\(A=\left(x+5\right)^2-62\ge-62\)

\(B=\left(\frac{1}{2}x^2+1-\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)

\(C=\left(x-3y+2\right)^2+\left(x-5\right)^2-9\ge-9\)

\(D=\left(x-y+1\right)^2+\left(y-4\right)^2\ge0\)

\(A=-\left(x-3\right)^2+12\le12\)

\(B=-2x^2-5x+3=-2\left(x+\frac{5}{4}\right)^2+\frac{49}{8}\le\frac{49}{8}\)

\(C=\frac{1}{\left(x-2\right)^2+5}\le\frac{1}{5}\)

\(A=x^2+10x-37\)

     \(=\left(x+5\right)^2-62\) 

Có \(\left(x+5\right)^2\ge0\forall x\in R\) 

 \(\Rightarrow\left(x+5\right)^2-62\ge-62\forall x\in R\) 

Dấu = xảy ra \(\Leftrightarrow x+5=0\Leftrightarrow x=-5\) 

Vậy A đạt GTNN là -62 tại x=-5

Bài 1.

a) Do hai phân thức bằng nhau , ta có :

( x +2)P( x² - 2²) = ( x - 1)Q( x -2)

=( x + 2)P( x - 2)( x + 2) = ( x - 1)Q( x - 2)

Suy ra : P = x - 1 ; Q = ( x + 2)²

b) Do hai phân thức bằng nhau , ta có :

( x + 2)P(x² - 2x + 1) = ( x - 2)Q( x² - 1)

= ( x + 2)P( x - 1)² = ( x - 2)Q( x - 1)( x + 1)

Suy ra : P = ( x - 2)( x + 1) = x² - x - 2

Q = ( x + 2)( x - 1) = x² + x + 2

Bài 2. a) Do : \(\dfrac{P}{Q}=\dfrac{R}{S}=>PS=QR\)

Xét : ( P + Q)S= PS + QS = QR + QS = Q( R + S)

-> \(\dfrac{P+Q}{Q}=\dfrac{R+S}{S}\)

b) Do : \(\dfrac{P}{Q}=\dfrac{R}{S}=>PS=QR\)

Xét : ( S - R)P = PS - PR = QR - PR = R( Q - P)

-> \(\dfrac{R-S}{R}=\dfrac{Q-P}{P}\)

- > \(\dfrac{R}{R-S}=\dfrac{P}{Q-P}\)

ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)

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

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

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

Để R=0 thì \(x^2+1=0\)(vô lý)

b: Ta có: |x|=1

=>x=1(loại) hoặc x=-1(loại)

a, ĐKXĐ : \(x-1\ne0\)

=> \(x\ne1\)

TH₁ : \(x-2\ge0\left(x\ge2\right)\)

=> \(\left|x-2\right|=x-2=1\)

=> \(x=3\left(TM\right)\)

- Thay x = 3 vào biểu thức P ta được :

\(P=\frac{3+2}{3-1}=\frac{5}{2}\)

TH₂ : \(x-2< 0\left(x< 2\right)\)

=> \(\left|x-2\right|=2-x=1\)

=> \(x=1\left(KTM\right)\)

Vậy giá trị của P là \(\frac{5}{2}\) .

a) \(P=\frac{x+2}{x-1}\) \(\left(ĐKXĐ:x\ne1\right)\)

Ta có: \(\left|x-2\right|=1\text{⇔}\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) (loại x = 1 vì x ≠ 1)

Thay \(x=3\) vào P, ta có:

\(P=\frac{3+2}{3-2}=\frac{5}{1}=5\)

Vậy P = 5 tại x = 3.

b) \(Q=\frac{x-1}{x}+\frac{2x+1}{x^2+x}=\frac{x-1}{x}+\frac{2x+1}{x\left(x+1\right)}=\frac{x^2-1}{x\left(x+1\right)}+\frac{2x+1}{x\left(x+1\right)}\) (ĐKXĐ: x ≠ 0, x ≠ -1)

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

Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :

\(y^2+2y+4^x-2^{x+1}+2=0\)

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

\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)

\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)

Dấu = xảy ra khi :

\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)

CHÚC BẠN HỌC TỐT........... 

mk chịu

Thanks bạn nha, chữ bạn đẹp quá

a)

$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$

$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$

$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$

$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$

$=2005^2-1002.2005=2005(2005-1002)=2011015$

b)

$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^{32}-1)(2^{32}+1)-2^{64}$

$=2^{64}-1-2^{64}=-1$

c) Do $x=16$ nên $x-16=0$

$R(x)=x^4-17x^3+17x^2-17x+20$

$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$

$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$

$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$

d) Do $x=12$ nên $x-12=0$. Khi đó:

$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$

$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$

$=(x-12)(x^9-x^8+x^7-....+x)-x+10$

$=0-x+10=-x+10=-12+10=-2$