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a) (2x+1)2=25
(2x+1)2= (+-5)2
=> 2x+1 = 5 hoặc 2x + 1 = -5
2x = 4 hoặc 2x = -6
x= 2 hoặc x=-3
b) (x-1)3=-125
(x-1)3= (-5)3
=> x-1 = -5
x= -4
c) 2x+2-2x=96
2x.22 - 2x = 96
2x( 4-1) = 96
2x = 96 : 3
2x= 32
2x = 25
=> x= 5
d) 7x+2+2.7x-1=345
7x-1 . 73 + 2.7x-1=345
7x-1( 73 +2) = 345
7x-1 . 345 = 345
7x-1 =1
=> x-1 = 0
=> x= 1
6) \(\frac{1}{2}.2x+2^{x+2}=2^8+5\)
\(\Rightarrow x+2^{x+2}=2^8+2^5=288\)
- Nếu x < 6 thì x + 2x+2 < 262
- Nếu x > 6 thì x + 2x+2 > 519
Vậy không có giá trị nào của x thỏa mãn
b) \(7^{x+2}+2.7^{x-1}=7^{x-1}.\left(7^3+2\right)=7^{x-1}.345=345\)
\(\Rightarrow7^{x-1}=1\Rightarrow x-1=0\Rightarrow x=1\)
Vậy x = 1 thỏa mãn
\(\left(2x+1\right)^2=25\)
\(\Rightarrow2x+1\in\left\{-5;5\right\}\)
\(\Rightarrow2x\in\left\{-6;4\right\}\)
\(\Rightarrow x\in\left\{-3;2\right\}\)
Vậy..
\(\left(x-1\right)^3=-125\)
\(\left(x-1\right)^3=-5^3\)
\(x-1=-5\)
\(x=-4\)
Vậy...
\(7^{x+2}.2.7^{x-1}=345\)
\(7^x.\left(7^2+\dfrac{2}{7}\right)=345\)
\(7x=7\)
\(x=1\)
Vậy...
a.
\(7^{x+2}+2\times7^{x-1}=345\)
\(7^x\times7^2+2\times7^x\div7=345\)
\(7^x\times\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\times\frac{345}{7}=345\)
\(7^x=345\div\frac{345}{7}\)
\(7^x=345\times\frac{7}{345}\)
\(7^x=7\)
\(x=1\)
b.
\(2^{x+2}-2^x=96\)
\(2^x\times\left(2^2-1\right)=96\)
\(2^x\times3=96\)
\(2^x=\frac{96}{3}\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
\(a,7^{x+2}+2.7^{x-1}=345\Rightarrow7^x.49+\frac{2}{7}.7^x=345\Rightarrow7^x\left(49+\frac{2}{7}\right)=345\Rightarrow7^x.\frac{345}{7}=345\Rightarrow7^x=345:\frac{345}{7}=7^1\Rightarrow x=1\)
\(b,2^{x+2}-2^x=96\Rightarrow2^x.4-2^x=96\Rightarrow2^x\left(4-1\right)=96\Rightarrow2^x.3=96\Rightarrow2^x=96:3=32\Rightarrow2^x=2^5\Rightarrow x=5\)
\(a,7^{x+2}+2.7^{x-1}=345=>7^{x-1+3}+2.7^{x-1}=345=>7^{x-1}.7^3+2.7^{x-1}=345\)
\(=>\left(7^3+2\right).7^{x-1}=345=>345.7^{x-1}=345=>7^{x-1}=1=7^0=>x-1=0=>x=1\)
\(b,2^{x+2}-2^x=96=>2^x.2^2-2^x=96=>2^x.\left(4-1\right)=96=>2^x.3=96=>2^x=32=2^5=>x=5\)
c) 5x+5x+2=650
=> 5x+5x52 = 650
=> 5x ( 1+ 25 ) = 650
=> 5x = 650 / 26
=> 5x = 25 = 52
=> x = 2
\(\Leftrightarrow7^x\cdot49+2\cdot7^x\cdot\dfrac{1}{7}=345\)
\(\Leftrightarrow7^x=7\)
=>x=1
2.3x + 3x - 1 = 7 . (32 + 2 . 62)
=> 2.3x + 3x - 1 = 567
=> 7 . 3x - 1 = 567
=> 3x - 1 = 567 : 7 = 81
=> x - 1 = 4
=> x = 5
a)2*3x+3x-1=7(32+2*62)
2*3x+3x-1=7(9+72)=7*81
2*3x+3x/3=567
2*3x+3x*1/3=567
(2+1/3)*3x=567
7/3*3x=567
3x=567:7/3
3x=243=35
=>x=5
b) mk ko hiểu đề mấy, cái chỗ 7x+2 là nhân vs 2 ak