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\(\frac{1}{2}\cdot2^x+2^x\cdot2^2=2^8+2^5\)
\(2^x\left(\frac{1}{2}+4\right)=2^8+2^5\)
\(2^x\cdot\frac{9}{2}=288\)
\(2^x=64\)
\(2^x=2^6\)
\(x=6\)
\(9^x:3^x=3^7\)
\(3^{2x}:3^x=3^7\)
\(3^x=3^7\)
\(x=7\)
\(7^{x+2}+2\cdot7^{x-1}=345\)
\(7^x\cdot7^2+2\cdot7^x:7=345\)
\(7^x\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\cdot\frac{345}{7}=345\)
\(7^x=7\)
\(x=1\)
a) 1/2.2^x + 2^x+2 = 256 + 32
1/2.2^x + 2^2.2^x=288
2^x(1/2+4)= 288
2^x.4,5=288
2^x= 288:4,5
2^x=64=2^6
x=6
7x + 2 + 2 . 7x - 1 = 345
7x - 1 . ( 73 + 2 ) = 345
7x - 1 . 345 = 345
7x - 1 = 345 : 345
7x - 1 = 1
7x - 1 = 70
x - 1 = 0
x = 0 + 1
x = 1
219 - 7(x + 1) = 10²
7(x + 1) = 219 - 10²
7(x + 1) = 219 - 100
7(x + 1) = 119
x + 1 = 119 : 7
x + 1 = 17
x = 17 - 1
x = 16
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(3x - 2⁴) . 7³ = 2.7⁴
(3x - 16).7³ = 2.7⁴
3x - 16 = 2.7⁴:7³
3x - 16 = 2.7
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 30 : 3
x = 10
Lời giải:
$10+2x=4^5:4^3=4^2=16$
$2x=16-10=6$
$x=6:2=3$
---------
$2x-6^2:18=3.2^2$
$2x-2=12$
$2x=14$
$x=14:2=7$
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$70-5(x-3)=3.2^5=96$
$5(x-3)=70-3.2^5=-26$
$x-3=\frac{-26}{5}$
$x=3+\frac{-26}{5}=\frac{-11}{5}$
------------------
$19-7(x+1)=10^2=100$
$7(x+1)=19-100=-81$
$x+1=\frac{-81}{7}$
$x=\frac{-81}{7}-1=\frac{-88}{7}$
------------
$(3x-2^4).7^3=2.7^4$
$3x-16=2.7^4:7^3=2.7=14$
$3x=14+16=30$
$x=30:3=10$
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Rightarrow2x-1=0\)hoặc \(2x-1=1\)hoặc \(2x-1=-1\)
\(\Rightarrow x=\frac{1}{2}\)hoặc \(x=1\)hoặc \(x=0\)
1. tìm x
kết quả x là : 2010
2. tìm x
a) x = 13
b) x = 6
đúng ko
câu 3 số lớn quá
1.
1(1+2+3+4+....+x)=2021055
1+2+3+4+....+x=2021055:1=2021055
x+(x+1)/1=2021055
=>x(x+1)=1*2021055=2021055=2011*2010 =>x=2010
\(7^{x-1+3}+2.7^{x-1}=345\Leftrightarrow7^{x-1}\left(7^3+2.1\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\Leftrightarrow7^{x-1}=1\Leftrightarrow7^{x-1}=7^0\Leftrightarrow x-1=0\Rightarrow x=1\)
Vậy x= 1
=> \(7^{x-1}.7^3+2.7^{x-1}=345\)
=> \(7^{x-1}.\left(7^3+2\right)=345\)
=> \(7^{x-1}.345=345\Rightarrow7^{x-1}=1\Rightarrow x-1=0\) => x = 1