a,1/2.2n +4.2n =9.5n
b.1/9. 27n =3n
c,(x+1)3 =125
d,2x+2 -2x =96
e,(2x+1)3 =343
K
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PD
28 tháng 9 2018
nhé
a)(2x-1)6=(2x-1)8
=> (2x-1)8-(2x-1)6=0
=> (2x-1)6.((2x-1)2-1)=0
+)th1(2x-1)6=0
+)th2((2x-1)2-1)=0
TH
28 tháng 9 2018
a) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)\in\left\{\pm1;0\right\}\)
TH1 : \(2x-1=0\) TH2 : \(2x-1=-1\) TH3 : \(2x-1=1\)
\(2x=1\) \(2x=0\) \(2x=2\)
\(x=\frac{1}{2}\) \(x=0\) \(x=1\)
Vậy \(x\in\left\{\frac{1}{2};0;1\right\}\)
b) Tương tự
KL
2
17 tháng 1 2021
2: Tìm x
a) Ta có: x+25=40
nên x=40-25=15
Vậy: x=15
b) Ta có: 198-(x+4)=120
\(\Leftrightarrow x+4=198-120=78\)
hay x=78-4=74
Vậy: x=74
c) Ta có: \(\left(2x-7\right)\cdot3=125\)
\(\Leftrightarrow2x-7=\dfrac{125}{3}\)
\(\Leftrightarrow2x=\dfrac{125}{3}+7=\dfrac{125}{3}+\dfrac{21}{3}=\dfrac{146}{3}\)
\(\Leftrightarrow x=\dfrac{146}{3}:2=\dfrac{146}{6}=\dfrac{73}{3}\)
Vậy: \(x=\dfrac{73}{3}\)
d) Ta có: \(x+16⋮x+1\)
\(\Leftrightarrow x+1+15⋮x+1\)
mà \(x+1⋮x+1\)
nên \(15⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(15\right)\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
Vậy: \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
17 tháng 1 2021
\(a,x+25=40\\ \Rightarrow x=40-25\\ \Rightarrow x=15\\ b,198-\left(x+4\right)=120\\ \Rightarrow-\left(x+4\right)=120-198\\ \Rightarrow-\left(x+4\right)=-78\\ \Rightarrow x+4=78\\ \Rightarrow x=78-4\\ \Rightarrow x=74\\ c,\left(2x-7\right).3=125\\ \Rightarrow2x-7=\dfrac{125}{3}\\ \Rightarrow2x=\dfrac{125}{3}+7\\ \Rightarrow2x=\dfrac{146}{3}\\ \Rightarrow x=\dfrac{146}{3}:2\Rightarrow x=\dfrac{73}{3}\\ d,\left(x+16\right)⋮\left(x+1\right)\\ \Rightarrow\left[\left(x+1\right)+15\right]⋮\left(x+1\right)\\ mà:\left(x+1\right)⋮\left(x+1\right)\\ \Rightarrow15⋮\left(x+1\right)\\ \Rightarrow\left(x+1\right)\inƯ\left(15\right)\\ \Rightarrow\left(x+1\right)\in\left\{-15;-1;1;15\right\}\\ \Rightarrow x\in\left\{-16;-2;0;14\right\}\)
Tự kết luận nhé bạn
14 tháng 10 2018
Bài toán 4: Viết các số sau dưới dạng tổng các luỹ thừa của 10.
213 = 2 . 100 + 1 . 10 +3 = 2. 10^2 + 1.10 + 3 . 10^0
421=4.100 + 2.10 + 1 = 4.10^2 + 2.10 + 1. 10^0
2009; = 2. 1000 + 9 = 2. 10^3 + 9 . 10^0
abc = a . 100 + b . 10 + c = a.10^2 + b.10 + c.10^0
abcde = a.10000 + b . 1000 + c . 100 + d . 10 + e = a . 10^4 + b. 10^3 + c.10^2 + d .10 + e . 10 ^0
16 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
2 tháng 4 2021
a)ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
Suy ra: \(x^2-1+x-2x+1=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
2 tháng 4 2021
b) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{5}{x-3}-\dfrac{2x-3}{x+3}=\dfrac{2x\left(1-x\right)}{x^2-9}\)
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(5x+15-2x^2+6x+3x-9-2x+2x^2=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow12x=-6\)
hay \(x=-\dfrac{1}{2}\)(thỏa ĐK)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
NH
1 tháng 9 2020
\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)
\(\Rightarrow-x=5\)
\(\Rightarrow x=-5\)
\(\text{Vậy x=-5}\)
\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)
\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)
\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)
\(\Rightarrow-16x-8=7\)
\(\Rightarrow-16x=15\)
\(\Rightarrow x=\frac{-15}{16}\)
\(\text{Vậy }x=\frac{-15}{16}\)
\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)
\(\Rightarrow-9+8x-1=8\)
\(\Rightarrow8x=18\)
\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)
\(\text{Vậy }x=\frac{9}{4}\)
\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)
26 tháng 10 2021
a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
19 tháng 10 2020
a)
$(2x+1)^2-(2x+1)(2x-1)=(2x+1)[(2x+1)-(2x-1)]$
$=2(2x+1)$
b)
$(4x+3)(x-1)-2x(2x+1)=4x^2-x-3-4x^2-2x=-3x-3=-3(x+1)$
c)
$(2x+3)^2-(4x+1)(x+5)=(4x^2+12x+9)-(4x^2+21x+5)$
$=-9x+4$
AH
Akai Haruma
Giáo viên
19 tháng 10 2020
d)
$(x+2)^3-(x-1)(x^2+x+1)=(x^3+6x^2+12x+8)-(x^3-1)$
$=6x^2+12x+9$
e)
$(x+2)(x^2-2x+1)-(x+3)(x-3)=(x^3-3x+2)-(x^2-9)$
$=x^3-x^2-3x+11$
f)
$(x+3)(x^2-3x+9)-(x^2+2x+4)(x-2)$
$=x^3+3^3-(x^3-2^3)=3^3+2^3=35$