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Tìm xx biết: \left(x^{4}\right)^{3}=\dfrac{x^{19}}{x^{6}}(x4)3=x6x19
Trả lời: x=x=
a, (-4) . (+125) . (-25) . 6 . (-8)
=[ (-4) . (-25) ] . [ (+125) . (-8) ] . 6
= 100 . (-1000) . 6
= -100 000 . 6
= -600 000
b, 122 . (-12345) + 12345 . (-78)
= 12345 . [ (-122) + (-78) ]
= 12345 . (-200)
= -2 469 000
`@` `\text {Ans}`
`\downarrow`
`a)`
`x + 10 = 20`
`=> x = 20 -10`
`=> x = 10`
Vậy, `x = 10`
`b)`
`2 * x + 15 = 35`
`=> 2x = 35 - 15`
`=> 2x = 20`
`=> x = 20 \div 2`
`=> x = 10`
Vậy, `x = 10`
`c)`
`3 * ( x + 2 ) = 15`
`=> x + 2 = 15 \div 3`
`=> x + 2 = 5`
`=> x = 5 - 2`
`=> x = 3`
Vậy, `x = 3`
`d)`
`10 * x + 15 * 11 = 20 * 10`
`=> 10x + 165 = 200`
`=> 10x = 200 - 165`
`=> 10x = 35`
`=> x = 35 \div 10`
`=> x = 3,5`
Vậy,` x = 3,5`
`e)`
`4 * ( x + 2 ) = 3 * 4`
`=> x + 2 = 12 \div 4`
`=> x + 2 = 3`
`=> x = 3 - 2`
`=> x = 1`
Vậy,` x = 1`
`f)`
`33 x + 135 = 26 * 9`
`=> 33x + 135 = 234`
`=> 33x = 234 - 135`
`=> 33x = 99`
`=> x = 99 \div 33`
`=> x = 3`
Vậy, `x = 3`
`g)`
`2 * x + 15 + 16 + 17 = 100`
`=> 2x + 48 = 100`
`=> 2x = 100 - 48`
`=> 2x = 52`
`=> x = 52 \div 2`
`=> x =26`
`h)`
`2 * (x + 9 + 10 + 11) = 4 . 12 . 25`
`=> 2 * (x + 9 + 10 + 11) = 4*25*12`
`=> 2 * (x + 9 + 10 + 11) = 100*12`
`=> x + 9 + 10 + 11 = 100*12 \div 2`
`=> x + 30 = 600`
`=> x = 600 - 30`
`=> x = 570`
Vậy, `x = 570.`
a) \(x+10=20\Leftrightarrow x=10\)
b) \(2x+15=35\Leftrightarrow2x=20\Leftrightarrow x=10\)
c) \(3.\left(x+2\right)=15\Leftrightarrow x+2=5\Leftrightarrow x=3\)
d) \(10x+15.11=20.10\Leftrightarrow10x+165=200\Leftrightarrow10x=35\Leftrightarrow x=\dfrac{35}{10}=\dfrac{7}{2}\)
e) \(4.\left(x+2\right)=3.4\Leftrightarrow x+2=3\Leftrightarrow x=1\)
f) \(35x+135=26.9\Leftrightarrow35x=234-135\Leftrightarrow35x=99\Leftrightarrow x=\dfrac{99}{35}\)
g) \(2x+15+16+17=100\Leftrightarrow2x+48=100\Leftrightarrow2x=52\Leftrightarrow x=26\)
h) \(2.\left(x+9+10+11\right)=4.12.25\)
\(\Leftrightarrow x+30=2.12.25\)
\(\Leftrightarrow x=600-30\)
\(\Leftrightarrow x=570\)
a: 2x(x+1)-135=-200
=>2(x^2+x)=-65
=>2x^2+2x+65=0
=>x^2+x+32,5=0
=>x^2+x+0,25+32,25=0
=>(x+0,5)^2+32,25=0(vô lý)
b: 4x-5(x-1)+15=13
=>4x-5x+5=-2
=>5-x=-2
=>x=5+2=7
c: 2/3x-1/4=3/5-7/8
=>2/3x=3/5-7/8+1/4=24/40-35/40+10/40=-1/40
=>x=-1/40:2/3=-1/40*3/2=-3/80
d: 1/2(2x-3)+105/2=-137/2
=>1/2(2x-3)=-137/2-105/2=-242/2=-121
=>2x-3=-242
=>2x=-239
=>x=-239/2
a/ 2130-(x+130)+72=-64
\(2130-\left(x+130\right)=-64-72\) \(2130-\left(x+130\right)=-136\)\(x+130=2130+136\)\(x+130=2266\)\(x=2266-130\)\(x=2136\)vậy x=2136
a/ 2130-(x+130)+72=-64
<=> -x-130+72+2130=-64
<=> x=2130+72-130+64=2136
b/135-|9-x|=35
<=> |9-x|=135-35=100
<=> \(\left[\begin{array}{nghiempt}9-x=100\\9-x=-100\end{array}\right.\)
<=>\(\left[\begin{array}{nghiempt}x=-91\\x=109\end{array}\right.\)
c/10-|x+3|=-4-(-10)
<=> |x+3|=10+4+(-10)
<=>|x+3|=4
<=>\(\left[\begin{array}{nghiempt}x+3=4\\x+3=-4\end{array}\right.\)
<=>\(\left[\begin{array}{nghiempt}x=1\\x=-7\end{array}\right.\)
a) 135 - 5 (x + 4) = 35
5(x+4)=135-35
5(x+4)=100
x+4=100:5
x+4=20
x=20-4
x=16
b)25 + 3 (x - 8) = 106
3 (x - 8)=106-25
3 (x - 8)=81
x-8=81:3
x-8=27
x=27+8
x=35
a,5(x+4)=135-35
5(x+4)=100
x+4=100:5
x+4=20
x=20-4
x=16
b,3(x-8)=106-25
3(x-8)=81
x-8=81:3
x-8=27
x=27+8
x=35
\(135-5\cdot\left(x+4\right)=35\)
\(5\cdot\left(x+4\right)=135-35\)
\(5\cdot\left(x+4\right)=100\)
\(x+4=100:5\)
\(x+4=20\)
\(x=20-4\)
\(x=16\)
b) \(25+3\cdot\left(x-8\right)=106\)
\(3\cdot\left(x-8\right)=106-25\)
\(3\cdot\left(x-8\right)=81\)
\(x-8=81:3\)
\(x-8=27\)
\(x=27+8\)
\(x=35\)
c) \(3^2\cdot\left(x+4\right)-5^2=5\cdot2^2\)
\(9\cdot\left(x+4\right)-25=5\cdot4\)
\(9\cdot\left(x+4\right)-25=20\)
\(9\cdot\left(x+4\right)=20+25\)
\(9\cdot\left(x+4\right)=45\)
\(x+4=45:9\)
\(x+4=5\)
\(x=5-4\)
\(x=1\)
Chúc bạn học tốt!!!!!!!!!!!!!!!!!!!!
a) \(4^{x^{ }}=64\)
\(\Rightarrow4^x=4^3\)
\(\Rightarrow x=3.\)
b) Vì \(35⋮x\)
\(\Rightarrow x\inƯ\left(35\right)=\left\{1;5;7;35\right\}\)
Vậy \(x\in\left\{1;5;7;35\right\}\)
c) \(135-\left(x-5\right)=105:3\)
\(\Leftrightarrow\)\(135-\left(x-5\right)=35\)
\(\Leftrightarrow x-5=135-35\)
\(\Leftrightarrow x-5=100\)
\(\Leftrightarrow x=100+5\)
\(\Leftrightarrow\) \(x=105.\)
d) Vì \(x⋮25\)
\(\Rightarrow x\in B\left(25\right)=\left\{0;25;50;75;100;...\right\}\)
Mà \(x< 100\)
\(\Rightarrow x\in\left\{0;25;50;75\right\}\)