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Có \(5^x+5^{x+1}=750\)
=> \(5^x\left(1+5\right)=750\)
=> \(5^x=\frac{750}{6}=125\)
=> \(x=3\)
Hok tốt nhen ^_^
Lời giải:
a)
$1+3+5+...+(2x-1)=750$
$\frac{(2x-1+1)x}{2}=750$
$x^2=750$
$x=\sqrt{750}$ (vô lý?!!)
b)
$1+5+9+13+...+x=501501$
$\frac{1}{2}(\frac{x-1}{4}+1)(x+1)=501501$
$(x+1)(x+3)=4012008$$x(x+4)=4012005=2001.2005$
$\Rightarrow x=2001$
5x + 5x+1 = 750 => 5x + 5x . 5 =750 => 5x . (1 + 5) = 750 => 5x . 6 = 750 => 5x = 750 : 6 => 5x = 125 => 5x = 53 => x = 3
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
a 25 + 8 + 12 + 25 = 70 b 87 + [ - 750 ] + 2018 - 13 + 750 = 2842 c [ 45 + 12 - 13 ] - [ 45 - 13 + 12 ] = 90 minh tinh gom bn ah
Ta có \(5^x+5^{x+1}=750\)
\(\Rightarrow5^x+5^x.5=750\)
\(\Rightarrow5^x.\left(1+5\right)=750\)
\(\Rightarrow5^x.6=750\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
\(5^x\) + \(5^{x+1}\) = \(750\)
=> \(5^x\) + \(5^x\) . \(5\)= \(750\)
=> \(5^x\). \(\left(1+5\right)\)=\(750\)
=> \(5^x\). \(6\)= \(750\)
=> \(5^x\)= \(125\)
=> \(5^x\) = \(5^3\)
=> \(x=3\)
Gọi d là UCLN của 2n+1 và 3n+1
Ta có :
\(2n+1⋮d\)
\(3n+1⋮d\)
\(\Rightarrow3\left(2n+1\right)⋮d\)
\(\Rightarrow2\left(3n+1\right)⋮d\)
\(\Rightarrow\left(6n+3\right)-\left(6n+2\right)⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Ta có: 5x+1=750-5x
\(\Leftrightarrow\)5x+1+5x=750
\(\Leftrightarrow\)5x.6=750
\(\Leftrightarrow\)5x=125
\(\Leftrightarrow\)5x=53
\(\Leftrightarrow\)x=3
Vậy .......
Hok tốt
k mk nha
\(5^x+5^{x-1}=750\)
\(\Rightarrow5^x+5^x.\frac{1}{5}=750\)
\(\Rightarrow5^x.\left(1+\frac{1}{5}\right)=750\)
\(\Rightarrow5^x.\frac{6}{5}=750\)
\(\Rightarrow5^x=750:\frac{6}{5}\)
\(\Rightarrow5^x=625\)
\(\Rightarrow5^x=5^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
Ta có : 5x + 5x - 1 = 750
\(\Rightarrow\) 5x + 5x - 1 = 625 + 125
\(\Rightarrow\) 5x + 5x - 1 = 54 + 53
\(\Rightarrow\) x = 4
Vậy x = 4