1) \(x^2-y^2-5x+5y\)

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

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

2) \(x^3-3x^2-3x+1\)

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

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

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

=\(\left(x-\left(2+\sqrt{3}\right)\right).\left(x-\left(2-\sqrt{3}\right)\right).\left(x+1\right)\)

3) \(3x^2-7x-10\)

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

=\(3x\left(x+1\right)-10\left(x+1\right)\)

=\(\left(3x-10\right)\left(x+1\right)\)

4) \(x^2-3x+2\)

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

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

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

Mặt phẳng đi qua trục hình trụ, cắt hình trụ theo thiết diện là hình vuông cạnh a. Thể tích khối trụ đó bằng

a) \(x^2\)\(+3x+7\)

=\(x^2\)\(+2.x.\frac{3}{2}\)\(+\frac{9}{4}\)\(+\frac{19}{4}\)

=\(\left(x+\frac{3}{2}\right)^2\)\(+\frac{19}{4}\)

Vì \(\left(x+\frac{3}{2}\right)^2\)\(\ge0\)

Nên \(\left(x+\frac{3}{2}\right)^2\)\(+\frac{19}{4}\)\(\ge\frac{19}{4}\)

Dấu "=" xảy ra khi:

 \(x+\frac{3}{2}\)\(=0\)

\(\Rightarrow x=-\frac{3}{2}\)

Vậy GTNN của \(x^2\)\(+3x+7\) là \(\frac{19}{4}\) khi \(x=-\frac{3}{2}\)

b) \(-9x^2+12x-15\)

=\(-\left(9x^2-12x+15\right)\)

=\(-\left(\left(3x\right)^2-2.3x.2+4+11\right)\)

=\(-\left(\left(3x-2\right)^2+11\right)\)

=\(-\left(3x-2\right)^2-11\)

Vì \(\left(3x-2\right)^2\)\(\ge0\)

Nên \(-\left(3x-2\right)^2-11\le-11\)

Dấu "=" xảy ra khi:

\(3x-2=0\)

\(\Rightarrow x=\frac{2}{3}\)

Vậy GTLN của \(-9x^2+12x-15\) là \(-11\) khì \(x=\frac{2}{3}\)

c) \(11-10x-x^2\)

=\(-\left(x^2+10x-11\right)\)

=\(-\left(x^2+2.x.5+25-36\right)\)

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

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

Vì \(\left(x+5\right)^2\ge0\)

Nên \(-\left(x+5\right)^2+36\le36\)

Dấu "=" xảy ra khi:

 \(x+5=0\)

\(\Rightarrow x=-5\)

Vậy GTLN \(11-10x-x^2\) là \(36\) khi \(x=-5\)

d)\(x^4+x^2+2\)

=\(\left(x^2\right)^2+2.x^2.\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\)

=\(\left(x^2+\frac{1}{2}\right)^2+\frac{7}{4}\)

Vì \(\left(x^2+\frac{1}{2}\right)^2\ge0\)

Nên \(\left(x^2+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)

Dấu "=" xảy ra khi:

 \(x^2+\frac{1}{2}=0\)

\(\Rightarrow x=\frac{1}{\sqrt{2}}\)

Vậy GTNN của \(x^4+x^2+2\) là \(\frac{7}{4}\) khi \(x=\frac{1}{\sqrt{2}}\)

\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow-5x^2-9x+2=0\)
\(\Leftrightarrow-5x^2-10x+x+2=0\)
\(\Leftrightarrow-5x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(-5x+1\right)\left(x+2\right)=0 \)
\(\Leftrightarrow\left(-5x+1\right)=0\) Hoặc \(x+2=0\)
\(\Leftrightarrow x=\frac{1}{5}\)Hoặc \(x=-2\)
 

Gọi \(A\) là tử (\(14x^2-8x+9\))

      \(C\) là mẫu (\(3x^2+6x+9\))

Ta có:\(A\) =\(14x^2-8x+9\)

\(\Rightarrow A_{min}\)=\(\frac{55}{7}\)

Ta có: \(C\)=\(3x^2+6x+9\)

\(\Rightarrow C_{min}\)=6

Suy ra \(B_{min}\)=\(\frac{\left(\frac{55}{7}\right)}{6}\)=\(\frac{55}{42}\)

Vậy GTNN của B là \(\frac{55}{42}\)

giờ mình đi học mai sẽ làm nốt phần còn lại

a) \(x^3\)+\(x^2\)=36

\(\Leftrightarrow\)\(x^3\)+\(x^2\)\(-36=0\)

\(\Leftrightarrow\)\(x^3\)\(-3x^2\)\(+4x^2\)\(-12x\)\(+12x-36=0\)

\(\Leftrightarrow\)\(x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2+4x+12\right)=0\)

Suy ra: \(x-3=0\) hoặc \(x^2+4x+12=0\)

Vậy \(x=3\)

a) \(\left(x-3\right)\)\(\left(x^2+3x+9\right)\)+\(x\left(x+2\right)\left(x-2\right)\) =1

\(\Leftrightarrow x^3\)\(-27\)+\(x\left(x^2-4\right)\) =1

\(\Leftrightarrow\)\(x^3\)\(-27\)\(+x^3\)\(-4x\) =1

\(\Leftrightarrow\)\(2x^3\)\(-4x-27\) = 1

Suy ra \(x\) =2,685673906