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a)TH1 x>=3 \(\left|x-3\right|\)=x-3
pttt: x-3-2x=1 suy ra x=-4 <3 -> loại
TH2 x=< 3 pttt 3-x-2x=1 suy ra x =2/3 thỏa mãn
b) VT=\(\dfrac{4^{x+2}+4^{x+1}+4^x}{21}=\dfrac{4^x\left(4^2+4+1\right)}{21}=4^x\)
VP= \(\dfrac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}=\dfrac{9^x\left(1+3+27\right)}{31}=9^x\)
vậy pt đã cho tương đương với 4^x=9^x \(\Leftrightarrow\left(\dfrac{4}{9}\right)\)^x =1 suy ra x =0
a/ \(\left|-x\right|=1,5\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\end{matrix}\right.\)
Vậy .....
b/ \(\left|x+\dfrac{1}{2}\right|=2\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{5}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy ....
c/ \(\left|0,5-x\right|=\left|-0,5\right|\)
\(\left|0,5-x\right|=0,5\)
\(\Leftrightarrow\left[{}\begin{matrix}0,5-x=0,5\\0,5-x=-0,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy ...
a/ \(\left|-1,3\right|-\left|-3,7\right|+\left|-\dfrac{1}{2}\right|\)
\(=1,3-3,7+\dfrac{1}{2}\)
\(=-2,4+\dfrac{1}{2}\)
\(=-2,9\)
b/ \(\left|\dfrac{2}{5}\right|-\left|-0,2\right|.\left|-7\right|\)
\(=\dfrac{2}{5}-0,2.7\)
\(=\dfrac{2}{5}.1,4\)
\(=0,56\)
c/ \(-15:\left|-3\right|+\left|0,5\right|\)
\(=-15:2+0,5\)
\(=-7,5+0,5\)
\(=-8\)
a/ \(x+\dfrac{3}{5}=\dfrac{4}{7}\)
\(x=\dfrac{4}{7}-\dfrac{3}{5}\)
\(x=-\dfrac{1}{35}\)
Vậy ....
b/ \(x-\dfrac{5}{6}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=1\)
Vậy ....
c/\(-\dfrac{5}{7}-x=\dfrac{-9}{10}\)
\(x=\dfrac{-5}{7}-\dfrac{-9}{10}\)
\(x=\dfrac{13}{70}\)
Vậy .....
d/ \(\dfrac{5}{7}-x=10\)
\(x=\dfrac{5}{7}-10\)
\(x=\dfrac{-65}{7}\)
Vậy ...
e/ \(x:\left(\dfrac{1}{9}-\dfrac{2}{5}\right)=\dfrac{-1}{2}\)
\(x:\dfrac{-13}{45}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}.\dfrac{-13}{45}\)
\(x=\dfrac{13}{90}\)
Vậy ....
f/ \(\left(\dfrac{-3}{5}+1,25\right)x=\dfrac{1}{3}\)
\(0,65.x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:0,65\)
\(x=\dfrac{20}{39}\)
Vậy ....
g/ \(\dfrac{1}{3}x+\left(\dfrac{2}{3}-\dfrac{4}{9}\right)=\dfrac{-3}{4}\)
\(\dfrac{1}{3}x+\dfrac{2}{9}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{-35}{36}\)
\(\Leftrightarrow x=\dfrac{-35}{12}\)
Vậy ...
|2x-1|=1,5
TH(1)2x-1=1,5
2x =1,5+1
2x =2,5
x =2,5 :2
x =1,25
TH(2) 2x-1=-1,5
2x =-1,5+1
2x =-0,5
x =-0,5:2
x =-0,25
các câu khác cứ tương tự bạn nhé
b) \(7,5-\left|5-2x\right|=-4,5\)
\(\left|5-2x\right|=7,5+4,7\)
\(\left|5-2x\right|=12\)
th1 :\(5-2x=12\)
\(2x=5-12\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
th2: \(5-2x=-12\)
\(2x=5+12\)
\(2x=17\)
\(x=17:2\)
\(x=8,5\)
c) \(-3+\left|x\right|=-1\)
\(\left|x\right|=-1+3\)
\(\left|x\right|=2\)
th1: \(x=-2\)
th2 : \(x=2\)
d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)
\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)
th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)
\(x=\dfrac{7}{3}-\dfrac{1}{2}\)
\(x=\dfrac{11}{6}\)
th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)
\(x=\dfrac{7}{3}+\dfrac{1}{6}\)
\(x=\dfrac{-5}{2}\)
e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{9}{14}\)
th1 :\(x+1=\dfrac{9}{14}\)
\(x=\dfrac{9}{14}-1\)
\(x=\dfrac{-5}{14}\)
th2 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{5}-\dfrac{7}{10}x=\dfrac{-3}{4}\cdot\dfrac{5}{3}=\dfrac{-5}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)
\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)
c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)
\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)
hay \(x=-\dfrac{26}{21}\)
a: \(\Leftrightarrow\left(3x-2\right):\dfrac{7}{5}=\dfrac{17}{7}:\dfrac{13}{5}=\dfrac{85}{91}\)
\(\Leftrightarrow3x-2=\dfrac{85}{91}\cdot\dfrac{7}{5}=\dfrac{17}{13}\)
=>3x=43/13
hay x=43/39
b: \(\Leftrightarrow9x+207=121-8x\)
=>19x=-86
hay x=-86/19
c: \(\Leftrightarrow x^2-9=16\)
=>x2=25
=>x=5 hoặc x=-5
d: \(\Leftrightarrow\left|x\right|=\dfrac{1.64\cdot3.11}{8.51}\simeq0,6\)
=>x=0,6 hoặc x=-0,6
a: \(=0.5\cdot10-\dfrac{1}{7}+15=20-\dfrac{1}{7}=\dfrac{139}{7}\)
b: \(=6\cdot\dfrac{-2}{3}+12\cdot\dfrac{4}{9}+18\cdot\dfrac{-8}{27}\)
\(=-4+\dfrac{16}{3}-\dfrac{16}{3}=-4\)
c: \(=\left(\dfrac{5}{2}+\dfrac{3}{8}-\dfrac{5}{8}+\dfrac{2}{3}\right):\left(\dfrac{17}{2}+\dfrac{49}{4}-\dfrac{17}{8}+\dfrac{34}{15}\right)\)
\(=\dfrac{35}{12}:\dfrac{2507}{120}=\dfrac{350}{2507}\)
a: \(\Leftrightarrow\left|x-3\right|=12-5x-8=-5x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(-5x+4\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(5x-4-x+3\right)\left(5x-4+x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(4x-1\right)\left(6x-7\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\)
b: \(\left(\sqrt{x}+3\right)^{10}=1024\cdot125^2\cdot25^2\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)^{10}=2^{10}\cdot5^6\cdot5^4=10^{10}\)
\(\Leftrightarrow\sqrt{x}+3=10\)
hay x=49
c: \(\dfrac{3-0.2x}{5}=\dfrac{7}{15}+1.4x\)
\(\Leftrightarrow\dfrac{9-0.6x}{15}=\dfrac{7}{15}+\dfrac{21x}{15}\)
=>21x+7=9-0,6x
=>21,6x=-2
hay x=-5/54
d: \(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{5^9\cdot7^9\left(4\cdot7-5^2\right)}{5^9\cdot7^9\cdot4}\)
\(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{28-25}{4}=\dfrac{3}{4}\)
=>3x=-1
hay x=-1/3