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\(\left(x-2\right)^{x+2}=\left(x-2\right)^{x+4}\)
\(\left(x-2\right)^{x+2}-\left(x-2\right)^{x+2}.\left(x-2\right)^2=0\)
\(\left(x-2\right)^{x+2}.\left[1-\left(x-2\right)^2\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^{x+2}=0\\1-\left(x-2\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\x-2=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\x=3\end{cases}}\)
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\(\frac{x+5}{3}=\frac{x-1}{4}\)
\(\Rightarrow\left(x+5\right).4=\left(x-1\right).3\)
\(\Rightarrow4x+20=3x-3\)
\(\Rightarrow4x-3x=-3-20\Rightarrow x=-23\)
\(\frac{x+5}{3}=\frac{x-1}{4}\)
\(\Rightarrow\left(x+5\right)\cdot4=\left(x-1\right)\cdot3\)
\(4x+20=3x-3\)
\(4x-3x=-3-20\)
\(x=-23\)
Vậy \(x=-23\)
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Ta có b/c=3/4 nên b/3=c/4 =>b/12=c/16(1)
mà a/1=b/4=>a/3=b/12(2)
Từ (1)và (2) suy ra :
a/3=b/12=c/16=>4a/12=b/12=c/16
áp dụng tính chất dãy tỉ số bằng nhau ta có :
4a/12=b/12=c/16=4a+b-c/12+12-16=8/8=1
=>a/3=1=>a=1.3=3
=>b/12=1=>b=1.12=12
=>c/16=1=>c=1.16=16
Vậy a=3,b=12,c=16
theo đề bài ta có :\(\frac{b}{3}=\frac{c}{4}\)và\(\frac{3}{4}=\frac{b}{c}\)
từ trên \(\Rightarrow\frac{b}{12}=\frac{c}{16}\)
\(\frac{a}{1}=\frac{b}{4}\Rightarrow\frac{a}{3}=\frac{b}{12}\)
như vậy từ đây áp dụng tính chất dãy tỉ số bằng nhau ,ta có:
\(\frac{4a}{12}=\frac{b}{12}=\frac{c}{16}=\frac{4a+b-c}{12+12-16}=\frac{8}{8}hay1\)
\(\frac{a}{3}=1\Rightarrow a=3\)
\(\frac{b}{12}=1\Rightarrow b=12\)
\(\frac{c}{16}=1\Rightarrow c=16\)
vậy ta có a=3 ; b=12 ; c=16
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\(\frac{69}{157}-\left(2+\left(3+4+5^{-1^{-1^{-1^{-1}}}}\right)\right)\)
\(=\frac{69}{157}-\left(2+\left(3+4+\frac{1}{5}\right)\right)\)
\(=\frac{69}{157}-\left(2+\frac{36}{5}\right)\)
\(=\frac{69}{157}-\frac{46}{5}\)
\(=\frac{-6877}{785}\)
chúc bạn học tốt
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\(\left|x-3,2\right|+\left|2x-\frac{1}{5}\right|=x+3.\)
ĐK : \(x+3\ge0\Leftrightarrow x\ge-3\)
Th1 : \(x-3,2+2x-\frac{1}{5}=x+3\)
\(x-3,2+2x=x+\frac{16}{5}\)
\(x+2x=x+\frac{32}{5}\)
\(2x=\frac{32}{5}\)
\(\Leftrightarrow x=3,2\)(tm)
\(x-3,2+2x-\frac{1}{5}=3-x\)
\(x-3,2+2x=3-x+\frac{1}{5}\)
\(x-3,2+2x=\frac{16}{5}-x\)
\(x+2x=\frac{16}{5}-x+3,2\)
\(x+2x=\frac{32}{5}-x\)
\(2x=\frac{32}{5}-x-x\)
\(2x=\frac{32}{5}-2x\)
\(4x=\frac{32}{5}\)
\(x=1,6\)(tm)
Vậy \(x=1,6\)hoặc \(x=3,2\)
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(x - 2/7)(x + 1/4) > 0
Xét 2 trường hợp:
- \(\hept{\begin{cases}x-\frac{2}{7}>0\\x+\frac{1}{4}>0\end{cases}\Rightarrow\hept{\begin{cases}x>\frac{2}{7}\\x>-\frac{1}{4}\end{cases}\Rightarrow}x>\frac{2}{7}}\)
- \(\hept{\begin{cases}x-\frac{2}{7}< 0\\x+\frac{1}{4}< 0\end{cases}\Rightarrow\hept{\begin{cases}x< \frac{2}{7}\\x< -\frac{1}{4}\end{cases}\Rightarrow}x< -\frac{1}{4}}\)
Vậy x > 2/7 hoặc x < -1/4
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