1:

a: A=(x+4)^3=10^3=1000

b: B=(x-2)^3=20^3=8000

1

a) \(A=x^3+3.x^2.4+3x.4^2+4^3=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

b) \(B=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

2

\(VT=\left(x-y\right)^2+4xy=x^2-2xy+y^2+4xy=x^2+2xy+y^2=\left(x+y\right)^2=VP\)

a) 7,89

b) 2398 

c) 58,85 

d) 14 

\(\text{a. 78,9 : 2,5 : 4. }\)

\(=31,56:4=7,89.\)

\(\text{ b. 56,75 + 8,4 : 4}.\)

\(=56,75+2,1=58,85.\)

\(\text{c. 23,6 x 100 + 2,9 : 0,1. }\)

\(=2360+29=2389.\)

\(\text{ d. 58 - 4,4 : 0,1}.\)

\(=58-44=14.\)

a) 7,89

b) 2398 

c) 58,85 

d) 14 

a) Với a = 3,05 thì ta có:

\(a\times2,46\) \(=3,05\times2,46=7,503\)

b) Với \(a=\dfrac{15}{8}\) thì ta có:

\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\) \(=\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\) \(=\left(\dfrac{10}{12}+\dfrac{7}{12}\right)\times\dfrac{8}{15}\) \(=\dfrac{17}{12}\times\dfrac{8}{15}\) \(=\dfrac{34}{45}\)

a) Thay a=3,05 vào 2,46a, ta được:

\(2.46\cdot3.05=7.503\)

b) Thay \(a=\dfrac{15}{8}\) vào biểu thức \(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):a\), ta được:

\(\left(\dfrac{5}{6}+\dfrac{7}{12}\right):\dfrac{15}{8}\)

\(=\dfrac{17}{12}\cdot\dfrac{8}{15}=\dfrac{34}{45}\)

|(5/8-5/14)+(3/8-9/14)|:4/7

=|(5/8+3/8)+(-5/14-9/14)|:4/7

=|1+(-1)|:4/7

=0

`A = ( 5 / 8 - 5 / 14 ) : 4 / 7 + ( 3 / 8 - 9 / 14 ) : 4 / 7`

`A = ( 5 / 8 - 5 / 14 + 3 / 8 - 9 / 14 ) : 4 / 7`

`A = [ ( 5 / 8 + 3 / 8 ) - ( 5 / 14 + 9 / 14 ) ] . 7 / 4`

`A = ( 8 / 8 - 14 / 14 ) . 7 / 4`

`A = ( 1 - 1 ) . 7 / 4 = 0 . 7 / 4 = 0`

\(A=\left(\dfrac{5}{8}-\dfrac{5}{14}\right)\times\dfrac{7}{4}+\left(\dfrac{3}{8}-\dfrac{9}{14}\right)\times\dfrac{7}{4}\)

\(A=\left(\dfrac{5}{8}-\dfrac{5}{14}+\dfrac{3}{8}-\dfrac{9}{14}\right)\times\dfrac{7}{4}\)

\(A=\left(1-1\right)\times\dfrac{7}{4}=0\)

a, Thay x = 3 và y = -6 vào bt ta đc

\(5.3-4.\left(-6\right)=15-\left(-24\right)=39\\ b,\\ 2.\left(-2\right)^2-5.4=8-20=\left(-12\right)\\ c,\\ 5.\left(-1\right)^2+3.\left(-1\right)-1=5+\left(-3\right)-1=1\)

a) Thay x=3; y=-6

\(5x-4y=5.3-4.\left(-6\right)=15+24=39\)

b) Thay x=-2; y=4

\(2x^4-5y=2.\left(-2\right)^4-5.4=32-20=12\)

c, Thay x=0

\(5x^2+3x-1=5.0+3.0-1=-1\)

+) x=-1

\(5x^2+3x-1=5.\left(-1\right)^2+3.\left(-1\right)-1=5-3-1=1\)

+) \(x=\dfrac{1}{3}\)

\(5x^2+3x-1=5.\left(\dfrac{1}{3}\right)^2+3.\dfrac{1}{3}-1\)

\(=\dfrac{5}{9}+1-1=\dfrac{5}{9}\)

\(a^2-2a+6b+b^2=-10\\ \Leftrightarrow a^2-2a+1+b^2+6b+9=0\\ \Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-3\end{matrix}\right.\)

Vậy \(\left(a;b\right)=\left(1;-3\right)\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Leftrightarrow xy+yz+zx=0\\ \Rightarrow\left\{{}\begin{matrix}xy+yz=-zx\\xy+zx=-yz\\yz+zx=-xy\end{matrix}\right.\)

Ta có: 

\(A=\dfrac{xz+yz}{z^2}+\dfrac{xy+yz}{y^2}+\dfrac{xy+xz}{x^2}\\ =\dfrac{-xy}{z^2}+\dfrac{-xz}{y^2}+\dfrac{-yz}{x^2}\\ =-xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\\ =-xyz\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}\right)\\ =0\)

\(a,256+789:3-178\)

\(=256+263-178\)

\(=341\)

\(b,1485+\left(735-47\cdot8\right)\)

\(=1485+\left(735-376\right)\)

\(=1485+359\)

\(=1844\)

\(c,5635:5-8172:9\)

\(=1127-908\)

\(=219\)

a) \(256+789:3-178\)

\(=256+263-178\)

\(=519-178\)

\(=341\)

b) \(1485+\left(735-47\times8\right)\)

\(=1485+\left(735-376\right)\)

\(=1485+359\)

\(=1844\)

`A=sqrt{8+2sqrt7}-sqrt{8-2sqrt7}`

`=sqrt{7+2sqrt7+1}-sqrt{7-2sqrt7+1}`

`=sqrt{(sqrt7+1)^2}-sqrt{(sqrt7-1)^2}`

`=sqrt7+1-sqrt7+1=2`

`B=sqrt{11-6sqrt2}+sqrt{6-4sqrt2}`

`=sqrt{9-2.3.sqrt2+2}+sqrt{4-2.2.sqrt2+2}`

`=sqrt{(3-sqrt2)^2}+sqrt{(2-sqrt2)^2}`

`=3-sqrt2+2-sqrt2=5-2sqrt2`

a) 78 060 : (10 – 7) + 300 045

= 78 060 : 3 + 300 045

= 26 020 + 300 045

= 326 065

b) 26 000 + 9 015 × 6

= 26 000 + 54 090

= 80 090