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a) \(\left(x-1\right)^2=49\)
\(\Rightarrow\left(x-1\right)^2=7^2=\left(-7\right)^2\)
\(\Rightarrow x-1=7\) hoặc \(x-1=-7\)
\(x=7+1=8\) \(x=-7+1=-6\)
Vậy x = 8 hoặc x = - 6
b) \(3\cdot\left(13-x\right)^2=27\)
\(\left(13-x\right)^2=27\div3=9\)
\(\Rightarrow\left(13-x\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow13-x=3\) hoặc \(13-x=-3\)
\(x=13-3=10\) \(x=13+3=16\)
Vậy x = 10 hoặc x = 16
c) \(164-\left(15-x\right)^3=100\)
\(\left(15-x\right)^3=164-100=64\)
\(\Rightarrow\left(15-x\right)^3=4^3\)
\(\Rightarrow15-x=4\)
\(x=15-4=11\)
Vậy x = 11
d) \(\left(x+3\right)^3-15=210\)
\(\left(x+3\right)^3=210+15=225\)
\(\Rightarrow\left(x+3\right)^3=...\)
Tương tự mũ lẻ cậu nhé
e) \(x^2\div4=16\)
\(x^2=16\cdot4=64\)
\(\Rightarrow x^2=8^2=\left(-8\right)^2\)
Vậy x = 8 hoặc x = - 8
a/\(\left(x-1\right)^2\)=49
\(\left(x-1\right)^2\)=\(7^2\)
=>x-1=7
x=7+1
x=8
b/3.\(\left(13-x\right)^2\)=27
\(\left(13-x\right)^2\)=27:3
\(\left(13-x\right)^2\)=9
\(\left(13-x\right)^2\)=\(3^2\)
=>13-x=3
x=13-3
x=10
c/164-\(\left(15-x\right)^3\)=100
\(\left(15-x\right)^3\)=164-100
\(\left(15-x\right)^3\)=64
\(\left(15-x\right)^3\)=\(4^3\)
=>15-x=4
x=15-4
x=11
d/\(\left(x+3\right)^3\)-15=210
\(\left(x+3\right)^3\)=210+15
\(\left(x+3\right)^3\)=225
sai đề bài câu d hay sao ý bạn ạ
chỉ có \(\left(x+3\right)^2\)thì mới tính được
e/\(x^2\):4=16
\(x^2\)=16.4
\(x^2\)=64
\(x^2\)=\(8^2\)
=>x=8
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a) 2x : 4 = 128
=> 2x = 512
=> 2x = 29
=> x = 9
b) x15 = x
=> x15 = x15
=> x = 1 hoặc 0
c) ( 2x + 1 )3 = 125
=> ( 2x + 1 )3 = 53
=> 2x + 1 = 5
=> 2x = 4
=> x = 2
a) 2x : 4 = 128
=> 2x = 128 . 4
=> 2x = 512
=> 2x = 29
=> x = 9
b) x15 = x
=> x15 = x15
=> x = 1 hoặc 0
c) ( 2x + 1 )3 = 125
=> ( 2x + 1 )3 = 53
=> 2x + 1 = 5
=> 2x = 5 - 1
=> 2x = 4
=> x = 4 : 2
=> x = 2
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(x+3)^2 + (x-15)^2 = 0
co (x + 3^2) > 0 va (x-15)^2 > 0
=> (x+3)^2 = 0 va (x - 15)^2 = 0
=> x + 3 = 0 va x - 15 = 0
=> x = -3 va x = 15
vay x thuoc tap hop rong :v
Bạn phuong uyen không phải và đâu mà là hoặc đấy chỉ cần 1 trong hai cái =0
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a) (3x+1 + 3x) : 2 = 18
3x.(3+1) = 36
3x = 9 = 32
=> x= 2
b) (x+3)2 + (y-5)2 = 0
mà \(\left(x+3\right)^2\ge0;\left(y-5\right)^2\ge0.\)
=> x = - 3; y = 5
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1,D = -33-|x+2|:
Ta có:\(|x+2|\ge0\)
\(\Rightarrow-|x+2|\le0\)
\(\Rightarrow-33-|x+2|\le-33\)
Dấu = xảy ra khi:\(|x+2|=0\)
\(\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)
VẬY .....
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Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3
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1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
\(2\cdot3^{x-1}+15=33\\ \)
\(\Rightarrow2\cdot3^{x-1}=33-15\\ \Rightarrow2\cdot3^{x-1}=18\\ \Rightarrow3^{x-1}=18:2\\ \Rightarrow3^{x-1}=9\\ \Rightarrow3^{x-1}=3^2\\ \Rightarrow x-1=2\\ \Rightarrow x=2+1=3\)
chú ý chọn đúng lớp!