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\(1.x^2+11x=0\)
\(\Leftrightarrow x\left(x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+11=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-11\end{cases}}\)
\(2.\left(x^2-1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+9\right)\left(x-9\right)=0\)
chia thành 4 TH :
\(TH1:X-1=0\)
\(\Leftrightarrow x=1\)
\(TH2:x+1=0\)
\(\Leftrightarrow x=-1\)
\(TH3:X+9=0\)
\(\Leftrightarrow X=-9\)
\(TH4:x-9=0\)
\(\Leftrightarrow x=9\)
Kết luận ....
\(3.\left(\left|x+1\right|-5\right)\left(x^2-9\right)\)
\(\Leftrightarrow\left(\left|x+1\right|-5\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|-5=0\\x-3=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|=5\\x=3\\x=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+1=+_-5\Leftrightarrow x+1=5,x+1=-5\Leftrightarrow x=4,x=-6\\x=3x\\x=-3\end{cases}}\)
kết luận x=.....
\(4.\left(3x-16\right)⋮\left(x+2\right)\)
\(\Leftrightarrow\left(3x+6\right)-22\)
\(\Leftrightarrow3\left(x+2\right)-22⋮\left(x+2\right)\)
Vì\(\left(x+2\right)⋮\left(x+2\right)\)
\(\Rightarrow\left(3x-16\right)⋮\left(x+2\right)\)
Kết luận x=.....
Bài 2:
a)|x| < 3
x\(\in\){-2;-1;0;1;2}
b)|x - 4 | < 3
x\(\in\){ 6 ; 5 ; 4 ; 3 ; 2 }
c) | x + 10 | < 2
x\(\in\){ -2 ; -10 }
Bài 1:
A = 1 + 2 - 3 + 4 + 5 - 6 +...+98 - 99
A = (1 + 4 + 7 +...+97) + [(2-3)+(5-6)+...+(98-99)]
A = 1617 + [(-1)+(-1)+...+(-1)]
A = 1617 + (-49)
A = +(1617-49) = A = 1568
B = - 2 - 4 + 6 - 8 + 10 + 12 - .... + 60
B =
2)
a) \(x\in\left\{2;1;0;-1;-2\right\}\)
b) \(x\in\left\{6;-6;5;-5;4\right\}\)
c) \(x\in\left\{-9;-11;-10\right\}\)
3)
\(\left(a;b\right)\in\left\{\left(0;1\right);\left(0;-1\right);\left(1;0\right);\left(-1;0\right)\right\}\)
a) \(2\left|x\right|-5=3\)
=> \(2\left|x\right|=3+5\)
=> \(2\left|x\right|=8\)
=> \(\left|x\right|=4\)
=> x = 4 hoặc x = -4
b) \(x-5=\left(-14\right)+2^3\)
=> \(x-5=\left(-14\right)+8\)
=> \(x-5=-6\)
=> \(x=-6+5=-1\)
c) \(10+2x=4^5:4^3\)
=> \(10+2x=4^{5-3}\)
=> \(10+2x=4^2\)
=> \(2x=4^2-10=16-10=6\)
=> \(2x=6\)
=> \(x=3\)
d) (x + 7) - 13 = 4
=> x + 7 = 17
=> x = 17 - 7 = 10
e) \(2x-10=2^4:2^2\)
=> \(2x-10=2^2\)
=> \(2x-10=4\)
=> \(2x=14\)
=> \(x=7\)
a) \(3^2.x+2^3.x=51\)
\(\Leftrightarrow x\left(3^2+2^3\right)=51\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\)
Vậy
b) \(6^2.2-\left(84-3^2.x\right):7=69\)
\(\Leftrightarrow\left(84-3^2.x\right):7=3\)
\(\Leftrightarrow84-3^2.x=21\)
\(\Leftrightarrow3^2.x=63\)
\(\Leftrightarrow x=7\)
Vậy
\(a,\left|x\right|-2=7-\left(-8\right)\)
\(\Rightarrow\left|x\right|-2=15\)
\(\Rightarrow\left|x\right|=17\)
\(\Rightarrow\orbr{\begin{cases}x=-17\\x=17\end{cases}}\)
\(b,5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}-50=75\)
\(\Rightarrow5^{2x-3}=125\)
\(\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
Bài 1 :
a) 72x-1 = 343
=> 72x-1 = 73
=> 2x - 1 = 3 => 2x = 4 => x = 2
b) (7x - 11)3 = 25.32 + 200
=> (7x - 11)3 = 32.9 + 200
=> (7x - 11)3 = 488
xem kĩ lại đề này :vvv
c) 174 - (2x - 1)2 = 53
=> (2x - 1)2 = 174 - 53
=> (2x - 1)2 = 174 - 125 = 49
=> (2x - 1)2 = (\(\pm\)7)2
=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)
Bài 2 :
a) x5 = 32 => x5 = 25 => x = 2
b) (x + 2)3 = 27
=> (x + 2)3 = 33
=> x + 2 = 3 => x = 3 - 2 = 1
c) (x - 1)4 = 16
=> (x - 1)4 = 24
=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)
d) (x - 1)8 = (x - 1)6
=> (x - 1)8 - (x - 1)6 = 0
=> (x - 1)6 [(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)
+) x - 1 = 1 => x = 2 ( tm)
+) x - 1 = -1 => x = 0 ( tm)
Vậy x = 1,x = 2,x = 0
.
42 . (x-6)=1
16(x-6)=1
x-6=1:16
x-6=1/16
x=1/16+6
x=97/16