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Bài 1: Tìm x
a) Ta có: \(\dfrac{4}{3}:0.8=\dfrac{2}{3}:\left(0.1\cdot x\right)\)
\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}:\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow x\cdot\dfrac{1}{10}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\cdot\dfrac{3}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{2}{5}\cdot10=\dfrac{20}{5}=4\)
Vậy: x=4
b) Ta có: \(\left|x\right|=-1.2\)
mà \(\left|x\right|\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
Bài 2: Tính
a) Ta có: \(\left(-2.5\right)\cdot\left(-4\right)\cdot\left(-7.9\right)\)
\(=\left(2.5\cdot4\right)\cdot\left(-7.9\right)\)
\(=-7.9\cdot10=-79\)
b) Ta có: \(\left(-0.375\right)\cdot\dfrac{13}{3}\cdot\left(-2\right)^3\)
\(=\dfrac{3}{8}\cdot8\cdot\dfrac{13}{3}\)
\(=3\cdot\dfrac{13}{3}=13\)
a)\(\dfrac{11}{12}-\left(\dfrac{2}{3}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{3}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{3}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{3}\)
\(x=\dfrac{-5}{12}\)
Vậy x=\(\dfrac{-5}{12}\)
b)\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1.x\)
\(\dfrac{4}{3}:0,8=\dfrac{20}{3}.x\)
\(\dfrac{5}{3}=\dfrac{20}{3}.x\)
\(x=\dfrac{5}{3}.\dfrac{3}{20}\)
\(x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4}\)
a, \(\dfrac{11}{12}-\left(\dfrac{2}{3}+x\right)=\dfrac{2}{3}\Rightarrow x=\dfrac{11}{12}-\dfrac{2}{3}-\dfrac{2}{3}=\dfrac{-5}{12}\)
b, \(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1x\Rightarrow\dfrac{4}{3}:\dfrac{4}{5}=\dfrac{2}{3}:\dfrac{1}{10}x\Rightarrow\dfrac{4}{3}.\dfrac{1}{10}x=\dfrac{4}{5}.\dfrac{2}{3}\Rightarrow\dfrac{4}{3}.\dfrac{1}{10}x=\dfrac{8}{15}\Rightarrow x=\dfrac{8}{15}.\dfrac{3}{4}.10=4\)
Đặt GTBT là A, ta có:
A=0,5+0,(3)−0,1(6)2,5+1,(6)−0,8(3)
A=12 +13 −16 52 +53 −56
A=12 +13 −16 5(12 +13 −16 ) =15
bạn tự điền mấy cái dấu gach p/s nha
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chúc bạn học tốt
a) \(\frac{x}{4}=\frac{16}{x^2}\)\(=>x^3=16.4\)\(=>x^3=64\)\(=>x=4\)
b) \(\frac{4}{3}:\frac{4}{5}=\frac{2}{3}.\left(\frac{1}{10}.x\right)\)\(=>\frac{4}{3}.\frac{5}{4}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}:\frac{2}{3}=\frac{1}{10}x\)\(=>\frac{5}{3}.\frac{3}{2}=\frac{1}{10}x\)\(=>\frac{5}{2}=\frac{1}{10}x\)\(=>x=\frac{5}{2}:\frac{1}{10}\)\(=>x=\frac{5}{2}.10\)\(=>x=25\)
vậy x=25
1.
a) \(\frac{x}{4}=\frac{16}{x^2}\)
\(\Rightarrow x^3=64\)
\(\Rightarrow x^3=4^3\)
\(\Rightarrow x=4\)
b) \(1\frac{1}{3}:0,8=\frac{2}{3}.\left(0,1.x\right)\)
\(\frac{5}{3}=\frac{2}{3}.\frac{x}{10}\)
\(\frac{x}{10}=\frac{5}{2}\)
\(\Rightarrow x=\frac{5.10}{2}=25\)
2.
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
4/3 : 0,8 = 2/3 : ( 0,1 . x )
4/3 : 4/5 = 2/3 : ( 1/10 . x )
5/3 = 2/3 : ( 1/10 . x )
( 1/10 . x ) = 2/3 : 5/3
( 1/10 . x ) = 2/5
x = 2/5 : 1/10
x = 4
4/3 : 0,8 = 2/3 : (0,1 . x)
=> 0,1 . x = 2/3 : (4/3 : 0,8)
=> 0,1 .x =2/3 : 5/3
=> 0,1 . x = 2/5
=> x = 2/5 : 0,1
=> x = 4