Cho A = 1 + 4+ 4\(^2\)+4\(^3\) +.....................+4 mũ 99
Tìm x biết 3.A+1=4 mũ x
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a)100-7.[x-5]=58
7.[x-5]=100-58=42
x-5=42:7=6
x=6+5
x=11
b)12.[x-1]:3=43+23
12 [x-1]:3=64+8
12.[x-1]:3=72
[x-1]:3=72:12=6
[x-1]=6.3=18
x=18-1
x=17
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a) 3/4 + 1/4: x = -3
1/4: x = -3 - 3/4
1/4 : x = -15/4
x = 1/4 : -15/4
x = -1/15
Bài 3
a) x^3 = x^5 => x^5 - x^3 =0 => x^3( x^2 - 1 ) = 0
=> x^3(x - 1 ) ( x+ 1 )= 0
=> x^3 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
=> x = 0 hoặc x = 1 hoặc x -1
tick đúng cho mình nha
a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
ko bt có đúng ko nữa, mấy câu kia tui ko bt lm
\(f\left(x\right)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3\)
\(f\left(x\right)=\left(x^4+3x^4\right)-\left(x^3-2x^3\right)-\left(3x^2+x^2\right)+x-1\)
\(f\left(x\right)=4x^4+x^3-4x^2+x-1\)
\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)
\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)
\(g\left(x\right)=-2x^4+4x^3+x-5\)
`@` `\text {Ans}`
`\downarrow`
`a,`
\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)
`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`
`= 4x^4 + x^3 -4x^2 + x -1`
\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)
`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`
`= -2x^4 + 4x^3 +x - 5`
M = 40 + 41 + 42 + 43 + ... + 498
=> 4M = 41 + 42 + 43 + 44 + ... + 499
Khi đó 4M - M = (41 + 42 + 43 + 44 + ... + 499) - (40 + 41 + 42 + 43 + ... + 498)
=> 3M = 499 - 40 = 499 - 1
Khi đó 2x = 3M + 1
<=> 2x = 499 - 1 + 1
=> 2x = 499
=> 2x = (22)99
=> 2x = 22.99
=> 2x = 2198
=> x = 198
Vậy x = 198
M = 40 + 41 + 42 + 43 + ... + 498
4M = 4( 40 + 41 + 42 + 43 + ... + 498 )
= 41 + 42 + 43 + 44 + ... + 499
=> 3M = 4M - M
= 41 + 42 + 43 + 44 + ... + 499 - ( 40 + 41 + 42 + 43 + ... + 498 )
= 41 + 42 + 43 + 44 + ... + 499 - 40 - 41 - 42 - 43 - ... - 498
= 499 - 1
2x = 3M + 1
<=> 2x = 499 - 1 + 1
<=> 2x = 499
<=> 2x = (22)99 = 2198
<=> x = 198
Bài 8:
a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)
b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
A = 1 + 4 + 42 + ... + 499
4A = 4 + 42 + ... + 4100
4A - A = 4100 - 1
3A = 4100 - 1
=> 4100 - 1 + 1 = 4x
=> 4100 = 4x
=> x = 100