75 . ( 3x - 23 ) = 74 . 72
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\(A=\dfrac{7^5}{7+7^2+7^3+7^4}=\dfrac{7^5}{\left(7+7^4\right)+\left(7^2+7^3\right)}=\dfrac{7^5}{7^5+7^5}=7^5\)
\(B=\dfrac{5^5}{5+5^2+5^3+5^4}=\dfrac{5^5}{\left(5+5^4\right)+\left(5^2+5^3\right)}=\dfrac{5^5}{5^5+5^5}=5^5\)
Vì 7 > 5 nên \(7^5>5^5\)
Vậy A > B
(Nhớ cho mik một tick nha cảm ơn bạn nhìu :3)
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45 x 63 = 2835
23 x 74 = 1702
36 x 29 = 1044
75 x 3 = 225
Ủng hộ mik nha
45 x 63 =2835
23 x 74 = 1702
36 x 29 = 1044
75 x 3 = 225
Đó là đáp án đúng
Chúc bạn học giỏi và gặp nhiều may mắn
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b) ( 3x – 2 3 ) . 7 = 7 4
3x – 8 = 7 4 : 7
3x – 8 = 7 3
3x – 8 = 343
3x = 343 + 8
3x = 351
x = 351 : 3 = 117
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71 + 72 + 73 + 74 + 75 + 76 + 77 + 78 + 79
= ( 71 + 79 ) + ( 72 + 78 ) + ( 73 + 77 ) + ( 74 + 76 ) + 75
= 150 + 150 + 150 + 150 + 75
= 150 x 4 + 75
= 600 + 75
= 675
Tìm x :
X x 7 + X x 2 = 108
X x ( 7 + 2 ) = 108
X x 9 = 108
X = 108 : 9
X = 12
\(71+72+73+74+75+76+77+78+79\)
\(=\left(71+79\right)+\left(72+78\right)+\left(73+77\right)+\left(74+76\right)+75\)
\(=150+150+150+150+75\)
\(=600+75\)
\(=675\)
\(X\)x\(7+X\)x\(2=108\)
\(X\)x\(\left(7+2\right)=108\)
\(X\)x\(9=108\)
\(X=108:9\)
\(X=12\)
Vậy \(X=12\)
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Sửa đề: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)Ta có: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
mà \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}>0\)
nên 100-x=0
hay x=100
Vậy: S={100}
Ta có : \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}+5=0\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
Thấy : \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\ne0\)
\(\Rightarrow100-x=0\)
\(\Leftrightarrow x=100\)
Vậy ...
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\(\left(-67\right)+125+\left(-33\right)+75\)
\(=\left[\left(-67\right)+\left(-33\right)\right]+\left(125+75\right)\)
\(=100+200=300\)
_______
\(86.\left(-108\right)+86.9-86\)
\(=86.\left[\left(-108\right)+9-1\right]\)
\(=86.\left(-100\right)=-8600\)
_______
\(23.\left(-16\right)-23.84+300\)
\(=23.\left[\left(-16\right)-84\right]+300\)
\(=23.\left(-100\right)+300\)
\(=-2300+300\)
\(=-2000\)
______
\(235-5\left[\left(5^3-3^3\right):14\right]\)
\(=235-5\left[\left(125-27\right):14\right]\)
\(=235-5\left[98:14\right]\)
\(=235-5.7\)
\(=235-35\)
\(=200\)
_______
\(95-\left(129-74\right):5+2022^0\)
\(=95-55:5+1\)
\(=95-11+1\)
\(=84+1=85\)
\(#NqHahh\)
75.(3x - 23) = 74.72
=> 75.(3x - 8) = 76
=> 3x - 8 = 7
=> 3x = 15
=> x = 5
Vậy x = 5
\(7^5.\left(3x-8\right)=7^6\)
\(3x-8=7\)
\(3x=15\)
\(x=5\)