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\(\Rightarrow\left[\begin{array}{nghiempt}x-9=15k\\y-12=20k\\z-24=40k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=15k+9\\y=20k+12\\z=40k+24\end{array}\right.}\)
ta có:
x.y=1200\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\Rightarrow\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}=k\)
=> (15k+9)(20k+12)=1200
=> 3.4(5k+3)(5k+3)=1200
=> (5k+3)2=100
=> 5k+3=\(\pm\)10
=> \(\left[\begin{array}{nghiempt}5k+3=10\\5k+3=-10\end{cases}\Rightarrow\left[\begin{array}{nghiempt}5k=7\\5k=-13\end{cases}\Rightarrow}\left[\begin{array}{nghiempt}k=\frac{7}{5}\\k=-\frac{13}{5}\end{array}\right.}\)
* với k=7/5
x=7/5x15+9=30
y=7/5x20+12=40
z=7/5x40+24=80
* với k=-13/5
x=-13/5x15+9=-30
y=-13/5x20+12=-40
z=-13/5x40+24=-80
b)
\(\frac{40}{x-30}=\frac{20}{y-50}=\frac{28}{z-21}\Rightarrow\frac{x-30}{40}=\frac{y-50}{20}=\frac{z-21}{28}k=\)
=>\(\left[\begin{array}{nghiempt}x-30=40k\\y-50=20k\\z-21=28k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=40k+30\\y=20k+50\\z=28k+21\end{array}\right.}\)
ta có:
x.y.z=22400
=> (40k+30)(20k+50)(28k+21)=22400
c) 15x=-10y=6z
\(\Rightarrow\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\Rightarrow\frac{x}{2}=-\frac{y}{3}=\frac{z}{5}=k\)
=> \(\left[\begin{array}{nghiempt}x=2k\\y=-3k\\z=5k\end{array}\right.\)
ta có:
x.y.z=30000
=> 2k.(-3k).5k=30000
=> k3=1000
=> k=10
ta có: x=10x2=20
y=10.(-3)=-30
z=10.5=50

\(\dfrac{3x-40}{50}+\dfrac{3x-10+2x+60+4x-360}{40}=0\)
=> \(\dfrac{3x-40}{50}+\dfrac{9x-310}{40}=0\)
=> \(\dfrac{3x-40}{50}=\dfrac{-9x+310}{40}\)
=> \(40\left(3x-40\right)=50\left(-9x+310\right)\)
=> \(120x-1600=-450x+15500\)
=> \(120x+450x=15500+1600\)
Hay \(570x=17100\)
=>x = 30
Hơi dài nhé bạn
\(\dfrac{3x-40}{50}\)+\(\dfrac{3x-10+2x+60+4x-360}{40}\)=0
⇒\(\dfrac{3x-40}{50}\)+\(\dfrac{9x-310}{40}\)=0
⇒\(\dfrac{3x-40}{50}\)=\(\dfrac{9x-310}{40}\)
⇒40(3x -40) = 50(-9x+310)
⇒120x - 1600 = -450x + 15500
⇒120x + 450x = 15500 + 1600
Mặt khác: 570x = 17100
⇒x = 30

c) Ta có: 450 = (22)50 = 2100
mà 2100 < 2101
Vậy 450 < 2101
e) Ta có: Mọi số thực âm có số mũ lẻ sẽ vẫn giữ nguyên dấu của nó
Vậy 277 > -815

a) \(2^{100}=\left(2^2\right)^{50}\)
\(2^2=4< 5\)
\(2^{100}< 5^{50}\)
b) \(4^{30}=\left(4^3\right)^{10}\)
\(4^3=8^2\)
\(4^{30}=8^{20}\)
\(8^{20}=\left(8^2\right)^{10}\)

Kêu người ta giúp mà ói vào mặt người ta vậy à?


a/ Ta có: 2100 = 22.50 = (22)50 = 450
550 = 550
Vì 4 < 5 nên 450 < 550
Vậy 2100 < 550
b/ Ta có: 430 = (22)30 = 260
820 = (23)20 = 260
Vì 260 = 260
Nên 430 = 820
\(a.\:2^{100}=\left(2^2\right)^{50}=4^{50}< 5^{50}\)
\(b.\:\left\{{}\begin{matrix}4^{30}=64^{10}\\8^{20}=64^{10}\end{matrix}\right.\Rightarrow4^{30}=8^{20}\)

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
= 120!
bạn phải viết rõ 120!=?