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Câu 9 cần bs điều kiện $x,y,z\neq 0$
$\frac{x}{3}=\frac{y}{4}\Rightarrow \frac{x}{15}=\frac{y}{20}$
$\frac{y}{5}=\frac{z}{6}\Rightarrow \frac{y}{20}=\frac{z}{24}$
$\Rightarrow \frac{x}{15}=\frac{y}{20}=\frac{z}{24}$ và đặt $=t$ (đk: $t\neq 0$)
$\Rightarrow x=15t; y=20t; z=24t$
Khi đó:
$M=\frac{2.15t+3.20t+4.24t}{3.15t+4.20t+5.24t}=\frac{186t}{245t}=\frac{186}{245}$
Đáp án B.
Câu 10:
Giả sử số $A$ được chia thành 3 phần $a,b,c$ sao cho
$a:b:c=\frac{2}{5}: \frac{3}{4}: \frac{1}{6}$
Đặt $a=\frac{2}{5}t; b=\frac{3}{4}t; c=\frac{1}{6}t$
$A=a+b+c=\frac{2}{5}t+\frac{3}{4}t+\frac{1}{6}t=\frac{79}{60}t$
Có:
$a^2+b^2+c^2=(\frac{2}{5}t)^2+(\frac{3}{4}t)^2+(\frac{1}{6}t)^2=24309$
$t^2=32400$
$t=\pm 180$
$\Rightarrow A=\frac{79}{60}t=\frac{79}{60}\pm 180=\pm 237$
Đáp án D.
tổng góc toi + góc pxa = 90+30 =120
góc giua mat pxa va pnam ngang = 30+(180-120)/2 = 60o
Ta có: 2a = 3b = 4c
\(\Rightarrow\frac{2a}{12}=\frac{3b}{12}=\frac{4c}{12}\) \(\Rightarrow\frac{a}{6}=\frac{b}{4}=\frac{c}{3}\)
\(\Rightarrow\frac{a}{6}=\frac{b}{4}=\frac{c}{3}=\frac{a+b+c}{6+4+3}=\frac{26}{13}=2\)
\(\Rightarrow a=6.2=12;b=2.4=8;c=2.3=6\)
\(\frac{x}{3}=\frac{y}{9}\) và \(y-x=12\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{9}=\frac{y-x}{9-3}=\frac{12}{6}=2\)
Do đó:
\(\frac{x}{3}=2\Rightarrow x=3.2=6\)
\(\frac{y}{9}=2\Rightarrow y=9.2=18\)
Vậy \(x=6;y=18\)
\(\frac{x}{3}\)=\(\frac{y}{9}\) và x-y=12
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{3}\)=\(\frac{y}{9}\)=\(\frac{x-y}{3-9}\)=\(\frac{12}{-6}\)=\(\frac{-2}{1}\)
==>x=\(\frac{3.-2}{1}\)=-6
y=\(\frac{9.-2}{1}\)=-18
Hok tốt!
1: \(75^3:\left(-25\right)^3=\left(\dfrac{75}{-25}\right)^3=\left(-3\right)^3=-27\)
2: \(\left(-60\right)^2:\left(-5\right)^2=\dfrac{60^2}{5^2}=12^2=144\)
3: \(169^2:\left(-13\right)^2=\dfrac{169^2}{13^2}=\left(\dfrac{169}{13}\right)^2=13^2=169\)
4: \(\left(\dfrac{1}{2}\right)^2:\left(\dfrac{3}{2}\right)^2=\left(\dfrac{1}{2}:\dfrac{3}{2}\right)^2=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
5: \(\left(\dfrac{2}{3}\right)^3:\left(\dfrac{8}{27}\right)^3=\left(\dfrac{2}{3}:\dfrac{8}{27}\right)^3=\left(\dfrac{2}{3}\cdot\dfrac{27}{8}\right)^3=\left(\dfrac{9}{4}\right)^3=\dfrac{729}{64}\)
6: \(\left(\dfrac{5}{4}\right)^4:\left(\dfrac{15}{2}\right)^4=\left(\dfrac{5}{4}:\dfrac{15}{2}\right)^4=\left(\dfrac{5}{4}\cdot\dfrac{2}{15}\right)^4=\left(\dfrac{1}{6}\right)^4=\dfrac{1}{1296}\)
7: \(\left(\dfrac{7}{8}\right)^5:\left(\dfrac{21}{16}\right)^5\)
\(=\left(\dfrac{7}{8}:\dfrac{21}{16}\right)^5\)
\(=\left(\dfrac{7}{8}\cdot\dfrac{16}{21}\right)^5=\left(\dfrac{2}{3}\right)^5=\dfrac{32}{243}\)
8: \(\left(\dfrac{5}{6}\right)^4:\left(\dfrac{25}{18}\right)^4=\left(\dfrac{5}{6}:\dfrac{25}{18}\right)^4=\left(\dfrac{5}{6}\cdot\dfrac{18}{25}\right)^4=\left(\dfrac{3}{5}\right)^4=\dfrac{81}{625}\)
9:
\(\left(-\dfrac{3}{4}\right)^3:\left(\dfrac{9}{8}\right)^3=\left(-\dfrac{3}{4}:\dfrac{9}{8}\right)^3=\left(-\dfrac{3}{4}\cdot\dfrac{8}{9}\right)^3\)
\(=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
10:
\(\left(\dfrac{9}{10}\right)^6:\left(\dfrac{27}{-20}\right)^6=\left(\dfrac{9}{10}:\dfrac{-27}{20}\right)^6\)
\(=\left(\dfrac{9}{10}\cdot\dfrac{20}{-27}\right)^6=\left(-\dfrac{2}{3}\right)^6=\dfrac{64}{729}\)
\(\frac{2}{5}x\) \(-\) \(x\) = \(\frac{7}{3}\)
\(-\frac{3}{5}x\) = \(\frac{7}{3}\)
\(x\) = \(\frac{7}{3}\) \(:\) \(-\frac{3}{5}\)
\(x\) = \(-\frac{35}{9}\)
- Hok T -
x.( 2/5-1)= 7/3
x. -3/5 = 7/3
x = 7/3: -3/5
x = -35/9