a: \(A=\left(4x+y\right)\left(4x-y\right)-8x\left(2x-1\right)\)

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

\(=8x-y^2\)

\(=8\cdot3-\left(-1\right)^2=24-1=23\)

b: \(B=16x\left(4x^2-5\right)-\left(4x+1\right)\left(16x^2-4x+1\right)\)

\(=64x^3-80x-\left(64x^3+1\right)\)

=-80x-1

Khi x=1/5 thì \(B=-80\cdot\frac15-1=-16-1=-17\)

c: \(C=3x^2-2x+3y^2-2y+6xy-100\)

\(=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)-100\)

\(=3\left(x+y\right)^2-2\left(x+y\right)-100\)

\(=3\cdot10^2-2\cdot10-100\)

=300-20-100

=300-120

=180

a: \(A=\left(4x+y\right)\left(4x-y\right)-8x\left(2x-1\right)\)

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

\(=8x-y^2\)

\(=8\cdot3-\left(-1\right)^2=24-1=23\)

b: \(B=16x\left(4x^2-5\right)-\left(4x+1\right)\left(16x^2-4x+1\right)\)

\(=64x^3-80x-\left(64x^3+1\right)\)

=-80x-1

Khi x=1/5 thì \(B=-80\cdot\frac15-1=-16-1=-17\)

c: \(C=3x^2-2x+3y^2-2y+6xy-100\)

\(=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)-100\)

\(=3\left(x+y\right)^2-2\left(x+y\right)-100\)

\(=3\cdot10^2-2\cdot10-100\)

=300-20-100

=300-120

=180

\(A=16x^2-y^2-16x^2+8x=8x-y^2\\ A=8\cdot3-\left(-1\right)^2=24-1=23\\ B=64x^3-80x-64x^3-1=-80x-1\\ B=-80\cdot\dfrac{1}{5}-1=-16-1=-17\)

a, \(\dfrac{3x-2}{3}-1>0\Leftrightarrow\dfrac{3x-5}{3}>0\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{5}{3}\\x\ne\dfrac{5}{3}\end{matrix}\right.\)

b, \(\dfrac{4x-1}{2}-\dfrac{2x-3}{5}< 0\Leftrightarrow\dfrac{20x-5-4x+6}{10}< 0\Leftrightarrow\left\{{}\begin{matrix}x< -\dfrac{1}{16}\\x\ne-\dfrac{1}{16}\end{matrix}\right.\)

a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

A=916/55>3/10

cách làm đâu