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B1. 2x + 3 + 22 = 72
=> 2x + 3 + 4 = 72
=> 2x + 3 = 72 - 4
=> 2x + 3 = 68
=> ko có gtri x
B2 : Ta có : A = 1 + 2 + 22 + 23 + 24 + 25 + 26 + ... + 22001 + 22002
= (1 + 2) + (22 + 23 + 24) + (25 + 26 + 27) + ... + (22000 + 22001 + 22002)
= 3 + 22.(1 + 2 + 22) + 25.(1 + 2 + 22 ) + ... + 22000 . (1 + 2 + 22)
= 3 + 22.7 + 25.7 + ... + 22000 . 7
= 3 + (22 + 25 + .... + 22000) . 7
=> Số dư của 7 là 3
\(5^x+5^{x+2}=650;5^x.26=650;5^x=25;x=2\)
\(2^x+2^{x+3}=144;2^x.9=144;2^x=16;x=4\)
\(3^{x-1}+5.3^{x-1}=162;3^{x-1}.6=162;3^{x-1}=27;x=4\)
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\rightarrow x-5=0\&x-5=1\) hoặc x - 5 = - 1
\(x-5=1;x=6;x-5=0;x=5;x-5=-1;x=4\)
\(\left(2^2:4\right).2^n=4;2^n=2^2;n=2\)
Ta thấy : \(\left(x-y^2+z\right)^2\ge0\forall x,y,z\)
\(\left(y-2\right)^2\ge0\forall y\)
\(\left(z+3\right)^2\ge0\forall z\)
Do đó : \(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2\ge0\forall x,y,z\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y^2+z\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x-y^2+z=0\\y-2=0\\z+3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2^2+\left(-3\right)=0\\y=2\\z=-3\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=7\\y=2\\z=-3\end{cases}}\)
Vậy : \(\left(x,y,z\right)=\left(7,2,-3\right)\)
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
\(\left(10^2+6^2\cdot2\right):\left(43\cdot x\right)=2^0\)
\(\Rightarrow172:\left(43\cdot x\right)=1\)
\(\Rightarrow43x=172\)
\(\Rightarrow x=172:43\)
\(\Rightarrow x=4\)
ta có
((102+62*2)/(43*x)=20
=1
=>102+62*2=43*x
100+36*2=43*x
172=43*x
x=172/43
x=4
a ) số đối : -9 = 9 ; 17 = 17
b ) 63 = 216 ; 70 = 1
Bài 2 :
a) 42
b) 30
c) 3800
Bài 3 :
a) x = -31
b) x = 2
c) x = { -2 ;-1 ; 0 ;1;2;3;4 }
1/a.9;-17.
b.216;1
2/a.42;b.30;c.3800
3/a.x=-31;b.x=2;c.x=-2;-1;0;1;2;3;4.
chuyen ve phai sang trai roi dat thua so chung nhe
Bài giải
\(\left(x-6\right)^3=\left(x-6\right)^2\)
\(\Rightarrow\text{ }\left(x-6\right)^3-\left(x-6\right)^2=0\)
\(\Rightarrow\text{ }\left(x-6\right)^2\left[\left(x-6\right)-1\right]=0\)
\(\Rightarrow\text{ }\left(x-6\right)^2\left[x-6-1\right]=0\)
\(\Rightarrow\text{ }\left(x-6\right)^2\left[x-7\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-6\right)^2=0\\x-7=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x-6=0\\x=7\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{6\text{ ; }7\right\}\)