2/5 x X - 1/4 x X = 2 1/3
2 1/3 là 2 và 1/3 nha!
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1: =>x=-5/3-1/2=-13/6
2: =>x=1/3-3/5=-4/15
4: =>x=-7/9+4/3=-7/9+12/9=5/9
5: =>x=5/6-7/3=5/6-14/6=-9/6=-3/2
6: =>x=9/10+1/5=11/10
7: =>x=3/8-5/12=36/96-40/96=-1/24
8: Đề sai rồi bạn
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a)\(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\text{⇔}\dfrac{7}{4}x-5=\dfrac{10}{3}\text{⇔}\dfrac{7}{4}x=\dfrac{25}{3}\text{⇔}x=\dfrac{100}{21}\)
b)\(\dfrac{2}{3}x+\dfrac{1}{4}=\dfrac{7}{12}\text{⇔}\dfrac{2}{3}x=\dfrac{1}{3}\text{⇔}x=\dfrac{1}{2}\)
c)\(\dfrac{1}{3}+\dfrac{2}{5}\left(x+1\right)=1\text{⇔}\dfrac{2}{5}\left(x+1\right)=\dfrac{2}{3}\text{⇔}x+1=\dfrac{5}{3}\text{⇔}x=\dfrac{2}{3}\)
d)\(\dfrac{1}{4}+\dfrac{1}{3}:3x=-5\text{⇔}\dfrac{1}{3}:3x=-\dfrac{21}{4}\text{⇔}\dfrac{1}{9x}=-\dfrac{21}{4}\text{⇔}9x=-\dfrac{4}{21}\text{⇔}x=-\dfrac{4}{189}\)
a, \(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\)
\(\Rightarrow\dfrac{7}{4}x=5+\dfrac{10}{3}=\dfrac{25}{3}\)
\(\Rightarrow x=\dfrac{25}{3}:\dfrac{7}{4}=\dfrac{100}{21}\)
Vậy ...
b, \(PT\Leftrightarrow\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{1}{4}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
Vậy ....
c, \(PT\Leftrightarrow\dfrac{2}{5}\left(x+1\right)=1-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Rightarrow x+1=\dfrac{2}{3}:\dfrac{2}{5}=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1=\dfrac{2}{3}\)
Vậy ...
d, \(PT\Leftrightarrow\dfrac{1}{3}:3x=-5-\dfrac{1}{4}=\dfrac{1}{9}x=-\dfrac{21}{4}\)
\(\Rightarrow x=-\dfrac{189}{4}\)
Vậy ...
1) 22x + 1 = 32
=> 22x + 1 = 25
=> 2x + 1 = 5
=> 2x = 5 - 1
=> 2x = 4
=> x = 2
(2) 3.x3 - 100 = 275
=> 3x3 = 275 + 100
=> 3x3 = 375
=> x3 = 375 : 3
=> x3 = 125
=> x3 = 53
=> x = 5
(4) (x - 1)3 - 25 = 72
=> (x - 1)3 = 49 + 32
=> (x - 1)3 = 81
(xem lại đề)
5) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{x-y}{3-5}=\frac{-4}{-2}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.5=10\end{cases}}\)
Vậy ...
6) Ta có: \(\frac{x}{2}=\frac{y}{3}\) => \(\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{4}\) => \(\frac{y}{15}=\frac{z}{12}\)
=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=\frac{-49}{37}\)
=> \(\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}}\) => \(\hept{\begin{cases}x=-\frac{49}{37}\cdot10=\frac{-490}{37}\\y=-\frac{49}{37}\cdot15=-\frac{735}{37}\\z=-\frac{49}{37}\cdot12=-\frac{588}{37}\end{cases}}\)
Vậy ...
mk lm bài mà mk cho là ''khó'' nhất thôi nha
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y+z=-49\)
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\left(1\right)\)
\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
ADTC dãy tỉ số bằng nhau ta có
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=-\frac{49}{37}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{49}{37}.10=-\frac{490}{37}\\y=-\frac{49}{37}.15=-\frac{735}{37}\\z=-\frac{49}{37}.12=-\frac{588}{37}\end{cases}}}\)
a: \(\dfrac{96}{\left(x-4\right)\left(x+4\right)}+\dfrac{7+x}{4-x}=\dfrac{2x-1}{x+4}-3\)
\(\Leftrightarrow\dfrac{96}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(x+7\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{3\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(96-x^2-11x-28=2x^2-9x+4-3\left(x^2-16\right)\)
\(\Leftrightarrow-x^2-11x+68=2x^2-9x+4-3x^2+48\)
\(\Leftrightarrow-x^2-11x+68=-x^2-9x+52\)
=>-11x+68=-9x+52
=>-2x=-16
hay x=8(nhận)
b: \(\dfrac{2}{x-1}+\dfrac{3}{x-2}=\dfrac{3}{x-3}\)
\(\Leftrightarrow2\left(x-2\right)\left(x-3\right)+3\left(x-1\right)\left(x-3\right)=3\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(x^2-5x+6\right)+3\left(x^2-4x+3\right)=3\left(x^2-3x+2\right)\)
\(\Leftrightarrow2x^2-10x+12+3x^2-12x+9=3x^2-9x+6\)
\(\Leftrightarrow5x^2-22x+21-3x^2+9x-6=0\)
\(\Leftrightarrow2x^2-13x+15=0\)
\(\Leftrightarrow2x^2-10x-3x+15=0\)
=>(x-5)(2x-3)=0
=>x=5(nhận) hoặc x=3/2(nhận)
\(\frac{2}{5}x-\frac{1}{4}x=\frac{7}{3}\)
\(x\left(\frac{2}{5}-\frac{1}{4}\right)=\frac{7}{3}\)
\(x\frac{3}{20}=\frac{7}{3}\)
\(x=\frac{7}{3}:\frac{3}{20}=\frac{140}{9}\)
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