Đáp án D

Ta có lim x → 2 − f x = lim x → 2 − 2 x 2 − 7 x + 6 x − 2 = lim x → 2 − 2 x 2 − 7 x + 6 x − 2 = lim x → 2 − − 2 x − 3 = − 1  

Và lim x → 2 − f x = lim x → 2 − a + 1 − x 2 + x = a − 1 4 ; f 2 = a − 1 4 .  

Theo bài ra, ta có lim x → 2 + f x = lim x → 2 − f x = f 2 ⇒ a = − 3 4  

Do đó, bất phương trình − x 2 + a   x + 7 4 > 0 ⇔ − x 2 − 3 4 x + 7 4 > 0 ⇔ − 7 4 < x < 1.  

a) \(P=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\)

\(MinP=4\Leftrightarrow x-1=0\Rightarrow x=1\)

b) \(Q=2x^2-6x\)

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

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

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

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

\(MinQ=\frac{-9}{2}\Leftrightarrow x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

M=x^2+y^2-x+6y+10

M=(x^2-x+1/4)+(y^2+6y+9)+3/4

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

\(minM=\frac{3}{4}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

\(2x^2+6x-5=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{19}{2}\ge-\dfrac{19}{2}\)

Dấu "=" xảy ra khi \(x=-\dfrac{3}{2}\)

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

Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)

nhanh giùm mình được không

Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

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

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

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

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

a: \(P=x^2+y^2-6x-2y+17\)

\(=x^2-6x+9+y^2-2y+1+7\)

\(=\left(x-3\right)^2+\left(y-1\right)^2+7\ge7\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y-1=0

=>x=3 và y=1

b: \(Q=x^2+xy+y^2-3x-3y+999\)

\(=x^2+x\left(y-3\right)+y^2-3y+999\)

\(=x^2+2\cdot x\cdot\left(\frac12y-\frac32\right)+\left(\frac12y-\frac32\right)^2+y^2-3y-\left(\frac12y-\frac32\right)^2+999\)

\(=\left(x+\frac12y-\frac32\right)^2+y^2-3y-\left(\frac14y^2-\frac32y+\frac94\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34y^2-\frac32y-\frac94+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y^2-2y-3\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y^2-2y+1-4\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y-1\right)^2+996\ge996\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ x+\frac12y-\frac32=0\end{cases}\Rightarrow\begin{cases}y=1\\ x=-\frac12y+\frac32=-\frac12+\frac32=\frac22=1\end{cases}\)

c: \(R=2x^2+2xy_{}+y^2-2x+2y+15\)

\(=x^2-4x+4+x^2+2xy+y^2+2x+2y+11\)

\(=\left(x-2\right)^2+x^2+2xy+y^2+2x+2y+1+10\)

\(=\left(x-2\right)^2+\left(x+y+1\right)^2+10\ge10\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ x+y+1=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=-x-1=-2-1=-3\end{cases}\)

d: \(S=x^2+26y^2-10xy+14x-76y+59\)

\(=x^2-10xy+25y^2+14x-70y+y^2-6y+59\)

\(=\left(x-5y\right)^2+14\left(x-5y\right)+49+y^2-6y+9+1\)

\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-3=0\\ x-5y+7=0\end{cases}\Rightarrow\begin{cases}y=3\\ x=5y-7=5\cdot3-7=15-7=8\end{cases}\)

e: \(T=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2-4xy+4y^2+10x-20y+y^2-2y+28\)

\(=\left(x-2y\right)^2+10\left(x-2y\right)+25+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ x-2y+5=0\end{cases}\Rightarrow\begin{cases}y=1\\ x=2y-5=2\cdot1-5=2-5=-3\end{cases}\)

a) P(x)=4x²-6x+a; Q(x)=x-3

Lấy P(x):Q(x)=4x-6 dư a+30

Vậy để P(x)⋮Q(x) ⇒ a+30=0 ⇒ a=-30

b) P(x)=2x²+x+a; Q(x)=x+3

Lấy P(x):Q(x)=2x-7 dư a+21

Vậy để P(x)⋮Q(x) ⇒ a+21=0 ⇒ a=-21

c) P(x)=x³+ax²-4; Q(x)=x²+4x+4

Lấy P(x):Q(x)=x+a-4 dư -4(a-5)x+12

Vậy để P(x)⋮Q(x) ⇒ -4(a-5)x+12=0 ⇒ (a-5)x=3

⇒ a-5 ϵ {-1;1;-3;3} (a ϵ Z)

⇒ a ϵ {4;6;2;8}

d) P(x)=2x²+ax+1; Q(x)=x-3

Lấy P(x):Q(x)=2x+a+6 dư 3a+19

Vậy để P(x)⋮Q(x) ⇒ 3a+19=0 ⇒ a=-19/3

e) P(x)=ax⁵+5x⁴-9; Q(x)=x-1

Lấy P(x):Q(x)=ax⁴+(a-5)x³+(a-5)x²+(a-5)x+1 dư a-4

Vậy để P(x)⋮Q(x) ⇒ a-4=0 ⇒ a=4

f) P(x)=6x³-x²-23x+a; Q(x)=2x+3

Lấy P(x):Q(x)=3x²-5x-4 dư a+12

Vậy để P(x)⋮Q(x) ⇒ a+12=0 ⇒ a=-12

g) P(x)=x³-6x²+ax-6 Q(x)=x-2

Lấy P(x):Q(x)=x²-2x+a-4 dư 2(a-4)-6

Vậy để P(x)⋮Q(x) ⇒ 2(a-4)-6=0 ⇒ a=7

Bài h có a,b bạn xem lại đề

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

Để B nhận giá trị nguyên thì \(x-2\inƯ_{\left(4\right)}=\left\{\pm1;\pm2;\pm4\right\}\)

Với \(x+2=-4\Rightarrow x=-6\)

      \(x+2=-2\Rightarrow x=-4\)

      \(x+2=-1\Rightarrow x=-3\)

      \(x+2=1\Rightarrow x=-1\)

      \(x+2=2\Rightarrow x=0\)

      \(x+2=4\Rightarrow x=2\)

ta có : \(\frac{6x+2}{x+2}=\frac{6}{x+2}+1\)

Để B nguyên thì \(6⋮x+2\) \(\Rightarrow\left(x+2\right)\inƯ\left(6\right)=\left\{-1;-2;-3;-6;1;2;3;6\right\}\)

ta có :
 

a) x=0

b) x=0

c) x=0

d)x=x

a b c d 

x=x