24:(9+x)=2
tính bằng cách thuận tiện: 4/9 x 19/24 + 19/24 x 7/9 -19/24 x 2/9?
giúp mình với?
( 4/9 + 7/9 - 2/9 ) x 19/24 = 1 x 19/24 = 19/24
\(=\dfrac{19}{24}\times\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)=\dfrac{19}{24}\times1=\dfrac{19}{24}\)
\(\dfrac{19}{24}\times\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{19}{24}\times1=\dfrac{19}{24}\)
a) 24 : x = 24 : 2 b) 9 : x > 9 : 2
GIÚP MÌNH VỚI
a) 24:x=24:2
24:x=12
x=24:12
x=2
b)9:x>9:2
\(\dfrac{9}{x}>\dfrac{9}{2}\Rightarrow x=1\)
1/2/24 x 5/2/5 x 2 x 3/7/9 x 2 x 2/17
2/2/17 x 1/1/24 x 5/2/5 x 3/7/9 x 2
bạn nào giải xong bài này trước thì mình sẽ tích
1) 10/2/5 hay 52/5
2) 90
x*(x+2)+9*(x+24)
a.x/60=-3/4
b. 2/5=12/x
c. x-5/7 =6/21
dx+7/8 =63/24
e x/6=24/x
f .x+5= 4/x+5
g. x+12/x+8=4/3
h. x+9/x+7=9/8
a)\(\dfrac{x}{60}=-\dfrac{3}{4}\)
\(\Rightarrow x\cdot4=60\cdot\left(-3\right)\)
\(x\cdot4=-180\)
x=45
b)\(\dfrac{2}{5}=\dfrac{12}{x}\)
\(\Rightarrow2x=5\cdot12\)
\(2x=60\)
x=30
c)\(x-\dfrac{5}{7}=\dfrac{6}{21}\)
\(x=\dfrac{2}{7}+\dfrac{5}{7}\)
x=1
d)\(x+\dfrac{7}{8}=\dfrac{63}{24}\)
\(x=\dfrac{21}{8}-\dfrac{7}{8}\)
\(\dfrac{14}{8}\)
a)\(\dfrac{x}{60}=\dfrac{-3}{4}\Rightarrow x=\dfrac{-3.60}{4}=-45\)
b)\(\dfrac{2}{5}=\dfrac{12}{x}\Rightarrow x=\dfrac{5.12}{2}=30\)
Tính nhanh
1 )152 x 9 + 152
2) 24 x 999 + 24
1) \(152\times9+152\times1=152\times\left(9+1\right)=152\times10=1520\)
2) \(24\times999+24\times1=24\times\left(999+1\right)=24\times1000=24000\)
1/ 152 x 9 + 152 x 1 = 152 x ( 9 + 1 ) = 152 x 10 = 1520
2/ 24 x 999 + 24 x 1 = 24 x ( 999 + 1 ) = 24 x 1000 = 24000
9 -(x-3)^2 -24(x-3) = 0
`9 -(x-3)^2 -24(x-3) = 0`
`<=> (x-3) (9-24-x+3)=0`
`<=> (x-3) ( -12-x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-12-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-12\end{matrix}\right.\)
\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\)
tìm x : a) (x + 1)^3 + (3 - 2)^3 = 2x^3 + 2(2x - 1)^2 - 9
b) (3x^3+24) : (x+2) + (2x^3−54) : (x^2+3x+9) = 6
a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)