Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

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

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

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

\(=-\left(x^2+2x+1-9\right)\)

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

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

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

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

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chúc bạn học ngu

a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)

\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)

\(A=9x\)

Thay x = 15 vào, ta có: 

\(A=9.15=135\)

b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(B=5x^2-20xy-4y^2+20xy\)

\(B=5x^2-4y\)

Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có: 

\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)

c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)

\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(C=9x^2y^2-xy^3-8x^3\)

Thay \(x=\frac{1}{2};y=2\) vào, ta có:

\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)

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

\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)

\(D=18x^2+12x-7\)

Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)

+) Với x = -2

\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)

+) Với x = 2

\(D=18.2^2+12.2-7=89\)

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

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

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

a) Tìm MTC:

\(\text{2x + 6 = 2(x + 3)}\)

\(\text{x2 – 9 = (x – 3)(x + 3)}\)

\(\text{MTC = 2(x – 3)(x + 3) = 2(x2 – 9)}\)

Nhân tử phụ:

\(\text{2(x – 3)(x + 3) : 2(x + 3) = x – 3}\)

\(\text{2(x – 3)(x + 3) : (x2 – 9) = 2}\)

Qui đồng:

b) Tìm MTC:

a) \(A=x^2+2x+2\)

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

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

b) \(B=4x^2-4x+11\)

\(=4x^2-4x+1+10\)

\(=\left(2x-1\right)^2+10>0\forall x\)

c) \(C=x^2-x+1\)

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

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

d) Ta có: \(D=x^2-2x+y^2+4y+6\)

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

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

e) Ta có: \(D=x^2-2xy+y^2+x^2-8x+20\)

\(=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4>0\forall x,y\)

\(A=x^2+2x+2=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)

\(B=4x^2-4x+11=\left(2x-1\right)^2+10\ge10>0\left(\forall x\right)\)

\(C=x^2-x+1=x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1>0\)

\(E=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4\ge4>0\)\(\left(\forall x,y\right)\)

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