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25 tháng 6 2018
Nhiều câu quá >.<
a/ \(2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20.\)
\(2x^2+10x=x^2+6x+9+x^2-2x+1+20.\)
\(10x=4x+30\)
\(6x=30\Rightarrow x=5\)
các câu còn lại tương tự
25 tháng 6 2018
\(a,2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20\)
\(\Leftrightarrow2x^2+10x=x^2+6x+9+x^2-2x+1+20\)
\(\Leftrightarrow2x^2+10x=2x^2+4x+30\)
\(\Leftrightarrow2x^2+10x-2x^2-4x=30\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
Vậy ...........
\(b,\left(2x-2\right)^2=\left(x+1\right)^2+3\left(x-2\right)\left(x+5\right)\)
\(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3x^2+15x-6x-30\)
\(\Leftrightarrow4x^2-8x+4=4x^2+11x-29\)
\(\Leftrightarrow4x^2-8x-4x^2-11x=-29-4\)
\(\Leftrightarrow-19x=-33\)
\(\Leftrightarrow x=\frac{33}{19}\)
Vậy...........
\(c,\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+1\right)+38\)
\(\Leftrightarrow x^2-2x+1+x^2+6x+9=2x^2+2x-4x-4+38\)
\(\Leftrightarrow2x^2+4x+10=2x^2-2x+34\)
\(\Leftrightarrow2x^2+4x-2x^2+2x=34-10\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy.............
\(d,\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-18\)
\(\Leftrightarrow x^3+6x+12x+8-\left(x^3-6x+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x+12x+8-x^3+6x-12x+8=12x^2-12x-8\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy............
MV
5 tháng 5 2017
2.
\(\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot15\cdot37\right)\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot5\cdot3\cdot37\right)\\=\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left[65\cdot111-\left(13\cdot5\right)\cdot\left(3\cdot37\right)\right]\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left[65\cdot111-65\cdot111\right]\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot0\\ =0\)
NQ
2
23 tháng 7 2015
a) \(2^{x-1}+2^{x+1}+2^{x+2}=104\)
=> \(2^{x-1}+2^x\cdot2+2^x\cdot2^2=104\)
=> \(2^x:2+2^x\cdot\left(2+2^2\right)=104\)
=> \(2^x\cdot\frac{1}{2}+2^x\cdot6=104\)
=> \(2^x\cdot\left(\frac{1}{2}+6\right)=104\Rightarrow2^x=104:\left(\frac{1}{2}+6\right)=104:\frac{13}{2}=16\)
=> \(x=4\)
AA
0
20 tháng 2 2020
a) 2x+2x+1+2x+2+2x+3=480
<=> \(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
<=> \(2^x.\left(1+2+2^2+2^3\right)=480\)
<=>\(2^x=\frac{480}{1+2+2^2+2^3}=32\)
=> x=5
b) (x2-49)*(x2-81)<0 Khi \(\hept{\begin{cases}x^2-49< 0\\x^2-81>0\end{cases}}\) hoặc \(\hept{\begin{cases}x^2-49>0\\x^2-81< 0\end{cases}}\)
TH1 \(\hept{\begin{cases}x^2-49< 0\\x^2-81>0\end{cases}}\)\(\Rightarrow81< x^2< 49\)(Vô lí)
TH2\(\hept{\begin{cases}x^2-49>0\\x^2-81< 0\end{cases}}\) \(\Rightarrow49< x^2< 81\)\(\Leftrightarrow7^2< x^2< 9^2\)Mà x nguyên \(\Rightarrow x=8\)
c) Làm giống câu a
18 tháng 10 2017
a, \(\left(2x+7\right)^4=10^{11}:10^7\)
\(\Rightarrow\left(2x+7\right)^4=10^4\)
\(\Rightarrow2x+7=10\)
\(\Rightarrow2x=10-7\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\) hay \(x=1,5\)
b, \(5^{x-1}.7^{x-1}=25.49\)
\(\Rightarrow\)\(5^{x-1}.7^{x-1}=5^2.7^2\)
\(\Rightarrow\left\{{}\begin{matrix}5^{x-1}=5^2\\7^{x-1}=7^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-1=2\\x-1=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)
18 tháng 10 2017
c, \(\left(x-5\right)^{2018}=9.\left(x-5\right)^{2016}\)
\(\Rightarrow\dfrac{\left(x-5\right)^{2018}}{\left(x-5\right)^{2016}}=9.\dfrac{\left(x-5\right)^{2016}}{\left(x-5\right)^{2016}}\)
\(\Rightarrow\left(x-5\right)^2=9\)
\(\Leftrightarrow\left(x-5\right)^2=3^2\)
\(\Rightarrow x-5=3\)
\(\Rightarrow x=3+5\)
\(\Rightarrow x=8\)
2 tháng 11 2016
a) 2x+1 = 64
2x . 2 = 64
2x = 64 : 2
2x = 32
2x = 25
=> x = 5
b) 3x + 3x+1 = 36
3x . 1 + 3x . 3 = 36
3x . ( 1 + 3 ) = 36
3x . 4 = 36
3x = 36 : 4
3x = 9
3x = 32
=> x = 2
c) 2x + 2x+1 + 2x+2 + 2x+3 = 120
2x . 1 + 2x . 2 + 2x . 22 + 2x . 23 = 120
2x . ( 1 + 2 + 22 + 23 ) = 120
2x . 15 = 120
2x = 120 : 15
2x = 8
2x = 23
=> x = 3
27 tháng 10 2016
a, 64=2^6=> x=5
b.3^x+3(x+1)=4.3^x=36=4.3^2=> x=2
c.2^x+........2^x3=(1+2+4+8).2^x=15.2^x=120=15.2^3 =>x=3
d.5^x+5^(x-2)=26.5^(x-2)=3150=126.5^2 có lẽ đề sai ở đâu đó
2 tháng 10 2016
b) \(3.2^{x+1}=12\)
\(2^{x+1}=12:3\)
\(2^{x+1}=4\)
\(2^{x+1}=2^2\)
\(x+1=2\)
\(x=2-1\)
\(x=1\)
Vậy \(x=1\)
c) \(2^{x-1}=2^3+2^4-2^3\)
\(2^{x-1}=8+16-8\)
\(2^{x-1}=16\)
\(2^{x-1}=2^4\)
\(x-1=4\)
\(x=5\)
Vậy \(x=5\)
d) \(x^{50}=x\)
\(x^{50}-x=0\)
\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
HC
2 tháng 10 2016
\(b.3.2^{x+1}=12\\ \Rightarrow2^{x+1}=4\\ \Rightarrow2^{x+1}=2^2\\ \Rightarrow x=1\\ \)
c) \(2^{x-1}=2^3-2^3+2^4\\ \Rightarrow2^{x-1}=0+16\\ \Rightarrow2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
d) \(x^{50}=x\\ \Rightarrow x=0;1\)
e) \(2\left(2x-1\right)^4=32\\ \Rightarrow\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\frac{3}{2}\)
g) Bí
21 tháng 10 2020
\(2^{x+2}-2^x=96\)
\(\Rightarrow2^x\cdot2^2-2^x=96\)
\(\Rightarrow2^x\left(2^2-1\right)=96\)
\(\Rightarrow2^x\left(4-1\right)=96\)
\(\Rightarrow2^x\cdot3=96\)
\(\Rightarrow2^x=96:3\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(5^x+5^{x+1}=750\)
\(\Rightarrow5^x+5^x\cdot5=750\)
\(\Rightarrow5^x\left(1+5\right)=750\)
\(\Rightarrow5^x\cdot6=750\)
\(\Rightarrow5^x=750:6\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
\(2^{x+3}+2^x=144\)
\(\Rightarrow2^x\cdot2^3+2^x=144\)
\(\Rightarrow2^x\left(2^3+1\right)=144\)
\(\Rightarrow2^x\cdot9=144\)
\(\Rightarrow2^x=144:9\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
\(\left(x-2\right)^{x+2}=\left(x-2\right)^{x+1}\)
\(\left(x-2\right)^{x+2}-\left(x-2\right)^{x+1}=0\)
\(\left(x-2\right)^{x+1}\left(x-2-1\right)=0\)
\(\left(x-2\right)^{x+1}\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^{x+1}=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x=3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy x = { 2; 3 }
\(\left(x-2\right)^{x+2}=\left(x-2\right)^{x+1}\)
\(\left(x-2\right)^{x+2}-\left(x-2\right)^{x+1}=0\)
\(\left(x-2\right)^{x+1}\left(x-2-1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=2\\x=3\end{cases}}\)