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a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
`@` `\text {Ans}`
`\downarrow`
`c)`
`( 34 - 2x ) . ( 2x - 6 ) = 0`
`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
Vậy, `x \in {17; 3}`
`d)`
`( 2019 - x ) . ( 3x - 12 ) =0` `?`
`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
Vậy, `x \in {2019; 4}`
`e) `
`57 . ( 9x - 27 ) = 0`
`=>`\(9x-27=0\div57\)
`=> 9x - 27 = 0`
`=> 9x = 27`
`=> x = 27 \div 9`
`=> x = 3`
Vậy, `x = 3`
`f)`
`25 + ( 15 - x ) = 30`
`=> 15 - x = 30 - 25`
`=> 15 - x = 5`
`=> x = 15 -5 `
`=> x = 10`
Vậy, `x = 10`
`g) `
`43 - ( 24 - x ) = 20`
`=> 24 - x = 43 - 20`
`=> 24 - x = 23`
`=> x = 24 - 23`
`=> x = 1`
Vậy, `x = 1`
`h) `
`2 . ( x - 5 ) - 17 = 25`
`=> 2 ( x - 5) = 25+17`
`=> 2 ( x - 5) = 42`
`=> x - 5 = 42 \div 2`
`=> x - 5 = 21`
`=> x = 21 + 5`
`=> x = 26`
Vậy, `x = 26`
`i)`
`3 . ( x + 7 ) - 15 = 27`
`=> 3(x + 7) = 27 + 15`
`=> 3(x + 7) = 42`
`=> x +7 = 42 \div 3`
`=> x + 7 = 14`
`=> x = 14 - 7`
`=> x = 7`
Vậy, `x = 7`
`j)`
`15 + 4 . ( x - 2 ) = 95`
`=> 4(x - 2) = 95 - 15`
`=> 4(x - 2) = 80`
`=> x - 2 = 80 \div 4`
`=> x - 2 = 20`
`=> x = 20 + 2`
`=> x = 22`
Vậy, `x = 22`
`k)`
`20 - ( x + 14 ) = 5`
`=> x + 14 = 20 - 5`
`=> x + 14 = 15`
`=> x = 15 - 14`
`=> x = 1`
Vậy, `x = 1`
`l) `
`14 + 3 . ( 5 - x ) = 27`
`=> 3(5 - x) = 27 - 14`
`=> 3(5 - x) = 13`
`=> 5 - x = 13 \div 3`
`=> 5 - x = 13/3`
`=> x = 5- 13/3`
`=> x = 2/3`
Vậy, `x = 2/3.`
`@` `\text {Kaizuu lv uuu}`
[545 - (45 + 4 x 25)] : 50 - 200 : 250 + 215 : 213
= [545 - (45 + 100)] : 50 - 200 : 250 + 22
= (545 - 4500) : 50 - 200 : 254
= (-3955) : 50 + 200 : 254
= [(-3955) + 200] : (50 + 254)
= (-3755) : 304
= bạn tự tính nha
[504 - (25 x 8 + 70)] : 9 - 15 + 190
= [504 - (200 + 70)] : 9 - 15 + 0
= (504 - 270) : 9 - 15
= 234 : 9 - 15
= 26 - 15
= 9
5 x {26 - [3 x 5 + 2 x 5 + 15] : 15}
= 5 x {26 - [(3 + 2) x 5) + 15] : 15}
= 5 x {26 - [5 x 5 + 15] : 15}
= 5 x [26 - (25 + 15) : 15]
= 5 x (26 + 25 - 15 : 15)
= 5 x (51 - 1)
= 5 x 50 = 250
d) [1104 - (25 x 8 + 40)] : 9 + 316 : 312
= [1104 - (200 + 40)] : 9 + 34
= (1104 - 240) : 9 + 81
= 864 : 9 + 81
= 96 + 81
= 177
Học tốt!!!
a) 35-{12-[-14+(-2)]}.
= 35 - 12 + 14 + 2
= 35 + 14 - (12 - 2)
= 49 - 10
= 39
b) 160:{17+[9×5-(14+2048:256)]}.
= 160: {17 + [45 - (14 + 8)]}
= 160: {17 + [45 - 22)]}
= 160:{17 + 23}
= 160: 40 = 4
c) 20210-{225:[20×15-8×25]-25}.
= 1 - {225:[300 - 200] - 25}
= 1 - {225 : 100 - 25}
= 1 - {2,25 - 25}
= 1 + 25 - 2,25
= 23,75
\(x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Lời giải:
a. $2^{50}.5^{45}=2^{45}.2^5.5^{45}=(2.5)^{45}.32$
$=32.10^{45}$
Số trên có tận cùng gồm $45$ chữ số $0$
b.
$4^{12}.25^{15}=(2^2)^{12}.(5^2)^{15}=2^{24}.5^{30}$
$=(2.5)^{24}.5^6=15625.10^{24}$ có tận cùng $24$ chữ số $0$
a) \(25-\left(25-x\right)=12+\left(42-65\right)\)
\(25-\left(25-x\right)=12+\left(-23\right)\)
\(25-\left(25-x\right)=-11\)
\(25-x=25-\left(-11\right)\)
\(25-x=36\)
\(x=25-36\)
\(x=-11\)
Làm 2 câu dài cho bạn nhá! Mấy câu ngắn tương tự!
a, \(25-\left(25-x\right)=12+\left(42-65\right)\)
\(\Rightarrow25-25+x=12+42-65\)
\(\Rightarrow x=-11\)
b, \(215+\left(-38\right)-54+90-\left(-48+x\right)=0\)
\(\Rightarrow215-38-54+90+48-x=0\)
\(\Rightarrow x=261\)
m, \(\left|x\right|-5=12\)
\(\Rightarrow\left|x\right|=17\Rightarrow\left\{{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
Chúc bạn học tốt!!!
Đáp án là C
Ta có: 1 0 = 60' ⇒ 0 , 25 0 = 0,25.60 = 15'
⇒ 15,25° = 15°15'