Olm chào em, em làm đúng rồi, em nhé!

10x +16y

A-B

\(=2x^3-4x^2y+131xy^2-y^4+1-\left(-2x^3-121x^2y-y^4-3\right)\)

\(=2x^3-4x^2y+131xy^2-y^4+1+2x^3+121x^2y+y^4+3\)

\(=4x^3+117x^2y+131xy^2+4\)

là sao

a) Ta có: \(2x^2+2x+3=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{5}{2}\)

\(=\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)

\(\Rightarrow S\le\frac{3}{\frac{5}{2}}=\frac{6}{5}\)

Vậy \(S_{max}=\frac{6}{5}\Leftrightarrow\sqrt{2}x+\frac{1}{\sqrt{2}}=0\Leftrightarrow x=-\frac{1}{2}\)

b) Ta có: \(3x^2+4x+15=\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.\frac{2}{\sqrt{3}}+\frac{4}{3}+\frac{41}{3}\)

\(=\left(\sqrt{3}x+\frac{2}{\sqrt{3}}\right)^2+\frac{41}{3}\ge\frac{41}{3}\)

\(\Rightarrow T\le\frac{5}{\frac{41}{3}}=\frac{15}{41}\)

Vậy \(T_{max}=\frac{15}{41}\Leftrightarrow\sqrt{3}x+\frac{2}{\sqrt{3}}=0\Leftrightarrow x=\frac{-2}{3}\)

c) Ta có: \(-x^2+2x-2=-\left(x^2-2x+1\right)-1\)

\(=-\left(x-1\right)^2-1\le-1\)

\(\Rightarrow V\ge\frac{1}{-1}=-1\)

Vậy \(V_{min}=-1\Leftrightarrow x-1=0\Leftrightarrow x=1\)

d) Ta có: \(-4x^2+8x-5=-\left(4x^2-8x+5\right)\)

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

\(=-\left(2x-2\right)^2-1\le-1\)

\(\Rightarrow X\ge\frac{2}{-1}=-2\)

Vậy \(X_{min}=-2\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

mệt rời o 

thông cảm 

hihi

Bài 7 

\(a,A=x^2-2x+5\)

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

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

GTNN \(A=4\) khi \(\left(x-1\right)^2=0\Rightarrow x=1\)

\(b,B=x^2-x+1\)

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

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

\(c,C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

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

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

Đặt \(x^2+5x=t\)

\(\Rightarrow C=\left(t-6\right)\left(t+6\right)\)

\(=t^2-36\)

\(\left(x^2+5x\right)^2-36\ge36\forall x\)

\(d,D=x^2+5y^2-2xy+4y-3\)

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

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

a) ( 5x³ - x +2 ) ( x-1 )
= 5x⁴ -5x³ - x² + 1 + 2x - 2
= 5x⁴ -5x³ - x² 2x - 1
b) ( 4x + 4 )(3 - x² - x³ )
= 12x - 8x³ - 4x⁴ + 12 - 4x² - 4x³
= -4x⁴ - 12x³ -4x² + 12x + 12

a) (5x³-x+2)(x-1) = 5x⁴-5x³-x²+3x-2

b) (4x+4)(3-x²-x³) = 12x-4x³-4x⁴+12-4x²-4x³ = -4x⁴ -8x³ - 4x² + 12x +12

\(a,x^2\left(x-2x^3\right)=x^3-3x^5\)

\(b,\left(x^2+1\right)\left(5-x\right)=5x^2-x^3+5-x\)

\(c,\left(x-2\right)\left(x^2+3x-4\right)=x^3+3x^2-4x-2x^2-6x+8\)

\(=x^3+x^2-10x+8\)

\(d,\left(x-2\right)\left(x-x^2+4\right)=x^2-x^3+4x-2x+2x^2-8\)

\(=x^3+3x^2+2x-8\)

Bài 3:

a: ĐKXĐ: x<>2

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

c: Khi x=4001/2000 thì \(M=\dfrac{3}{\dfrac{4001}{2000}-2}=3:\dfrac{1}{2000}=6000\)

Bài 1:

a/\(xy\ne0\), nhân cả tử và mẫu với \(xy\) ta được:

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

b/ \(x\ne\pm1\), nhân cả tử và mẫu với \(x^2-1=\left(x-1\right)\left(x+1\right)\) ta được:

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

c/ \(x\ne\pm1\), nhân cả tử và mẫu với \(\left(x-1\right)\left(x+1\right)\) ta được:

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

Bài 2:

a/ Xem lại đề, thấy có vẻ ko đối xứng lắm, \(\frac{2x+1}{2x-2}\) hay \(\frac{2x+1}{2x-1}\) bạn?

b/ \(x\ne\left\{-1;0;1\right\}\)

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

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

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

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