XL gtnn B = 19/4

GTNN = -1/4

GTNN B = 23/4

b) Ta có: \(B=x^2+2x+y^2-4y+6\)

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

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

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)

c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

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

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)

\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)

\(B=x^2+2x+y^2-4y+6\)

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

\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2

\(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)

dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)

\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)

=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)

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

\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)

dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

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

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

\(A=4x^2-4x+19=4x^2+4x+1+18=\left(2x+1\right)^2+18\ge18\)

\(minA=18\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

\(A=4x^2-4x+19\)

\(=4x^2-4x+1+18\)

\(=\left(2x-1\right)^2+18\ge18\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(A=-\left(4x^2-4x+1\right)-\left(y^2+6y+9\right)+11\\ A=-\left(2x-1\right)^2-\left(y+3\right)^2+11\le11\\ A_{max}=11\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

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

\(4x^2-4x+1-\left(y^2+8y+16\right)=\left(2x-1\right)^2-\left(y+4\right)^2\)

\(=\left(2x-1-y-4\right)\left(2x-1+y+4\right)=\left(2x-y-5\right)\left(2x+y+3\right)\)

câu 1 B 

câu 2 D

câu 3 ko bt 

câu 4 x=-1/2; x = -(căn bậc hai(3)*i-1)/4;x = (căn bậc hai(3)*i+1)/4;

câu 5 x=-5/3, x=0, x=1

Câu 1:  x² + 2 xy + y²   bằng:

A. x² + y²                   B.(x + y)²                  C. y² – x²                  D. x² – y²

Câu 2:  (4x + 2)(4x – 2)  bằng:

A. 4x² + 4                  B. 4x² – 4                 C. 16x² + 4                D. 16x² – 4

Câu 3: 25a²  + 9b²  - 30ab  bằng:

A.(5a-9b)²                  B.(5a – 3b)²              C.(5a+3b)²                D.(5a)² – (3b)²

Câu 4: 8x³ +1 bằng

A.(2x+1).(4x²-2x+1)      B. (2x-1).(4x²+2x+1)       C.(2x+1)³            D.(2x)³-1³

Câu 5:Thực hiện phép nhân  x(3x² + 2x - 5) ta được:

A.3x³ - 2x² – 5x          B. 3x³ + 2x² – 5x      C. 3x³ - 2x² +5x         D. 3x³ + 2x² + 5x

\(B=y^2-y+1\)

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

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

Ta thấy: \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)

\(\Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall y\)

Dấu \("="\) xảy ra \(\Leftrightarrow y-\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{1}{2}\)

Vậy \(B_{min}=\dfrac{3}{4}\) khi \(y=\dfrac{1}{2}\).

\(---\)

\(C=x^2-4x+y^2-y+5\)

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

\(=\left(x^2-2\cdot x\cdot2+2^2\right)+\left[y^2-2\cdot y\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{3}{4}\)

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

Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)

              \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;y\)

Dấu \("="\) xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(C_{min}=\dfrac{3}{4}\) khi \(x=2;y=\dfrac{1}{2}\).

\(Toru\)

\(B=y^2-y+1\)

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

Vì \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\Rightarrow B\ge\dfrac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)

\(C=x^2-4x+y^2-y+5\)

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

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

Vì \(\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(a,M=x^2-4x+5=\left(x-2\right)^2+5\\ \Rightarrow M\ge5\)

Dấu "=" xảy ra \(\Leftrightarrow x=2\)

\(b,N=y^2-y-3=\left(y-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\\ \Rightarrow N\ge-\dfrac{13}{4} \)

Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)

\(P=x^2+y^2-4x+y+7=\left(x-2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\\ \Rightarrow P\ge\dfrac{11}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)

a: M=x^2-4x+4+1

=(x-2)^2+1>=1

Dấu = xảy ra khi x=2

b: N=y^2-y+1/4-13/4

=(y-1/2)^2-13/4>=-13/4

Dấu = xảy ra khi y=1/2

c: P=x^2-4x+4+y^2+y+1/4+11/4

=(x-2)^2+(y+1/2)^2+11/4>=11/4

Dấu = xảy ra khi x=2 và y=-1/2