x\ne-6;x\ne6;x\ne0;x\ne3\)  ,Nên với \(x< 6\)và  \(x\ne-6;x\ne0;x\ne3\)  thì \(P< 0\)

Bài 1 : 

a ) Ta có : 

\(P=\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}\)

\(=\left(\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right):\frac{2x-6}{x\left(x+6\right)}\)

\(=\frac{x.x-\left(x-6\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{x\left(x+6\right)}{2x-6}\)

\(=\frac{x^2-x^2+12x-36}{x-6}.\frac{1}{2\left(x-3\right)}\)

\(=\frac{12\left(x-3\right)}{x-6}.\frac{1}{2\left(x-3\right)}\)

\(=\frac{6}{x-6}\)

b ) \(P=1\Leftrightarrow\frac{6}{x-6}=1\Leftrightarrow x-6=6\Leftrightarrow x=12\left(tm\right)\)

c ) \(p< 0\Leftrightarrow\frac{6}{x-6}< 0\Leftrightarrow x-6< 0\Rightarrow x< 6\)

\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

Dấu "=" \(\Leftrightarrow x=-1\)

b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)