\(119\times24-53\times23-24\times66\)

\(=\left(119-66\right)\times24-53\times23\)

\(=53\times24-53\times23\)

\(=53\times\left(24-23\right)\)

\(=53\times1\)

\(=53\)

\(119\cdot24-53\cdot23-24\cdot66\)

\(=24\left(119-66\right)-53\cdot23\)

\(=53\cdot24-53\cdot23=53\)

''_''  là sao vậy ?

\(\left[184:\left(96-124:31\right)-2\right].3651\)

\(=\left[184:\left(96-4\right)-2\right].3651\)

\(=\left[184:92-2\right].3651\)

\(=\left[2-2\right].3651\)

\(=0.3651\)

`= 3651`

a: \(5^{56}=\left(5^7\right)^8=78125^8\)

\(11^{24}=\left(11^3\right)^8=1331^8\)

mà 78125>1331

nên \(5^{56}>11^{24}\)

b: \(5^{23}=5\cdot5^{22}< 6\cdot5^{22}\)

c: 

\(7\cdot2^{13}< 8\cdot2^{13}=2^3\cdot2^{13}=2^{16}\)

a: \(\dfrac{21}{25}\cdot\dfrac{11}{9}\cdot\dfrac{5}{7}=\dfrac{21}{7}\cdot\dfrac{5}{25}\cdot\dfrac{11}{9}\)

\(=3\cdot\dfrac{11}{9}\cdot\dfrac{1}{5}=\dfrac{11}{3}\cdot\dfrac{1}{5}=\dfrac{11}{15}\)

b: \(\dfrac{5}{23}\cdot\dfrac{17}{26}+\dfrac{5}{23}\cdot\dfrac{9}{26}\)

\(=\dfrac{5}{23}\left(\dfrac{17}{26}+\dfrac{9}{26}\right)\)

\(=\dfrac{5}{23}\cdot1=\dfrac{5}{23}\)

c: \(\dfrac{7}{13}\cdot\dfrac{5}{19}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\dfrac{7}{19}\)

\(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}\right)-3-\dfrac{7}{19}\)

\(=\dfrac{7}{19}-3-\dfrac{7}{19}=-3\)

\(a,\dfrac{21}{25}.\dfrac{11}{9}.\dfrac{5}{7}\)

\(=\dfrac{21.11.5}{25.9.7}\)

\(=\dfrac{1.11.1}{5.3.1}\)

\(=\dfrac{11}{15}\)

\(b,\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{9}{26}\)

\(=\dfrac{5}{23}.\left(\dfrac{17}{26}+\dfrac{9}{26}\right)\)

\(=\dfrac{5}{23}.1\)

\(=\dfrac{5}{23}\)

\(c,\dfrac{7}{13}.\dfrac{5}{19}+\dfrac{7}{19}.\dfrac{8}{13}\)

\(=\dfrac{7}{19}.\dfrac{5}{13}+\dfrac{7}{19}.\dfrac{8}{13}\)

\(=\dfrac{7}{19}.\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\)

\(=\dfrac{7}{19}.1\)

\(=\dfrac{7}{19}\)

(x - 45) x 27 = 0

=> x - 45 = 0 : 27 

=> x - 45 = 0 

=> x = 0 + 45

=> x = 45 

Vậy: .. 

`(x - 45) . 27 = 0`

`=> x - 45 = 0 : 27`

`=> x - 45 = 0`

`=> x = 45`

Vậy `x = 45`

\(5^7-5^6+5^5\\ =5^5\cdot5^2-5^5\cdot5+5^5\\ =5^5\cdot\left(5^2-5+1\right)\\ =5^5\cdot\left(25-5+1\right)\\ =5^5\cdot21⋮21\)

=> `5^7-5^6+5^5` chia hết cho 21 

`5^7 - 5^6 + 5^5`

`= 5^5 . (5^2 - 5 + 1)`

`= 5^5 . (25 - 5 + 1)`

`= 5^5 . 21 vdots 21 (đpcm)`

`2x + 3x + 1 - 4x + 2 = 36`

`=> (2+3-4) x + 3 = 36`

`=> x + 3 = 36`

`=> x = 36-3`

`=> x = 33`

Tìm x biết.                                                180-(x_45):2=120

\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55=7^3.7.11.5=5.77.7^3\)

Do 77 chia hết 77 \(\Rightarrow5.77.7^3⋮77\)

Vậy \(\left(7^6+7^5-7^4\right)⋮77\)

\(\left(3x-2^4\right)\cdot7^{13}=2\cdot7^{14}\)

=>\(3x-16=2\cdot\dfrac{7^{14}}{7^{13}}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\dfrac{30}{3}=10\)

\(\left(3x-2^4\right).7^{13}=2.7^{14}\\ \Rightarrow\left(3x-16\right).7^{13}=2.7.7^{13}\\ \Rightarrow3x-16=2.7\\ \Rightarrow3x-16=14\\ \Rightarrow3x=30\\ \Rightarrow x=30:3\\ \Rightarrow x=10\)

Vậy: \(x=10\)

Áp dụng t/c dãy tỉ số bằng nhau: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

Từ \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có: \(VT=\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{bk^3+b^3}{dk^3+d^3}=\dfrac{b.\left(k+1\right)^3}{d.\left(k+1\right)^3}=\dfrac{b}{d}\)

\(VP=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\dfrac{b.\left(k+1\right)^3}{d.\left(k+1\right)^3}=\dfrac{b}{d}\)

Vậy \(VT=VP\left(đpcm\right)\)

____________

VT = vế trái

VP = vế phải

\(#NqHahh\)