Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x_1-1}{5}=\dfrac{x_2-2}{4}=\dfrac{x_3-3}{3}=\dfrac{x_4-4}{2}=\dfrac{x_5-5}{1}\)

\(=\dfrac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)

\(=\dfrac{\left(x_1+x_2+x_3+x_4+x_5\right)-\left(1+2+3+4+5\right)}{15}\)

\(=\dfrac{30-15}{15}=1\)

\(\Rightarrow x_1=x_2=x_3=x_4=x_5=6\)

Vậy...

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x1-1}{5}\)=\(\dfrac{x2-2}{4}\)\(\dfrac{x3-3}{3}\)=\(\dfrac{x4-4}{2}\)=\(\dfrac{x5-5}{1}\)=\(\dfrac{x1-1+x2-2+x3-3+x4-4+x5-5}{5+4+3+2+1}\)=\(\dfrac{x1+x2+x3+x4+x5-\left(1+2+3+4+5\right)}{15}\)=\(\dfrac{30-15}{15}\)=\(\dfrac{15}{15}\)=1

\(\dfrac{x1-1}{5}\)=1 => x1-1=5 => x1 =6

\(\dfrac{x2-2}{4}\)=1 => x2-2=4 => x2 =6

\(\dfrac{x3-3}{3}\)=1 => x3-3=3 => x3 =6

\(\dfrac{x4-4}{2}\)=1 => x4-4=2 => x4 =6

\(\dfrac{x5-5}{1}\)=1 => x5-5=1 => x5 = 6

Vậy x1=x2=x3=x4=x5 =6

10

9/8

1/4

35/12

`5/2xx4=5xx2=10`

`2-7/8=16/8-7/8=9/8`

`9/4-2=9/4-8/4=1/4`

`5/6:2/7=5/6xx7/2=35/12`

\(\dfrac{5}{2}\times4=10\)

\(2-\dfrac{7}{8}=\dfrac{9}{8}\)

\(\dfrac{9}{4}-2=\dfrac{1}{4}\)

\(\dfrac{5}{6}:\dfrac{2}{7}=\dfrac{35}{12}\)

\(a)\)\(\dfrac{a}{b}\times4+\dfrac{1}{6}=\dfrac{19}{6}\)

   \(\dfrac{a}{b}\times4=\dfrac{19}{6}-\dfrac{1}{6}\)

   \(\dfrac{a}{b}\times4=\dfrac{18}{6}\)

   \(\dfrac{a}{b}=\dfrac{18}{6}\div4\)

   \(\dfrac{a}{b}=\dfrac{18}{6}\times\dfrac{1}{4}\)

\(\dfrac{a}{b}=\dfrac{18}{24}\)

Lời giải:

a.

$=(x^2)^2+(\frac{1}{2}y^4)^2+2.x^2.\frac{1}{2}y^4-x^2y^4$

$=(x^2+\frac{1}{2}y^4)^2-(xy^2)^2$
$=(x^2+\frac{1}{2}y^4-xy^2)(x^2+\frac{1}{2}y^4+xy^2)$
b.

$=(\frac{1}{2}x^2)^2+(y^4)^2+2.\frac{1}{2}x^2.y^4-x^2y^4$
$=(\frac{1}{2}x^2+y^4)^2-(xy^2)^2$
$=(\frac{1}{2}x^2+y^4-xy^2)(\frac{1}{2}x^2+y^4+xy^2)$

c.

$=(8x^2)^2+(y^2)^2+2.8x^2.y^2-16x^2y^2$

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

d.

$=\frac{64x^4+y^4}{64}=\frac{1}{64}(8x^2+y^2-4xy)(8x^2+y^2+4xy)$

c: \(64x^4+y^4\)

\(=64x^4+16x^2y^2+y^4-16x^2y^2\)

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

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

a) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)

b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)

d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)

e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)

f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)

g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)

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

2) \(4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\)

3) \(4x^2-49=\left(2x\right)^2-7^2=\left(2x-7\right)\left(2x+7\right)\)

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

5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2\right)^2-\left(\dfrac{1}{2}\right)^2=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)

6) \(x^{32}-1\\ =\left(x^{16}\right)^2-1^2\\ =\left(x^{16}-1\right)\left(x^{16}+1\right)\\ =\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

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

2: \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

3: \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)

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

5: \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)

(a) \(9x^2+12x+4=0\)

\(\Leftrightarrow\left(3x+2\right)^2=0\Leftrightarrow3x+2=0\Leftrightarrow x=-\dfrac{3}{2}\)

(b) \(x^2+\dfrac{1}{4}=x\)

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

(c) \(4-\dfrac{12}{x}+\dfrac{9}{x^2}=0\left(x\ne0\right)\)

\(\Leftrightarrow\left(2-\dfrac{3}{x}\right)^2=0\Leftrightarrow2-\dfrac{3}{x}=0\Leftrightarrow x=\dfrac{3}{2}\)