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a, 36:{336:[200–(12+8.20)]}
= 36:{336:[200–(12+160)]}
= 36:{336:[200–172]}
= 36:{336:28}
= 36:12 = 3
b, {145–[130–(246–236)]:2}.5
= {145–[130–10:2]}.5
= {145–130}.5
= 20.5 = 100
c, 100:{250:[450–(4. 5 3 – 2 2 .25]}
= 100:{250:[450–400]}
= 100:{250:50}
= 100:5 = 20
d, 798+100:[16–2.( 5 2 –22)]
= 798+100:10
= 798+10 = 808
e, (6954+1525:5+47.19).(29–58.2)
= (6954+1525:5+47.19).0 = 0
f, 2 4 .157– 2 4 .58+16
= 16.(157–58+1) = 1600
a,70-5(x-3)=45
5(x-3) = 70 - 45
5(x-3) = 25
x-3 = 25 : 5
x - 3 = 5
x = 5 + 3
x = 8.
b,12+(5+x)=20
(5+x) = 20 - 12
(5+x) = 8
x = 8 - 5
x = 3
c,130-(100 + x)=25
100 + x = 130 - 25
100 + x = 105
x = 105 - 100
x = 5
Hok tốt !
a,70-5(x-3)=45
5(x-3) =70-45
5(x-3) =25
(x-3) =25:5
x-3 =5
x=5+3
x=8
a) \(70-5\left(x-3\right)=45\)
\(5\left(x-3\right)=25\)
\(x-3=5\)
\(\Rightarrow x=8\)
b) \(12+\left(5+x\right)=20\)
\(5+x=8\)
\(\Rightarrow x=3\)
c) \(130-\left(100+x\right)=25\)
\(100+x=105\)
\(\Rightarrow x=5\)
d) \(175+\left(30-x\right)=200\)
\(30-x=25\)
\(\Rightarrow x=5\)
e) \(5\left(x+12\right)+22=92\)
\(5\left(x+12\right)=70\)
\(x+12=14\)
\(\Rightarrow x=2\)
a. 70 - 5.(x - 3) = 45
5.(x - 3) = 70 - 45
x - 3 = 25 : 5
x = 5 + 3
x = 8
b. 12 + (5 + x) = 20
5 + x = 20 - 12
x = 8 - 5
x = 3
c. 130 - (100 + x) = 25
100 + x = 130 - 25
x = 105 - 100
x = 5
d. 175 + (30 - x) = 200
30 - x = 200 - 175
x = 30 - 25
x = 5
e. 5.(x + 12) + 22 = 92
5.(x + 12) = 92 - 22
x + 12 = 70 : 5
x = 14 - 12
x = 2
\(100-3x=-130\)
\(3x=100-\left(-130\right)\)
\(3x=230\)
\(x=230:3\)
\(x=\dfrac{230}{3}\)
Vậy .....
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
chia 2 dư 1-1,3,5,7,9
chia 5 dư 1-1,6
số đó có thể là 121.
121 bạn nhé
t ick mình nhé
Gọi số cần tìm là x(x ϵ N*), theo đề bài, ta có:
\(x⋮12\\ x⋮15\\ 100\le x\le150\) \(\Rightarrow x\in BC\left(12,15\right)\)
Ta có:
\(12=2^2.3\)
\(15=3.5\)
\(\Rightarrow BCNN\left(12,15\right)=2^2.3.5=60\)
\(\Rightarrow B\left(60\right)=\left\{0;60;120;180;240;300;...\right\}\) Mà \(100\le x\le150\)
\(\Rightarrow x=120\)
Vậy số cần tìm là 120.
\(30\cdot\left(x+2\right)-6\cdot\left(x-5\right)-24\cdot x=100\)
\(\Leftrightarrow30x+60-6x+30-24x=100\)
\(\Leftrightarrow0x+90=100\)
=> PT vô nghiệm
b) \(|7x+3|=66\)
\(\Leftrightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=63\\7x=-69\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=\frac{-69}{7}\end{cases}}}\)
\(\text{30.(x+2)-6.(x-5)-24.x=100}\)
\(\Leftrightarrow30x+60-6x+30-24x=100\)
\(\Leftrightarrow90=100\)(vô lí)
Vậy phương trình vô nghiệm
\(|7x+3|=66\)
TH1 \(7x+3=66\Leftrightarrow x=\frac{66-3}{7}=9\)
TH2 \(7x+3=-66\Leftrightarrow x=\frac{-66-3}{7}=-\frac{69}{7}\)
\(130-\left(100-7x\right)=-5\)
\(\left(100-7x\right)=130-\left(-5\right)=135\)
\(7x\) \(=100-135=-35\)
\(x\) \(=\left(-35\right):7=-5\)
\(\text{Hok tốt!}\)
@FF