Tìm x :
1+3+5+7+....+(2x-1) = 750
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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$
$\Rightarrow x=2001$
\(5^{x+2}+5^{x+3}=750\)
\(5^x.5^2+5^x.5^3=750\)
\(5^x.25+5^x\cdot125=750\)
\(5^x.\left(25+125\right)=750\)
\(5^x.150=750\)
\(5^x=750:150\)
\(5^x=5\)
\(5^x=5^1\)
\(\Rightarrow x=1\)
a) \(5^{x+2}\)+ \(5^{x+3}\)=625
\(5^x\). \(2^x\)+ \(5^x\) . \(3^x\)=625
\(5^x\). (\(2^x\)+ \(3^x\) ) =625
\(5^x\). \(5^x\) =625
\(25^x\) =625
\(25^x\)= \(25^2\)
vậy x=2
hình như câu a bn ghi nhầm 625 thành 750
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
1: Tìm x
a) Ta có: \(\left(2x-1\right)^3=-27\)
\(\Leftrightarrow2x-1=-3\)
\(\Leftrightarrow2x=-3+1=-2\)
hay x=-1
Vậy: x=-1
b) Ta có: \(\left(2x-3\right)^4=625\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=-5\\2x-3=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-5+3=-2\\2x=5+3=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;4\right\}\)
c) Ta có: \(\left(x-2\right)^5=\left(x-2\right)^7\)
\(\Leftrightarrow\left(x-2\right)^5-\left(x-2\right)^7=0\)
\(\Leftrightarrow\left(x-2\right)^5\left[1-\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)^5\cdot\left[1-\left(x-2\right)\right]\cdot\left[1+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)^5\cdot\left(1-x+2\right)\cdot\left(1+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)^5\cdot\left(-x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^5=0\\-x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-x=-3\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2;3\right\}\)
d) Ta có: \(5^{x+2}+5^{x+3}=750\)
\(\Leftrightarrow5^{x+2}\cdot1+5^{x+2}\cdot5=750\)
\(\Leftrightarrow5^{x+2}\left(1+5\right)=750\)
\(\Leftrightarrow5^{x+2}\cdot6=750\)
\(\Leftrightarrow5^{x+2}=125\)
\(\Leftrightarrow x+2=3\)
hay x=1
Vậy: x=1
Giải
( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ( x + 7 ) + ( x + 9 ) = 750
x x ( 1 + 3 + 5 + 7 + 9 ) = 750
x x 25 = 750
x = 750 : 25
x = 30
Đúng nha
x + 1 + x + 3 +x + 5 + x + 7 + x + 9 = 750
( x + x + x + x + x ) + ( 1 + 3 + 5 + + 7 + 9 ) = 750
x x 5 + 25 = 750
x x 5 = 750 - 25
x x 5 = 725
x = 725 :5
x = 145
Dãy: 1 + 3 + 5 +....+ 999 có: (999-1):2+1 = 500 (số hạng)
Tổng của dãy trên là: (999+1) x 500 : 2 = 250 000
=> 99 x 250 + 99 x 750 + ( 1 + 3 + .... + 999)
= 99 x (250 + 750) + 250 000
= 99 x 1000 + 250 000
= 99 000 + 250 000
= 349 000
99 x 250 x 99 x 750 + ( 1 + 3 + 5 + ...+ 999 )
Số các số hạng của 1 + 3 + 5 + ... + 999 là:
( 999 - 1 ) : 2 + 1 = 500 ( số hạng )
Ttoongr của 1 + 3 + 5 + ... + 999 là:
(999 + 1 ) x 500 : 2 = 250000
=> 99 x 250 + 99 x 750 + ( 1 + 3 + 5 + ... + 999 )
= 99 x ( 250 + 750 ) + 250000
=99 x 1000 + 250000
= 99 x 1000 + 250 x 1000
= 1000 x ( 99 + 250 )
= 1000 x 349
= 349000
a, (x + 2) + (x + 4) + (x + 6) + ... + (x + 50) = 750
=> x + 2 + x + 4 + x + 6 + ... + x + 50 = 750
=> (x + x + x + ... + x) + (2 + 4 + 6 + ... + 50) = 750
=> 25x + (50 + 2).25 : 2 = 750
=> 25x + 52.25 : 2 = 750
=> 25x + 650 = 750
=> 25x = 100
=> x = 4
a) ( x+x+...+x)+(2+4+6+...+50)= 750
( x*25)+ (50+2)*25:2 = 750
(x*25)+ 650 = 750
x* 25 = 750 - 650 = 100
x = 100 :25 = 4
Đặt S = 1+3+5+7+...+(2x−1)
S có: (2x−1−1):2+1=x (số hạng)
1+3+5+7+...+(2x−1) = (2x−1+1).x:2 = 225
(2x−1+1).x = 225.2
⇒(2x−1+1).x= 450
⇒2x^2 = 450
⇒x^2= 450:2
⇒x2= 225
⇒ x^= 15^
=>x=15
Vậy xx = 15
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