\(\dfrac{x}{-15}=-\dfrac{60}{x}\)

\(\Leftrightarrow x.x=-15.\left(-60\right)\)

\(\Leftrightarrow x^2=900\)

\(\Rightarrow x^2=\left(\pm30\right)^2\)

\(\Rightarrow x=30\) hoặc \(x=-30\)

a, \(8,5:3=2,8\left(3\right)\)

b, \(18,7:6=3,11\left(6\right)\)

\(c,58:11==5,\left(27\right)\)

d, \(14,2:3,3=4,\left(30\right)\)

Từ TLT: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{\left(a+c\right)}{\left(b+d\right)}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{ac}{bd}\left(dpcm\right)\)

Ta có: \(32^9=\left(2^5\right)^9=2^{45}\)

\(18^{13}< 16^{13}=\left(2^4\right)^{13}=2^{52}\)

\(\rightarrow2^{45}< 2^{52}< 18^{13}\)

\(\Rightarrow2^{45}< 18^{13}\)

Hay: \(32^9< 18^{13}\)

Vậy \(32^9< 18^{13}\)

C1: Ta có: \(\widehat{B_2}+\widehat{B_3}=180^0\) (kề bù)

\(Hay:50^0+\widehat{B_3}=180^0\)

\(\Rightarrow\widehat{B_3}=180^0-50^0=130^0\)

Vì \(\widehat{A_3}=\widehat{B_3}=130^0\)

Mà \(\widehat{A_3}\) và \(\widehat{B_3}\) ở vị trí đồng vị \(\Rightarrow a//b\)
C2: Có: \(\widehat{A_3}=\widehat{A_1}=130^0\) (hai góc đối đỉnh)

Vì \(\widehat{A_1}=\widehat{B_3}=130^0\) (theo cm ở C1)

Mà \(\widehat{A_1}\) và \(\widehat{B_3}\) ở vị trí đối đỉnh \(\Rightarrow a//b\)

C3: Ta có: \(\widehat{A_3}+\widehat{A_4}=180^0\) (kề bù)

\(Hay:130^0+\widehat{A_4}=180^0\)

\(\Rightarrow\widehat{A_4}=180^0-130^0=50^0\)

Do đó: \(\widehat{A_4}+\widehat{B_3}=50^0+130^0=180^0\) (theo cm C1)

Mà \(\widehat{A_4}\) và \(\widehat{B_3}\) ở vị trí trong cùng phía \(\Rightarrow a//b\)

Chúc bn hok tốt!

Ta có:

\(2a=3b=5c\Leftrightarrow\dfrac{30a}{15}=\dfrac{30b}{10}=\dfrac{30c}{6}\\ \Rightarrow\dfrac{a}{15}=\dfrac{b}{10}=\dfrac{c}{6}va`a-b+c=11\)

AD t/c DTS = nhau ta có:

\(\dfrac{a}{15}=\dfrac{b}{10}=\dfrac{c}{6}=\dfrac{a-b+c}{15-10+6}=\dfrac{11}{11}=1\)

+) \(\dfrac{a}{15}=1\Rightarrow a=15\)

+) \(\dfrac{b}{10}=1\Rightarrow b=10\)

+) \(\dfrac{c}{6}=1\Rightarrow c=6\)

Vậy \(a=15;b=10;c=6\)

a, \(\left(4.2\right)^5:\left(2^3.\dfrac{1}{16}\right)=8^5:\left(2^3.\dfrac{1^4}{2^4}\right)=\left(2^3\right)^5:\dfrac{2^3.1^4}{2^4}=2^{15}:\dfrac{1}{2}=2^{15}.2=2^{16}\)

\(b,\dfrac{2^2.4.32}{2^2.2^5}=\dfrac{2^2.2^4.2^5}{2^2.2^5}=2^4=16\)

a, \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\left(\dfrac{5}{3}-\dfrac{1}{4}\right).\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)

b, \(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3=2:\left(-\dfrac{1}{2}\right)^3=2:-\dfrac{1}{8}=2.-8=-16\)

Ta có: \(3x=-2y\Rightarrow\dfrac{3}{-2}=\dfrac{y}{x}\Rightarrow\dfrac{x}{-2}=\dfrac{y}{3}\)

Ta lại có: \(\dfrac{x}{-2}=\dfrac{y}{3}\Rightarrow\dfrac{2x}{-4}=\dfrac{y}{3}\)

AD t/c DTS = nhau ta có:

\(\dfrac{2x}{-4}=\dfrac{y}{3}=\dfrac{2x+y}{-4+3}=\dfrac{5}{-1}=-5\)

+) \(\dfrac{2x}{-2}=-5\Rightarrow2x=10\Rightarrow x=5\)

+) \(\dfrac{y}{3}=-5\Rightarrow y=-15\)

Chúc bn hok tốt!

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