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\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
a) \(\frac{7-8x}{6}=\frac{-4+2x}{5}\)
=> \(\left(7-8x\right).5=6\left(-4+2x\right)\)
=> 35 - 40x = -24 + 12x
=> 35 + 24 = 12x + 40x
=> 52x = 59
=> x = 59/52
b) \(\frac{1-3:x}{8}=\frac{8}{1-3:x}\)
=> (1 - 3: x)2 = 82
=> \(\orbr{\begin{cases}1-3:x=8\\1-3:x=-8\end{cases}}\)
=> \(\orbr{\begin{cases}3:x=-7\\3:x=9\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{7}\\x=\frac{1}{3}\end{cases}}\)
c) \(\left(x+1\right)\left(x-2\right)\ge0\)
=> \(\hept{\begin{cases}x+1\ge0\\x-2\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-2\le0\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\x\ge2\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\le2\end{cases}}\)
=> \(-1\le x\le2\)
h) \(\left(x+1\right)\left(x-3\right)\le0\)
=> \(\hept{\begin{cases}x+1\ge0\\x-3\le0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-3\ge0\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\x\le3\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\ge3\end{cases}}\) (loại)
= \(-1\le x\le3\)
\(a,\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)\(=\frac{3^3.5^3}{\left(-3\right)^5.5^2}\)\(=-\frac{5}{\left(3\right)^2}=-\frac{5}{9}\)
\(b,\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)\(=-\frac{6^3}{2^8.3^5}\)\(=-\frac{2^3.3^3}{2^8.3^5}\)\(=-\frac{1}{2^5.3^2}=-\frac{1}{288}\)
\(c,\frac{3^6.7^2-3^7.7}{3^7.21}\)\(=\frac{3^6.7\left(7-3\right)}{3^7.21}\)\(=\frac{3^6.7.4}{3^7.7.3}\)\(=\frac{4}{3.3}=\frac{4}{9}\)
\(a,\left(x-1,2\right)^2=4\)
\(\Rightarrow x-1,2=2\)
\(\Rightarrow x=3,2\)
\(b,\left(x+1\right)^3=-125\)
\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Rightarrow x+1=-5\Rightarrow x=-6\)
\(c,\left(x-5\right)^3=2^6\)
\(\Rightarrow\left(x-5\right)^3=4^3\)
\(\Rightarrow x-5=4\Rightarrow x=9\)
\(d,\left(2x+1\right)^{x+1}=5^{x+1}\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
1) \(\left|x+y-\frac{1}{4}\right|^2+\left|x-y+\frac{1}{5}\right|=0\)
Ta có : \(\hept{\begin{cases}\left|x+y-\frac{1}{4}\right|^2\ge0\\\left|x-y+\frac{1}{5}\right|\ge0\end{cases}}\Leftrightarrow\left|x+y-\frac{1}{4}\right|^2+\left|x-y+\frac{1}{5}\right|\ge0\)
Mà \(\left|x+y-\frac{1}{4}\right|^2+\left|x-y+\frac{1}{5}\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x+y-\frac{1}{4}\right|^2=0\\\left|x-y+\frac{1}{5}\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y=\frac{1}{4}\\x-y=-\frac{1}{5}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}-y\\\frac{1}{4}-y-y=\frac{-1}{5}\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}-y\\-2y=-\frac{9}{20}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}-\frac{9}{40}=\frac{1}{40}\\y=\frac{9}{40}\end{cases}}}\)
Vậy .........
2) \(\left|3x+8\right|-2x=5\)
\(\Leftrightarrow\left|3x+8\right|=2x+5\)( 1 )
Ta có : \(\left|3x+8\right|=\orbr{\begin{cases}3x+8\forall x\ge-\frac{8}{3}\\-3x-8\forall x< \frac{-8}{3}\end{cases}}\)
Để giải phương trình ( 1 ) ta quy về giải 2 phương trình sau :
+) \(3x+8=2x+5\) với \(x\ge\frac{-8}{3}\)
\(\Leftrightarrow3x-2x=5-8\)
\(\Leftrightarrow x=-3\left(KTM\right)\)
+) \(-3x-8=2x+5\)với \(x< \frac{-8}{3}\)
\(\Leftrightarrow-5x=13\Leftrightarrow x=\frac{-13}{5}\left(KTM\right)\)
Vậy phương trình vô nghiệm
c) \(\left|x-2\right|+\left|x+3\right|=6\)
+) với \(x\ge2\)
\(x-2+x+3=6\)
\(\Leftrightarrow2x+1=6\)
\(\Leftrightarrow x=\frac{5}{2}\left(tm\right)\)
+) Với x< -3
\(2-x-x-3=6\)
\(\Leftrightarrow-2x-1=6\)
\(\Leftrightarrow-2x=7\Leftrightarrow x=\frac{-7}{2}\left(tm\right)\)
Vậy .........