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a, \(5^x+5^{x+1}+5^{x-2}=151\)
\(\Rightarrow5^x.\left(1+5+5^{-2}\right)=151\)
\(\Rightarrow5^x.6,04=151\Rightarrow5^x=25=5^2\)
Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=2\)
b, \(5^{x-1}+5^{x-2}+5^{x-3}=155\)
\(\Rightarrow5^x.\left(5^{-1}+5^{-2}+5^{-3}\right)=155\)
\(\Rightarrow5^x.0,248=155\Rightarrow5^x=625=5^4\)
Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=4\)
c, \(5^{2+x}+5^{3+x}=750\) \(\Rightarrow5^x.\left(5^2+5^3\right)=750\) \(\Rightarrow5^x.150=750\Rightarrow5^x=5=5^1\) Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=1\) Chúc bạn học tốt!!!\(•5^x+5^{x+1}+5^{x-2}=151\\ 5^x\left(1+5+\dfrac{1}{25}\right)=151\\ 5^x=25\\ \Rightarrow x=2\)
\(•5^{x-1}+5^{x-2}+5^{x-3}=155\\ 5^x.\left(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}\right)=155\\ 5^x=625\\ \Rightarrow x=4\)
\(•5^{2+x}+5^{3+x}=750\\ 5^x\left(25+125\right)=750\\ 5^x=5\\ \Rightarrow x=1\)
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
5: Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\)
nên x=5k; y=3k
Ta có: \(x^2-y^2=4\)
\(\Leftrightarrow25k^2-9k^2=4\)
\(\Leftrightarrow k^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm\dfrac{5}{4}\\y=\pm\dfrac{3}{4}\end{matrix}\right.\)
Bài 1 :
\(3x+5=2\left(x-\frac{1}{4}\right)\)
\(\Leftrightarrow3x+5=2x-\frac{1}{2}\)
\(\Leftrightarrow5+\frac{1}{2}=2x-3x\)
\(\Leftrightarrow\frac{11}{2}=-x\)
\(\Leftrightarrow\frac{-11}{2}=x\)
Vậy \(x=\frac{-11}{2}\)
Bài 2:
a, \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{2018}{2019}\right|\ge0\\\left|z-3\right|\ge0\end{cases}}\)
Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)
\(\Rightarrow+,\left|x+\frac{19}{5}\right|=0\)
\(\Leftrightarrow x+\frac{19}{5}=0\)
\(\Leftrightarrow x=\frac{-19}{5}\)
\(\Rightarrow+,\left|y+\frac{2018}{2019}\right|=0\)
\(\Leftrightarrow y+\frac{2018}{2019}=0\)
\(\Leftrightarrow y=\frac{-2018}{2019}\)
\(\Rightarrow+,\left|z-3\right|=0\)
\(\Leftrightarrow z-3=0\)
\(\Leftrightarrow z=3\)
Vậy \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-2018}{2019}\\z=3\end{cases}}\)
b, Ta có : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)
Vì : \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y+4\right|\ge0\\\left|z-5\right|\ge0\end{cases}}\)
Mà : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)
\(\Rightarrow+,\left|x-\frac{1}{2}\right|\ge0\)
\(\Rightarrow x\inℚ\)
\(\Rightarrow+,\left|2y+4\right|\ge0\)
\(\Rightarrow y\inℚ\)
\(\Rightarrow+,\left|z-5\right|\ge0\)
\(\Rightarrow z\inℚ\)
Vậy chỉ cần \(\hept{\begin{cases}x\inℚ\\y\inℚ\\z\inℚ\end{cases}}\)thì thỏa mãn.
Bài 2:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
\(\frac{6-2x}{5}=\frac{-y-4}{4}=\frac{3z-15}{2}\)
\(\Rightarrow\hept{\begin{cases}4\left(6-2x\right)=5\left(-y-4\right)\\2\left(-y-4\right)=4\left(3z-15\right)\\2\left(6-2x\right)=5\left(3z-15\right)\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}8x-5y=44\\y+6z=26\\4x+15z=87\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{8x-44}{5}\\z=\frac{87-4x}{15}\end{cases}}\)
Mà x-y+z=-1 nên \(x-\frac{8x-44}{5}+\frac{87-4x}{15}=-1\)\(\Leftrightarrow15x-3\left(8x-44\right)+87-4x+15=0\)
\(\Leftrightarrow\)\(13x=234\Rightarrow x=18\Rightarrow y=\frac{8.18-44}{5}=20\Rightarrow z=\frac{87-4.18}{15}=1\)
Vậy x=18 ; y=20 ; z=1
dễ ợt bài này còn không biết làm mình học lớp 11 trường lê hồng phong
\(5^x+5^{x+1}+5^{x+2}=155\)
=> \(5^x+5^x\cdot5+5^x\cdot5^2=155\)
=> \(5^x\cdot\left(1+5+5^2\right)=155\)
=> \(5^x\cdot31=155\)
=> \(5^x=155:31=5\)
=> \(x=1\)
5x+5x+1+5x+2=155
5x.(1+5+52)=155
5x.31=155
5x=5
=>x=1