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\(\frac{x+110}{100}+\frac{x+8}{102}+\frac{x+6}{104}=\frac{x+4}{106}+\frac{x+2}{108}\)
\(\Leftrightarrow\frac{x+110}{100}+\left(\frac{x+8}{102}+1\right)+\left(\frac{x+6}{104}+1\right)=\left(\frac{x+4}{106}+1\right)+\left(\frac{x+2}{108}+1\right)\)
\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}=\frac{x+110}{106}+\frac{x+110}{108}\)
\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}-\frac{x+110}{106}-\frac{x+110}{108}=0\)
\(\Leftrightarrow\left(x+110\right)\left(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}+\frac{1}{106}+\frac{1}{108}\right)=0\)
\(\Leftrightarrow x+110=0\) (vì \(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}-\frac{1}{106}-\frac{1}{108}>0\))
\(\Leftrightarrow x=-110\)
Vậy \(x=-110\)
mình ko đáng cái j linh tinh hết đây là các bài toán mà mình ko giải đc
b. (x-7)x+1-(x-7)x+11=0
(x-7)x+1.[1-(x-7)10]=0
=> (x-7)x+1=0 hoặc 1-(x-7)10=0
• (x-7)x+1= 0 => x-7=0 => x=7
• 1-(x-7)10=0=> (x-7)10=1=>x-7=1 hoặc x-7=-1 => x=8 hoặc x=6
Vậy x thuộc {6;7;8}
1/53+-1/106+-1/159=|x|/318
6/318+-3/318+-2/318=|x|/318
1/318=|x|/318
=>|x|=1
x=1 hoặc x=-1
=> \(\frac{1}{53}\)+ \(\frac{-1}{106}\)+\(\frac{-1}{159}\)= \(\frac{\left|x\right|}{318}\)
=> \(\frac{1}{318}\)= \(\frac{\left|x\right|}{318}\)
=> x thuộc {1; -1}
\(\left(1-\frac{52}{53}\right)+\left(\frac{105}{106}-1\right)+\left(\frac{158}{159}-1\right)=\frac{\left|x\right|}{318}\)
\(\Rightarrow\frac{1}{53}+\frac{-1}{106}+\frac{-1}{159}=\frac{\left|x\right|}{318}\)
\(\Rightarrow\frac{1}{318}=\frac{\left|x\right|}{318}\)
\(\Rightarrow\left|x\right|=1\)
\(\Rightarrow x=\pm1\)
Vậy..............................
Bài 1:
a)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2\cdot4\right)^5}{\left(0,2\cdot2\right)^6}=\frac{\left(0,2\right)^5\cdot\left(2^2\right)^5}{\left(0,2\right)^6\cdot2^6}=\frac{\left(0,2\right)^5\cdot2^{10}}{\left(0,2\right)^6\cdot2^6}=\frac{2^4}{0,2}=\frac{16}{\frac{2}{10}}=80\)
b)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=256\)
Bài 2:
a)\(2^{x-1}=16\)
\(\Rightarrow2^{x-1}=2^4\)
\(\Rightarrow x-1=4\Rightarrow x=5\)
b)\(\left(x-1\right)^2=25\)
\(\Rightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow x-1=5\) hoặc \(x-1=-5\)
\(\Rightarrow x=6\) hoặc \(x=-4\)
Vậy \(x=6\) hoặc \(x=-4\)
c)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\1=\left(x-1\right)^4\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\\left(x-1\right)^4=\left(-1\right)^4=1^4\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x-1=1\\x-1=-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)
d)\(\left(x+20\right)^{100}+\left|y+4\right|=0\left(1\right)\)
Ta thấy: \(\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\left(2\right)\)
Từ (1) và (2) suy ra \(\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}\)
\(\Rightarrow\begin{cases}x+20=0\\y+4=0\end{cases}\)\(\Rightarrow\begin{cases}x=-20\\y=-4\end{cases}\)
Đề đoạn cuối hình như phải là x-400/100 chứ bạn
=> (x+5/105 - 1) + (x+6/106 - 1) + (x+10/110 - 1) + (x-400/100 + 3) = 0
=> x-100/105 + x-100/106 + x-100/110 + x-100/100 = 0
=> (x-100).(1/105 + 1/106 + 1/110 + 1/100) = 0
=> x-100 = 0 ( vì 1/105 + 1/106 + 1/110 + 1/100 > 0 )
=> x = 100
Vậy x = 100
Tk mk nha
cảm ơn bn nhìu