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1 tháng 5 2023

a. \(\dfrac{3}{4}\) + \(\dfrac{-4}{5}\) - \(\dfrac{1}{2}\) = \(\dfrac{-1}{20}\)  - \(\dfrac{1}{2}\) = \(\dfrac{-11}{20}\) 

b. (4 - \(\dfrac{5}{12}\) ): 2 + \(\dfrac{5}{24}\) 

\(\dfrac{43}{12}\) : 2 + \(\dfrac{5}{24}\) 

\(\dfrac{43}{24}\) + \(\dfrac{5}{24}\) 

=\(\dfrac{48}{24}\) = 2

2.2 

\(\dfrac{4}{7}\) .x - \(\dfrac{2}{3}\) = \(\dfrac{1}{5}\) 

  \(\dfrac{4}{7}\) x   = \(\dfrac{1}{5}\) + \(\dfrac{2}{3}\) 

  \(\dfrac{4}{7}\) . x = \(\dfrac{13}{15}\) 

        x = \(\dfrac{91}{60}\)

1 tháng 5 2023

2.1

\(a)\dfrac{3}{4}+\dfrac{-4}{5}-\dfrac{1}{2}\\ =\dfrac{3\times5}{4\times5}+\dfrac{-4\times4}{5\times4}-\dfrac{1\times10}{2\times10}\\ =\dfrac{15}{20}+\dfrac{-16}{20}-\dfrac{10}{20}\\ =\dfrac{15-16-10}{20}\\ =\dfrac{-11}{20}\)

\(b)\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{4\times12}{1\times12}-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{48}{12}-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{48-5}{12}\right):2+\dfrac{5}{24}\\ =\dfrac{43}{12}:2+\dfrac{5}{24}\\ =\dfrac{43}{12}\times\dfrac{1}{2}+\dfrac{5}{24}\\ =\dfrac{43}{24}+\dfrac{5}{24}\\ =\dfrac{43+5}{24}\\ =\dfrac{48}{24}\\ =2\)

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

12 tháng 7 2021

ai giúp mik vs

23 tháng 8 2018

1) = \(\frac{3}{5}\)

2) =\(\frac{6}{7}\)

3)\(\frac{9}{13}\)

4)\(\frac{4}{13}\)

20 tháng 3 2022

\(a,x=\dfrac{1}{2}-\dfrac{2}{5}\)

\(x=\dfrac{1}{10}\)

\(b,x+\dfrac{3}{7}=\dfrac{7}{10}\)

\(x=\dfrac{7}{10}-\dfrac{3}{7}\)

\(x=\dfrac{19}{70}\)

\(c,19-x=\dfrac{17}{20}\)

\(x=19-\dfrac{17}{20}\)

\(x=\dfrac{363}{20}\)

20 tháng 3 2022

a,1/10

b,19/70

c,18/4

26 tháng 8 2021

`a)sqrt{1-4x+4x^2}+5=x-2`

`<=>\sqrt{(2x-1)^2}=x-2-5`

`<=>|2x-1|=x-7(x>=7)`

`<=>[(2x-1=x-7),(2x-1=7-x):}`

`<=>[(x=-6(ktm)),(3x=8):}`

`<=>x=8/3(ktm)`

Vậy PTVN

`b)3sqrt{12+4x}+4/7sqrt{147+49x}=3/2sqrt{48+16x}+4(x>=-3)`

`<=>6sqrt{x+3}+4sqrt{x+3}=6sqrt{x+3}+4`

`<=>4sqrt{x+3}=4`

`<=>sqrt{x+3}=1<=>x+3=1`

`<=>x=-2(tm)`

Vậy `S={-2}`

26 tháng 8 2021

a) \(\sqrt{1-4x+4x^2}+5=x-2\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\Leftrightarrow\left|1-2x\right|=x-7\left(1\right)\)TH1: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow1-2x=x-7\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(không thỏa đk)

TH2: \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow2x-1=x-7\Leftrightarrow x=-6\)(không thỏa đk)

Vậy \(S=\varnothing\)

b) \(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}=6\sqrt{3+x}+4\Leftrightarrow4\sqrt{3+x}=4\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\)

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)