4.(5-x)-3.(1-x)=-32
3.|2x-3|=7-(-14)
22x-1-32=23
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\(a,\Rightarrow x+35=103\\ \Rightarrow x=68\\ b,\Rightarrow x+7=42\\ \Rightarrow x=35\\ c,\Rightarrow x-15=9\\ \Rightarrow x=24\\ d,\Rightarrow x+23=31\\ \Rightarrow x=8\\ e,\Rightarrow18-x=6\\ \Rightarrow x=12\\ f,\Rightarrow x+14=34\\ \Rightarrow x=20\\ g,\Rightarrow2x+1=21\\ \Rightarrow x=11\\ h,\Rightarrow3\left(x+1\right)=99\\ \Rightarrow x+1=33\Rightarrow x=32\\ i,5\left(x-3\right)=25\\ \Rightarrow x-3=5\\ \Rightarrow x=8\\ j,\Rightarrow8\left(2x+7\right)=88\\ \Rightarrow2x+7=11\Rightarrow x=2\\ k,\Rightarrow5\left(x+4\right)=85\\ \Rightarrow x+4=17\\ \Rightarrow x=13\)
17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
a) \(\left|-\frac{2}{11}+\frac{3}{22}x\right|-\frac{1}{2}=\frac{5}{7}\)
=> \(\left|-\frac{2}{11}+\frac{3}{22}x\right|=\frac{17}{14}\)
=> \(\orbr{\begin{cases}-\frac{2}{11}+\frac{3}{22}x=\frac{17}{14}\\-\frac{2}{11}+\frac{3}{22}x=-\frac{17}{14}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{215}{21}\\x=-\frac{53}{7}\end{cases}}\)
b) \(-\frac{7}{8}x-5\frac{3}{4}=3\)
=> \(-\frac{7}{8}x-\frac{23}{4}=3\)
=> \(-\frac{7}{8}x=3+\frac{23}{4}=\frac{35}{4}\)
=> \(x=\frac{35}{4}:\left(-\frac{7}{8}\right)=\frac{35}{4}\cdot\left(-\frac{8}{7}\right)=-10\)
c) \(2x+\left(-\frac{2}{7}\right)-7=-11\)
=> \(2x-\frac{2}{7}-7=-11\)
=> \(2x=-11+7+\frac{2}{7}=-\frac{26}{7}\)
=> \(x=\left(-\frac{26}{7}\right):2=-\frac{13}{7}\)
d) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)
=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)
=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)
x(x+2)=0
suy ra x=0 hoặc x+2=0
5-2x=-7
2x=-7+5
2x=-(7-5)
2x=-2
x=-2:2
x=-1
Vậy x=-1
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Ta có: \(4.\left(5-x\right)-3.\left(1-x\right)=-3^2\)
\(\Leftrightarrow20-4x-3+3x=9\)
\(\Leftrightarrow-4x+3x=9-20+3\)
\(\Leftrightarrow-x=-8\)
\(\Leftrightarrow x=8\)
Vậy: \(x=8\)
Ta có: \(3.\left|2x-3\right|=7-\left(-14\right)\)
\(\Leftrightarrow3.\left|2x-3\right|=7+14=21\)
\(\Leftrightarrow\left|2x-3\right|=\dfrac{21}{3}=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy: \(x=-2;5\)
Ta có: \(2^{2x-1}-3^2=23\)
\(\Leftrightarrow2^{2x-1}-9=23\)
\(\Leftrightarrow2^{2x-1}=32\)
\(\Leftrightarrow2^{2x-1}=2^5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy: \(x=3\)
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