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a, x . y = 2=> x=1, y=2 hoặc x=2,y=1
b, x . y = 10=>x=2, y=5 hoặc x=5, y=2
c, x . y = 12=>x=2, y=6 hoặc x=6, y=2
d, x . y = 40=> x=4, y=10 hoặc x=10,y=4 hoặc x=5, y=8 hoặc x=8, y=5
V )x+(x+1)+(x+2)+....+(x+30)=1240
31 . x + (1 + 2 + 3 + 4 +...+ 29 + 30) = 1240
31 . x + 31.15 = 1240
31 . x = 1240 - 31.15
31 . x = 775
x = 775 : 31
x = 25
a) 52 . x = 62 + 82
\(5^2\cdot x=36+64\)
\(5^2\cdot x=100\)
\(x=100\div5^2\)
\(x=100\div25\)
\(x=4\)
b) ( 22 + 42 ) . x + 24 . 5 . x = 102
\(\left(4+16\right)\cdot x+16\cdot5\cdot x=100\)
\(x\cdot\left(20+80\right)=100\)
\(x\cdot100=100\)
\(x=100\div100\)
\(x=1\)
c ) 24 . x = 26
\(x=2^6\div2^4\)
\(x=2^{6-4}\)
\(x=2^2\)
\(x=4\)
d) 33 . x + 23 . x = 102
\(x\cdot\left(23+27\right)=100\)
\(x\cdot50=100\)
\(x=100\div50\)
\(x=2\)
e) 78 . x = 710
\(x=7^{10}\div7^8\)
\(x=7^{10-8}\)
\(x=7^2\)
\(x=49\)
a) 3x.3 = 243
3x + 1 = 35
x + 1 = 5
x = 4
b) x2 = x
x2 - x = 0
x(x - 1) = 0
x = 0 hoặc x - 1 = 0
x = 1
d) 64.4x = 16
4x = 16 : 64
4x = 1/4
4x = 4-1
x = -1
Bài 1 :
a) (x-15 ) .32 = 32
<=> (x - 15 ) = 1
<=> x = 1 + 15
<=> x = 16
Vậy x = 90
b) ( x- 15 ) - 75 = 0
<=> x- 15 = 75
<=> x = 75 + 15
<=> x = 90
Vậy x= 90
c) 315 +(125 - x ) =435
<=> 125 -x = 435 - 315
<=> 125 - x = 120
<=> x = 125 - 120
<=> x = 5
Vậy x = 5
d) (x-78 ) . 2020 = 0
<=> x- 78 = 0
<=> x = 0 + 78
<=> x = 78
Vậy x = 78
e) 219 - 7. ( x + 1 )= 0
<=> 7.(x + 1 ) = 219
<=> 7x + 7 = 219
<=> 7x = 212
<=> x = 212 /7 ( L )
Vậy x \(\in\varnothing\)
g) 3x .3 = 243
<=> 3 x = 81
<=> 3x = 34
<=> x = 4
Vậy x = 4
Phần h) bạn làm tương tự
Bài 2 :
Mình đang làm ,lát mình làm xong rồi gửi,tầm 4:30 h gì đó ,vì mình đang học trực tuyến !
bài 2 :
a) Ta có : 2161 > 2160 = (24 )40 =1640 > 1340
Vậy 1340 < 2161
b) 10249 = (210)9 = 2 90
2100 = 2100
=> ta thấy 290 < 2100 => 10249 < 2100
c) tương tự d) nha bạn ,thực ra mình làm d) trước
d) 48 . ( 4 + 8 ) = 48 . 12 = 576
43 + 83 = 576
=> 48.(4 + 8 ) = 43 + 83
a) \(4^x=2^{x+1}\)
\(2^{2x}=2^{x+1}\)
\(\Rightarrow2x=x+1\)
\(\Rightarrow2x-x=1\)
\(\Rightarrow x=1\)
b) \(16=\left(x-1\right)^4\)
\(2^4=\left(x-1\right)^4\)
\(\Rightarrow x-1=2\)
\(\Rightarrow x=3\)
c) \(x^{10}=1^x\)
\(x^{10}=1\)
\(x^{10}=1^{10}\)
\(\Rightarrow x=1\)
d) \(x^{10}=x\)
\(x^{10}-x=0\)
\(x\left(x^9-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
e) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=\pm1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{15}{2}\\x=\left\{8;7\right\}\end{cases}}\)
\(A,4^X=2^{X+1}\)
\(\left(2^2\right)^X=2^{X+1}\)
\(\Rightarrow2^{2X}=2^{X+1}\)
\(\Rightarrow2X=X+1\)
\(\Rightarrow2X-X=1\Leftrightarrow X=1\)
\(B,16=\left(x-1\right)^4\)
\(\Rightarrow x-1=\hept{\begin{cases}2\\-2\end{cases}}\)
\(\Rightarrow x=\hept{\begin{cases}3\\-1\end{cases}}\)
1/
a. \(x^3-2=25\)
\(x^3=25+2\)
\(x^3=27\)
\(\Rightarrow x=3\)
b.\(\left(x-3\right)^2=25\)
\(\left(x-3\right)^2=5^2\)
\(\Rightarrow x-3=5\)
\(\Rightarrow x=8\)
1,a, x^3-2=25 b, (x-3)^2=25 c, x^3-x^2=55 d,[(8.x-12):4].3^7=3^10
x^3=27 (x-3)^2=5^2 không có giá trị x (8.x-12):4=3^3
x^3=3^3 x-3=5 8.x-12=108
x=3 x=8 8.x=120
x=15
2, a, \(7^6:7^4+3^4.3^2-3^7:3\) b, 1736-(21-16).32+6.7^2 c,56.17+17.44-4^3.5+6.(3^2-2)
=\(7^2+3^6-3^6\) =1736-5.32+6.49 =17.(56+44)-320+42
=\(49\) =1736-160+294 =17.10-278
=1736+134 =170-278
=1870 =-108
d, 3.10^2-[1200-(4^2-2.3)^3]
=300-[1200-(16-6)^3]
=300-(1200-10^3)
=300-(1200-1000)
=300-200
=100
a. ( 3x + 9 ).( 1 - 3x ) = 0
\(\Leftrightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-9\\3x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
Vậy \(x\in\left\{-3;\frac{1}{3}\right\}\)
b, \(\left(x^2+1\right)\left(81-x^2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\81-x^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x^2=81\end{cases}}\) ( vô lí ở trg hợp 1 nha )
<=> \(x^2=81\)
\(\Leftrightarrow\) \(x\in\left\{-9;9\right\}\)
Vậy \(x\in\left\{-9;9\right\}\)
a.(3x+9).(1-3x)=0
\(\Rightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=-9\Rightarrow x=-9:3=-3\\3x=1\Rightarrow x=\frac{1}{3}\end{cases}}\)
Vậy...........................................................
x=1 hoặc x=0
\(Ta\)có : \(x^7=x^6\)
\(\Leftrightarrow x^7-x^6=0\)
\(\Leftrightarrow x^6.\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^6=0\\\left(x-1\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(x=\left\{0;1\right\}\)