5x+2 + 5x+3 = 750
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\(5x^2+25x-750=0\)
\(\Leftrightarrow5\left(x^2+5x-150\right)=0\)
\(\Leftrightarrow5\left(x^2+15x-10x-150\right)=0\)
\(\Leftrightarrow5\left[\left(x^2+15x\right)-\left(10x+150\right)\right]=0\)
\(\Leftrightarrow5\left[x\left(x+15\right)-10\left(x+15\right)\right]=0\)
\(\Leftrightarrow5\left(x-10\right)\left(x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-10=0\\x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-15\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{10;-15\right\}\)
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
\(5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2\)
=>\(5^x+5^x\cdot5+5^x\cdot25+5^x\cdot125=88\cdot\dfrac{\left(88+1\right)}{2}-16\)
=>\(156\cdot5^x=44\cdot89-16=3900\)
=>\(5^x=\dfrac{3900}{156}=25\)
=>x=2
Bài 1:
Ta có: m>n
\(\Leftrightarrow8m>8n\)
\(\Leftrightarrow8m-2>8n-2\)
Bài 3:
a) Ta có: 2-5x<3(2-x)
\(\Leftrightarrow2-5x< 6-3x\)
\(\Leftrightarrow2-5x-6+3x< 0\)
\(\Leftrightarrow-4-2x< 0\)
\(\Leftrightarrow2x< -4\)
hay x<-2
b) Ta có: \(\frac{5x-2}{3}\ge x+1\)
\(\Leftrightarrow\frac{5x-2}{3}-x-1\ge0\)
\(\Leftrightarrow\frac{5x-2}{3}-\frac{3x}{3}-\frac{3}{3}\ge0\)
\(\Leftrightarrow5x-2-3x-3\ge0\)
\(\Leftrightarrow2x-5\ge0\)
\(\Leftrightarrow2x\ge5\)
hay \(x\ge\frac{5}{2}\)
Bài 4:
Ta có: |x+5|=3x-2
\(\Leftrightarrow\left[{}\begin{matrix}x+5=3x-2\\x+5=2-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+5-3x+2=0\\x+5-2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x+7=0\\4x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2x=-7\\4x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{-3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{7}{2};\frac{-3}{4}\right\}\)
1. Cho m > n, hãy so sánh 8m - 2 với 8n - 2
Ta có : \(m>n\)
\(\Rightarrow8m>8n\)
\(\Rightarrow8m-2>8n-2\)
Lời giải:
$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$
$5^x(1+5+5^2+5^3)=88.89:2-16$
$5^x.156=3900$
$5^x=3900:156=25=5^2$
$\Rightarrow x=2$
Ta có: \(5^{x+2}+5^{x+3}=750\)
\(\Leftrightarrow5^x\cdot\left(5^2+5^3\right)=750\)
\(\Leftrightarrow5^x\cdot150=750\)
\(\Leftrightarrow5^x=5\)
\(\Rightarrow x=1\)
5x+2 + 5x+3 = 750
<=> 5x . 52 + 5x . 53 = 750
<=> 5x (52 + 53) = 750
<=> 5x (25 + 125) = 750
<=> 5x . 150 = 750
<=> 5x = 750 : 150
<=> 5x = 5
<=> x = 1
#Học tốt!!!
~NTTH~