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\(A=\left|3,7-x\right|+2,5\)
\(\Rightarrow GTLN\)là 2,5
Khi 3,7 - x = 0
x = -3,7
a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
⇔\(7\left(x-3\right)=5\left(x+5\right)\)
⇔\(7x-21=5x+25\)
⇔\(7x-21-5x-25=0\)
⇔\(2x-46=0\)
⇔\(2x=46\)
⇔\(x=23\)
c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
⇔\(\left(x+4\right)\left(x+4\right)=100\)
⇔\(\left(x+4\right)^2=10^2\)
⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
a) |x - 1,7| = 2,3
Xét 2 trường hợp:
TH1: x - 1,7 = -2,3
x = -2,3 +1,7
x = -0,6
TH2: x - 1,7 = 2,3
x = 2,3 + 1,7
x = 4
Vậy: Tự kl :<
Trả lời:
\(\left(\frac{2}{3}x-\frac{4}{9}\right).\left[\frac{1}{2}+\left(-\frac{3}{7}\right)\div x\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\left(-\frac{3}{7}\right)\div x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{4}{9}\\\frac{-3}{7}\div x=\frac{-1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{2}{5},\frac{6}{7}\right\}\)
Học tốt nhé
Trả lời :
\(\left(\frac{2}{3}x-\frac{4}{9}\right)\times\left(\frac{1}{2}-\frac{3}{7}\div x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}\div x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{4}{9}\\\frac{3}{7}\div x=\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)
\(\left|x+1\right|và\left|x+2\right|\ge0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)+\left(x+2\right)=3\\\left(x+1\right)+\left(x+2\right)=-3\end{cases}}\)
\(\orbr{\begin{cases}2x+3=3\\2x+3=-3\end{cases}}\)
\(\orbr{\begin{cases}2x=0\\2x=-6\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
\(\left|x+1\right|+\left|x+2\right|=3\)
Xét \(x+1\ge0;x+2\ge0\Leftrightarrow x\ge-1;x\ge-2\Rightarrow x\ge-1\) ta có : \(\hept{\begin{cases}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\end{cases}}\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|=3\Leftrightarrow x+1+x+2=3\Leftrightarrow2x+3=3\Rightarrow x=0\)(TM)
Xét \(x+1\le0;x+2\ge0\Leftrightarrow-2\le x\le-1\) ta có : \(\hept{\begin{cases}\left|x+1\right|=-x-1\\\left|x+2\right|=x+2\end{cases}}\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|=3\Leftrightarrow-x-1+x+2=3\Leftrightarrow1=3\) (loại)
Xét \(x+1\le0;x+2\le0\Leftrightarrow x\le-1;x\le-2\Leftrightarrow x\le-2\) ta có : \(\hept{\begin{cases}\left|x+1\right|=-x-1\\\left|x+2\right|=-x-2\end{cases}}\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|=-x-1-x-2=-2x-3=3\Rightarrow x=-3\)(TM)
Vậy \(x=\left\{-3;0\right\}\)
a, Ta có
\(\left|x-1,7\right|=2,3\\ \Rightarrow\left[{}\begin{matrix}x-1,7=2.3\\x-1.7=-2,3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy....
b, Ta có :
\(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy...