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4a) \(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)
\(\Leftrightarrow x=\frac{1}{10}:\frac{-2}{3}=\frac{1}{10}.\frac{3}{-2}=\frac{3}{-20}\)
Vậy x=\(\frac{3}{-20}\)
b) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\Leftrightarrow\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)
\(\Leftrightarrow\frac{-5}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{12}:\frac{-5}{6}=\frac{5}{12}.\frac{6}{-5}=\frac{1}{-2}\)
Vậy x=\(\frac{1}{-2}\)
g)Sửa đề: \(\left|4x-1\right|=\left(-3\right)^2\)
\(\Leftrightarrow\left|4x-1\right|=9\)
\(\Rightarrow\left[{}\begin{matrix}4x-1=9\\4x-1=\left(-9\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{2};-2\right\}\)
i) \(\left(x-1^3\right)=125\)
\(\Leftrightarrow x-1=125\)
\(\Leftrightarrow x=125+1=126\)
Vậy x=126
k) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
Bài 1:
a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)
\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)
\(=\frac{2+3}{12}=\frac{5}{12}\)
b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{-7}{10}\)
\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)
Bài 2:
a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)
\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)
\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)
\(\Leftrightarrow-2x=-6-1=-7\)
hay \(x=\frac{7}{2}\)
Vậy: \(x=\frac{7}{2}\)
A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0
\(\frac{3}{4}.\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{3}{5}-x=\frac{2}{3}\)
\(x=\frac{3}{5}-\frac{2}{3}\)
\(x=-\frac{1}{15}\)
Vậy \(x=-\frac{1}{15}\)
\(x+\frac{1}{2}.\frac{2}{3}=\frac{3}{4}\)
\(x+\frac{1}{3}=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{1}{3}\)
\(x=\frac{5}{12}\)
vậy \(x=\frac{5}{12}\)
hơ hơ =v
\(\frac{3}{4}\times\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{3}{5}-x=\frac{2}{3}\)
\(x=\frac{3}{5}-\frac{2}{3}\)
\(x=\frac{-1}{15}\)
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
\(a,\frac{2}{3}.\left(3-x\right)+\frac{1}{2}=\frac{3}{4}.\left(2.x+1\right)
\)
\(2-\frac{2}{3}x+\frac{1}{2}=\frac{3}{2}.\frac{3}{4}x+\frac{3}{4}
\)
\(\frac{2}{3}x+2-\frac{1}{2}=\frac{9}{8}x+\frac{3}{4}\)
\(\frac{2}{3}x+\frac{3}{2}=\frac{9}{8}x+\frac{3}{4}\)
\(\frac{3}{2}-\frac{3}{4}=\frac{9}{8}x-\frac{2}{3}x\)
\(\frac{6}{4}-\frac{3}{4}=\frac{27}{24}x-\frac{16}{24}x\)
\(\frac{11}{24}x=\frac{3}{4}\)
\(x=\frac{3}{4}:\frac{11}{24}\)
\(x=\frac{3}{4}.\frac{24}{11}\)
\(x=\frac{18}{11}\)
\(Vậy
x=\frac{18}{11}\)
\(b,\frac{5-x}{3}=\frac{2x+1}{5}\)
\(\frac{\left(5-x\right).5}{15}=\frac{\left(2x+1\right).3}{15}\)
\(\Rightarrow\left(5-x\right).5=\left(2x+1\right).3\)
\(25-5x=6x+3\)
\(25-3=6x+5x\)
\(\Rightarrow11x=22\)
\(\Rightarrow x=22:11\)
\(\Rightarrow x=2\)
\(Vậy
x=2\)
a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
1. \(x=\frac{61}{42}\)
2. \(x=\frac{-36}{5}\)
3. \(x=\frac{13}{11}\)
4. \(x=\frac{1}{12}\)
5.\(x=\frac{-5}{2}\)
a) \(\left(\frac{11}{4}.\frac{-5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)
=\(\frac{11}{4}\left(-\frac{5}{9}-\frac{4}{9}\right).\frac{8}{33}\)
=\(\frac{11}{4}\cdot-1\cdot\frac{8}{33}\)
=\(-\frac{11}{4}\cdot\frac{8}{33}\)
=\(-\frac{2}{3}\)
b)\(-\frac{1}{4}\cdot\frac{152}{11}+\frac{68}{4}\cdot-\frac{1}{11}\)
=\(\frac{-1.152}{4.11}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1.152}{11.4}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1}{11}\cdot\frac{152}{4}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1}{11}\cdot\left(\frac{152}{4}+\frac{68}{4}\right)\)
=\(\frac{-1}{11}\cdot55=-5\)
c)\(\frac{-2}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{3}{5}\)
=\(-1\cdot\frac{2}{3}\left(\frac{4}{5}+\frac{3}{5}\right)\)
=\(-1\cdot\frac{2}{3}\cdot\frac{7}{5}\)
=\(-\frac{2}{3}\cdot\frac{7}{5}\)
=\(\frac{-14}{15}\)
d) chưa nghĩ ra nhé
e) bạn chép sai đề bài rồi
mk mới kiểm tra 45 phút nên biết
đề bài nè
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{899}{30^2}\)
=\(\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot...\cdot\frac{29.31}{30^2}\)
=\(\frac{1.3.2.4.3.5...29.31}{2.2.3^2.4^2...30.30}\)
=\(\frac{1.2.3^2.4^2.5^2....29^2.30.31}{2.2.3^2.4^2.5^2....29^2.30.30}\)
=\(\frac{1.31}{2.30}\)
=\(\frac{31}{60}\)
a)trong ngoac bn dat thau so chung la 11/4 rui tinh binh thuong b)bn tu lam nhe c)dat thua so chung d)tinh trong ngoac ra rui nhan vs e) mk bo tay
\(A=2.\left|x-\frac{1}{2}\right|-2019\)
Vì \(\left|x-\frac{1}{2}\right|\ge0,\forall x\)
\(\Rightarrow2.\left|x-\frac{1}{2}\right|\ge0,\forall x\)
\(\Rightarrow2.\left|x-\frac{1}{2}\right|-2019\ge-2019,\forall x\)
Dấu \("="\)xảy ra
\(\Leftrightarrow\left|x-\frac{1}{2}\right|=0\)
\(x-\frac{1}{2}=0\)
\(x=0+\frac{1}{2}\)
\(x=\frac{1}{2}\)
Vậy \(A_{min}=-2019\Leftrightarrow x=\frac{1}{2}\)
\(A=2.\left|x-\frac{1}{2}\right|-2019\)
Ta có : \(2.\left|x-\frac{1}{2}\right|\ge0\forall x\)
\(\Rightarrow2\left|x-\frac{1}{2}\right|-2019\ge-2019\)
Dấu "=" xảy ra \(\Leftrightarrow2.\left|x-\frac{1}{2}\right|=0\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy : \(A_{min}=-2019\) tại \(x=\frac{1}{2}\)
\(B=4\left|3x-2\right|+3\left|4x+1\right|-\frac{1}{3}\)
Ta có : \(4\left|3x-2\right|\ge0\forall x,3\left|4x+1\right|\ge0\forall x\)
\(\Rightarrow4\left|3x-2\right|+3\left|4x+1\right|\ge0\forall x\)
\(\Rightarrow4\left|3x-2\right|+3\left|4x+1\right|-\frac{1}{3}\ge-\frac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-2=0\\4x+1=0\end{cases}}\)
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