a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).

Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$

$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$

$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$

$\Leftrightarrow 13x+15=2$

$\Leftrightarrow 13x=2-15=-13$

$\Leftrightarrow x=-13:13=-1$

Bài 2:

$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:

$(y+4)y=5$

$\Leftrightarrow y^2+4y-5=0$

$\Leftrightarrow (y-1)(y+5)=0$

$\Leftrightarrow y=1$ hoặc $y=-5$

Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$

Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$

k mk đi

ai k mk 

mk k lại

thanks

không ai trả lời 

a,\(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)

\(< =>6x-2-5x+15-18x+36=24\)

\(< =>-29x+49=24< =>29x=25< =>x=\frac{25}{29}\)

b,\(2x^2+4\left(x^2-1\right)=2x\left(3x+1\right)\)

\(< =>2x^2+4x^2-4=6x^2+2x\)

\(< =>2x=-4< =>x=-\frac{4}{2}=-2\)

c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)

\(< =>10x-6x^2+6x^2-10x-3x+21=4\)

\(< =>-3x=4-21=-17< =>x=\frac{17}{3}\)

d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)

\(< =>5x^2+5x-4x^2-8x=1-x\)

\(< =>x^2-3x+x-1=0\)

\(< =>x^2-2x-1=0\)

\(< =>\left(x-1\right)^2=2\)

\(< =>\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{cases}}\)

Sai rồi bn:)

a, \(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)

\(\Leftrightarrow6x-2-5x+15-18x+36=24\)

\(\Leftrightarrow-17x+25=0\Leftrightarrow x=\frac{25}{17}\)

b, \(2x^2+4\left(x^2-1\right)=2x\left(3x+1\right)\)

\(\Leftrightarrow2x^2+4x^2-4=6x^2+2x\)

\(\Leftrightarrow-4-2x=0\Leftrightarrow x=-2\)

c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)

\(\Leftrightarrow10x-6x^2+6x^2-10x-3x+21=4\)

\(\Leftrightarrow-3x+17=0\Leftrightarrow x=\frac{17}{3}\)

d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)

\(\Leftrightarrow5x^2+5x-4x^2-8x=1-x\)

\(\Leftrightarrow x^2-3x-1+x=0\Leftrightarrow x^2-2x-1=0\)

\(\Leftrightarrow x^2-2x+1-2=0\Leftrightarrow\left(x-1\right)^2=2\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{cases}}}\)

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

tìm x biết:

(3x-1) [- 1/2x+5]=0

1/4+1/3:(2x-1)=-5

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

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3(3x-1/2)^3+1/9=0

\(x\div4\frac{1}{3}=-2,5\)

\(\Leftrightarrow x\div\frac{13}{3}=\frac{-5}{2}\)

\(\Leftrightarrow x=\frac{-5}{2}.\frac{13}{3}\)

\(\Leftrightarrow x=\frac{-65}{6}\)

\(x\div\frac{-3}{5}=\frac{-10}{21}\)

\(\Leftrightarrow x=\frac{-10}{21}.\frac{-3}{5}\)

\(\Leftrightarrow x=\frac{30}{105}\)

\(\Leftrightarrow x=\frac{2}{7}\)

\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}+\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{6}{10}\)

\(\Leftrightarrow x=\frac{6}{10}\div\frac{2}{3}\)

\(\Leftrightarrow x=\frac{18}{20}\)

\(\Leftrightarrow x=\frac{9}{10}\)

\(\frac{1}{2}x+\frac{1}{2}=\frac{5}{2}\)

\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=\frac{5}{2}\)

\(\Leftrightarrow\left(x+1\right)=\frac{5}{2}\div\frac{1}{2}\)

\(\Leftrightarrow\left(x+1\right)=5\)

\(\Leftrightarrow x=5-1\)

\(\Leftrightarrow x=4\)

\(\frac{-2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)

\(\Leftrightarrow\frac{-2}{3}-\frac{2x}{3}+\frac{5}{3}=\frac{3}{2}\)

\(\Leftrightarrow1-\frac{2x}{3}=\frac{3}{2}\)

\(\Leftrightarrow\frac{2x}{3}=1-\frac{3}{2}\)

\(\Leftrightarrow\frac{2x}{3}=\frac{-1}{2}\)

\(\Leftrightarrow x=\frac{-1}{2}\div\frac{2}{3}\)

\(\Leftrightarrow x=\frac{-3}{4}\)

\(\frac{2}{5}x+\frac{1}{2}=\frac{-3}{4}\)

\(\Leftrightarrow\frac{2}{5}x=\frac{-3}{4}-\frac{1}{2}\)

\(\Leftrightarrow\frac{2}{5}x=\frac{-5}{4}\)

\(\Leftrightarrow x=\frac{-5}{4}\div\frac{2}{5}\)

\(\Leftrightarrow x=\frac{-25}{8}\)