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a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
\(x⋮12;25;30\Rightarrow x\in BC\left(12;25;30\right)\)
\(12=2^2.3;25=5^2;30=2.3.5\)
\(BCNN\left(12;25;30\right)=2^2.3.5^2=300\)
\(BC\left(12;25;30\right)=B\left(300\right)\)
\(B\left(300\right)=\left\{0;300;600;....\right\}\)
\(0\le x\le500\Rightarrow x=300\)
\(\left(3x-2^4\right).7^3=2.7^4\)
\(\left(3x-16\right)=2.7\)
\(3x-16=14\)
\(3x=30\)
\(x=10\)
\(\left|x-5\right|=16+2.\left(-3\right)\)
\(\left|x-5\right|=10\)
\(\Leftrightarrow x-5=10\Rightarrow x=15\)
\(\Leftrightarrow x-5=-10\Rightarrow x=-5\)
\(5\frac{4}{7}:x=13\)
\(\frac{2}{3}\times x-0,5\times x=\frac{5}{12}\)
\(y-85\%\times y=1\frac{1}{5}\)
\(5\frac{4}{7}:x=13\)
=> \(x=\frac{39}{7}:13\)
=> \(x=\frac{3}{7}\)
\(\frac{2}{3}x-0,5x=\frac{5}{12}\)
\(\Leftrightarrow x\left(\frac{2}{3}-0,5\right)=\frac{5}{12}\)
\(\Leftrightarrow x\left(\frac{4}{6}-\frac{3}{6}\right)=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{12}\div\frac{1}{6}\)
\(\Leftrightarrow x=\frac{5}{2}\)
1. Tìm \(x\):
a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
\(\dfrac{x}{5}=\dfrac{1}{5}\)
\(\Rightarrow x=1\)
b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)
\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)
\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)
\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)
\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)
\(x=\dfrac{-17}{8}\)
c) \(2016^3.2016^x=2016^8\)
\(2016^x=2016^8:2016^3\)
\(2016^x=2016^{8-3}\)
\(2016^x=2016^5\)
\(\Rightarrow x=5\)
d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)
\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)
\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)
\(x+\dfrac{3}{4}=\dfrac{35}{4}\)
\(x=\dfrac{35}{4}-\dfrac{3}{4}\)
\(x=\dfrac{32}{4}=8\)
e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)
\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)
\(2,8.x-2^5=6\)
\(2,8.x=6+32\)
\(2,8.x=38\)
\(x=38:2,8\)
\(x=\dfrac{95}{7}\)
f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)
\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}:\dfrac{4}{7}\)
\(x=\dfrac{28}{15}\)
g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\)
\(x=\left(-6\right):3\)
\(x=-2\)
2. Thực hiện phép tính:
a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)
\(=\dfrac{7}{18}+\dfrac{9}{5}\)
\(=\dfrac{197}{90}\)
b) \(\dfrac{7.5^2-7^2}{7.24+21}\)
\(=\dfrac{7.25-7.7}{7.24+7.3}\)
\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)
\(=\dfrac{7.18}{7.27}\)
\(=\dfrac{2}{3}\)
c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
Tìm X:Y\(\in Z\)biết
a] [x-7]\(\times\left[xy+1\right]\)=9
b] \(\frac{x-4}{y-3}=\frac{4}{3}\)với x-y=5
a) tính bình thường thôi
b)\(\left(3x-4\right)\times\left(x-1\right)^3=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)
c) \(2^{2x-1}:4=8^3\)
\(\Leftrightarrow2^{2x-1}=2048\Leftrightarrow2^{2x-1}=2^{11}\Leftrightarrow2x-1=11\Leftrightarrow x=6\)
d) \(x^{17}=x\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
e) \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)
vậy........
a) (x : 23 + 45) . 37 - 22 = 24.105
=> (x : 23 + 45).37 - 22 = 1680
=> (x : 23 + 45).37 = 1702
=> x : 23 + 45 = 46
=> x : 23 = 1
=> x = 23
b) (3x - 4).(x - 1)3 = 0
=> \(\orbr{\begin{cases}3x-4=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)
Vậy \(x\in\left\{\frac{4}{3};1\right\}\)
c) 22x - 1 : 4 = 83
=> 22x - 1 : 22 = (23)3
=> 22x - 1 : 22 = 29
=> 22x - 1 = 211
=> 2x - 1 = 11
=> 2x = 12
=> x = 6
d) x17 = x
=> x17 - x = 0
=> x(x16 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^{16}-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^{16}=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
Vậy \(x\in\left\{0;1;-1\right\}\)
e) (x - 5)4 = (x - 5)6
=> (x - 5)6 - (x - 5)4 = 0
=> (x - 5)4[(x - 5)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1^2\end{cases}}\Rightarrow\orbr{\begin{cases}x-5=0\\x-5=\pm1\end{cases}}\Rightarrow x-5\in\left\{0;1;-1\right\}\)
=> \(x\in\left\{5;6;4\right\}\)
a;(- \(x+5\)).(3 - \(x\)) = 0
\(\left[\begin{array}{l}-x+5=0\\ 3-x=0\end{array}\right.\)
\(\left[\begin{array}{l}x=5\\ x=3\end{array}\right.\)
Vậy \(x\in\) {3; 5}
b; (\(x-1\)) x (\(x+2\)) x (- \(x-3\)) = 0
\(\left[\begin{array}{l}x-1=0\\ x+2=0\\ -x-3=0\end{array}\right.\)
\(\left[\begin{array}{l}x=1\\ x=-2\\ x=-3\end{array}\right.\)
Vậy \(x\) ∈ {-2; -3; 1}
chia trường hợp ra vd a) TH1: -x+5=0 TH2: 3-x=0
-x=-5 ,x=5 x=3-0,x=3 vậy x=5;x=3