Bài 3:

a) \(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)\

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\1+5x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-\frac{1}{5}\end{array}\right.\)

b)\(x+1=\left(x+1\right)^2\)

\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)

\(\Leftrightarrow-\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

c)\(x^3+x=0\)

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

<=> x=0 (Vì \(x^2+1>0\) )

goi so thoc 1la x(tan,x>0)(x<1400)

=)so thoc kho 2 la 1400-x(tan)

vi 2% so thoc kho 1 =3% so thoc kho 2 nen

2x/100=3(1400-x)/100

=)2x=4200-3x

=)5x=4200

=)x=840(tm)

=)so thoc kho thu nhat la 840(tan)

   so thoc kho thu hai la 560 (tan)

Bài 1:

a) \(5x-20y=5\left(x-4y\right)\)

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

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

55 ha 17 m2 = ...,...ha

17 m2 = 0,0017 ha

Vậy 55 ha 17 m2 = 55 + 0,0017 = 55,0017 ha

Vậy 55 ha 17 m2 = 55,0017 ha

a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)

\(=1+1+0,5=2,5\)

b)\(\frac{3}{7}\cdot19\frac{1}{3}-\frac{3}{7}\cdot33\frac{1}{3}\)

\(=\frac{3}{7}\left(19\frac{1}{3}-33\frac{1}{3}\right)=\frac{3}{7}\cdot\left(-14\right)=-6\)

c) \(9\left(-\frac{1}{3}\right)^3+\frac{1}{3}=9\cdot\left(\frac{-1}{27}\right)+\frac{1}{3}=-\frac{1}{3}+\frac{1}{3}=0\)

d) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)=-\frac{7}{5}:\left(15\frac{1}{4}-25\frac{1}{4}\right)=-\frac{7}{5}:\left(-10\right)=14\)

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

\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_8}{a_9}=\frac{a_9}{a_1}=\frac{a_1+a_2+...+a_8+a_9}{a_2+a_3+..+a_9+a_1}=1\)

=> \(\frac{a_1}{a_2}=1\Rightarrow a_1=a_2\)

     \(\frac{a_2}{a_3}=1\Rightarrow a_2=a_3\)

      .....

     \(\frac{a_8}{a_9}=1\Rightarrow a_8=a_9\)

     \(\frac{a_9}{a_1}=1\Rightarrow a_9=a_1\)

=> \(a_1=a_2=..a_9\)

Có:

\(VT:2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+4\cdot2-2\sqrt{6}\)

\(=1+8=9=VP\)

=>đpcm

a)\(\sqrt{75\cdot48}=\sqrt{25\cdot3\cdot48}=\sqrt{25\cdot144}=\sqrt{25}\cdot\sqrt{144}=5\cdot12=60\)

b) \(\sqrt{2,5\cdot14,4}=\sqrt{25\cdot144\cdot\frac{1}{100}}=\sqrt{25}\cdot\sqrt{144}\cdot\sqrt{\frac{1}{100}}=5\cdot12\cdot\frac{1}{10}=6\)

a)\(\sqrt{10}\cdot\sqrt{40}=\sqrt{10\cdot40}=\sqrt{400}=20\)

b) \(\sqrt{2}\cdot\sqrt{162}=\sqrt{2\cdot162}=\sqrt{2\cdot2\cdot81}=\sqrt{4}\cdot\sqrt{81}=2\cdot9=18\)

Có 2 cách chứng minh: tia Oz là tia p/g của góc xOy

+) Cách 1: tia Oz nằm giữa 2 tia Ox và Oy

          và góc xOz = zOy

+) Cách 2: góc xOz = zOy = xOy / 2