2^3 : 8 x ( 12^3 - 3^3 x 2^6 + 1284 )
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- Tính giá trị biểu thức:
a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145
b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7
c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11
- Tính giá trị biểu thức:
a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21
b) (5/17 x 21/32 x 47/24 x 0)
= 0
c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7
- Tìm x:
a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x
b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15
c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2
- Tìm x:
a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36
câu,b,không,đủ,thông,tin,nhan,bạn.

a)Để biểu thức vô nghĩa thì \(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1\right\}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne1\end{matrix}\right.\Leftrightarrow x\notin\left\{-2;1\right\}\)
b) Ta có: \(\dfrac{5x-2}{12}-\dfrac{2x^2+1}{8}=\dfrac{x-3}{6}+\dfrac{1-x^2}{4}\)
\(\Leftrightarrow\dfrac{2\left(5x-2\right)}{24}-\dfrac{3\left(2x^2+1\right)}{24}=\dfrac{4\left(x-3\right)}{24}+\dfrac{6\left(1-x^2\right)}{24}\)
\(\Leftrightarrow10x-4-6x^2-3=4x-12+6-6x^2\)
\(\Leftrightarrow-6x^2+10x-7+6x^2-4x+6=0\)
\(\Leftrightarrow6x-1=0\)
\(\Leftrightarrow6x=1\)
\(\Leftrightarrow x=\dfrac{1}{6}\)
Vậy: \(S=\left\{\dfrac{1}{6}\right\}\)

\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)
\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)

Lời giải:
$x=4$ thì $\frac{x}{2}=2=y$
$\Rightarrow y-\frac{x}{2}=0$
Do đó:
$(\frac{x}{2}-y^3)^3-6(y-\frac{x}{2})^2-12(y-\frac{x}{2})-8$
$=(\frac{x}{2}-y^3)^3-8=(2-2^3)^3-8=-224$

\(\left(3x-6y\right)\left(x^2+2xy+4y^2\right)-3\left(x^3-8y^3+12\right)\)
\(=3\left(x-2y\right)\left(x^2+2xy+4y^2\right)-3\left(x^3-8y^3+12\right)\)
\(=3\left(x^3-8y^3\right)-3\left(x^3-8y^3+12\right)\)
=-36

a | b | c | a x (b - c) | a x b - a x c |
---|---|---|---|---|
3 | 7 | 3 | 3 x (7 - 3) = 12 | 3 x 7 - 3 x 3 = 12 |
6 | 9 | 5 | 6 x (9 - 5) = 24 | 6 x 9 - 6 x 5 = 24 |
8 | 5 | 2 | 8 x (5 - 2) = 24 | 8 x 5 - 8 x 2 = 24 |

a | b | c | a x (b - c) | a x b - a x c |
---|---|---|---|---|
3 | 7 | 3 | 3 x (7 - 3) = 12 | 3 x 7 - 3 x 3 = 12 |
6 | 9 | 5 | 6 x (9 - 5) = 24 | 6 x 9 - 6 x 5 = 24 |
8 | 5 | 2 | 8 x (5 - 2) = 24 | 8 x 5 - 8 x 2 = 24 |
Tính giá trị biểu thức a x 12 + 2,1 x b + 5 với a = 19,36 và b = 7,9 ?
Đáp số:

\(a,\dfrac{4}{5}+\dfrac{7}{10}=\dfrac{8}{10}+\dfrac{7}{10}=\dfrac{15}{10}=\dfrac{3}{2}\\ b,\dfrac{7}{8}\times3+\dfrac{5}{12}=\dfrac{21}{8}+\dfrac{5}{12}=\dfrac{21\times3+5\times2}{24}=\dfrac{73}{24}\\ c,\dfrac{6}{9}:\left(\dfrac{2}{3}:3\right)=\dfrac{2}{3}:\dfrac{2}{3}\times3=1\times3=3\)
`2^3:8xx(12^3-3^3xx2^6+1284)`
`=8:8xx[12^3-3^3xx(2^2)^3+1284)`
`=8:8xx[12^3-(3xx2^2)^3+1284]`
`=8:8xx[12^3-(3xx4)^3+1284]`
`=8:8xx(12^3-12^3+1284)`
`=8:8xx(0+1284)`
`=8:8xx1284`
`=1xx1284`
`=1284`
Vậy: `...`
2^3 : 8 x (12^3 - 3^3 x 2^6 + 1284)
= 1 x (1728 - 1728 + 1284)
= 1284