320-y:10=5x48:24
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\(320\div x-10=5\times48\div24\)
\(\Leftrightarrow320\div x-10=10\)
\(\Leftrightarrow320\div x=20\)
\(\Leftrightarrow x=16\)
hok tốt!
xy = 320 <=> y = 320/x
(x-16)(y+10)=320 <=> (x-16)((320/x)+10)=320 <=> 320+10x-160-(5120/x)=320 <=> 10x - (5120/x)=160 <=> 10x^2 - 160x-5120=160
=> x1=32 =>y1=10
x2=-16 =>y2=-20
10 % của 320 là 32
5% của 320 là 16
2,5% của 320 là 8
Vậy 17,5 của 320 là : 56
T i ck mk vs
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a, -720 + ( 143 + 720 )
= -720 + 143 + 720
= [ ( -720 ) + 720 ] + 143
= 0 + 143
= 143
b, ( 14 - 97 ) - 14
= 14 - 97 - 14
= ( 14 - 14 ) - 97
= 0 - 97
= -97
c) 10 - [ 12 - ( -9 - 1 ) ]
= 10 - 12 + 9 - 1
= -2 + 9 - 1
= 7 - 1
= 6
a: \(=3\left[\left(m-5\right)^3+8\right]\)
\(=3\left(m-5+2\right)\left[\left(m-5\right)^2-2\left(m-5\right)+4\right]\)
\(=3\left(m-3\right)\left(m^2-10m+25-2m+10+4\right)\)
\(=3\left(m-3\right)\left(m^2-12m+39\right)\)
b: \(=5\left(1-6k\right)^3-5\cdot64\)
\(=5\left(1-6k-4\right)\left[\left(1-6k\right)^2+4\left(1-6k\right)+16\right]\)
\(=5\left(-6k-3\right)\left(36k^2-12k+1+4-24k+16\right)\)
\(=-15\left(2k+1\right)\left(36k^2-36k+21\right)\)
\(=-45\left(2k+1\right)\left(12k^2-12k+7\right)\)
c: \(=\left(a+b-2b+a\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(2b-a\right)+\left(2b-a\right)^2\right]\)
\(=\left(2a-b\right)\left(a^2+2ab+b^2+2ab-a^2+2b^2-ab+4b^2-4ab+b^2\right)\)
\(=\left(2a-b\right)\cdot b\cdot\left(6b+a\right)\)
Ta có:
`x/10=y/5 -> x/20=y/10` `(1)`
`y/2=z/3 -> y/10=z/15` `(2)`
Từ `(1)` và `(2)`
`-> x/20=y/10=z/15` `-> x/20=y/10=(4z)/60`
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
`x/20=y/10=(4z)/60=(x+4z)/(20+60)=320/80=4`
`-> x/20=y/10=z/15=4`
`-> x=20*4=80, y=10*4=40, z=15*4=60`.
Ta có:
\(\left\{{}\begin{matrix}\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}\\\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\end{matrix}\right.\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}=\dfrac{x+4z}{20+4.15}=\dfrac{320}{80}=4\)
Do đó:
\(\dfrac{x}{20}=4\Rightarrow x=80\)
\(\dfrac{y}{10}=4\Rightarrow y=40\)
\(\dfrac{z}{15}=4\Rightarrow z=60\)
\(\dfrac{x}{10}\) = \(\dfrac{y}{5}\) ⇒ \(x\) = \(\dfrac{y}{5}\) \(\times\) 10 = 2y
\(\dfrac{y}{2}\) = \(\dfrac{z}{3}\) = ⇒ \(\dfrac{4y}{8}\) = \(\dfrac{4z}{12}\) ⇒ 4z = \(\dfrac{4y}{8}\) \(\times\) 12 = 6y
Theo bài rat ta có:
\(x+4z\) = 2y + 6y = 320 ⇒ 8y = 320 ⇒ y = 320: 8 =40
\(x\) = 40 \(\times\) 2 = 80
z = \(\dfrac{y}{2}\) \(\times\) 3 = \(\dfrac{40}{2}\) \(\times\) 3 = 60
Vậy \(x\) = 80; y = 40; z = 60
320-y:10=5x48:24
320-y:10=10
y:10=310
y=310x10=3100