Bài 2 :

1) \(x-70=-45\) 2) \(\frac{4}{7}:x=\frac{12}{28}\)

\(\Rightarrow\) \(x=-45+70\) \(\Rightarrow x=\frac{4}{7}:\frac{12}{28}\)

\(\Rightarrow\) \(x=25\) \(\Rightarrow x=\frac{4}{3}\)

Vậy \(x=25\) Vậy \(x=\frac{4}{3}\)

3) Giống câu c) ở bài 1

4) \(x-50=-35\) 5) \(\frac{4}{7}.x=\frac{11}{18}\)

\(\Rightarrow x=-35+50\) \(\Rightarrow x=\frac{11}{28}:\frac{4}{7}\)

\(\Rightarrow x=15\) \(\Rightarrow x=\frac{77}{72}\)

Vậy \(x=15\) Vậy \(x=\frac{77}{72}\)

6) \(\left(\frac{2}{3}x+2,5\right):2\frac{2}{6}=6\)

\(\Rightarrow\)\(\left(\frac{2}{3}x+2,5\right):\frac{14}{6}=6\)

\(\Rightarrow\) \(\frac{2}{3}x+2,5=6.\frac{14}{6}\)

\(\Rightarrow\frac{2}{3}x+2,5=14\)

\(\Rightarrow\frac{2}{3}x=\frac{23}{2}\)

\(\Rightarrow x=\frac{23}{2}:\frac{2}{3}\)

\(\Rightarrow x=\frac{69}{4}\)

Vậy \(x=\frac{69}{4}\)

Bài 1:

1) \(\frac{7}{5}+\frac{-8}{5}=-\frac{1}{5}\)

2) \(-\frac{6}{5}.\frac{15}{24}=-\frac{3}{4}\)

3) \(\left(\frac{2}{3}+1,5\right)-3,5:7\frac{1}{2}=\)\(\frac{13}{6}-\frac{7}{15}=\frac{17}{10}\)

4) \(\frac{5}{8}-\frac{-7}{9}=\frac{5}{8}+\frac{7}{9}=\frac{101}{72}\)

5)\(\frac{-7}{3}.\frac{12}{28}=-1\)

Tìm \(x\) câu a:

\(\frac13.x\) + \(\frac25.\left(x+1\right)\) = 0

\(\frac{5}{15}x\) + \(\frac{6}{15}x\) + \(\frac25\) = 0

\(\frac{11}{15}x\) = - \(\frac25\)

\(x=-\frac25:\frac{11}{15}\)

\(x\) = - \(\frac25\times\frac{15}{11}\)

\(x\) = - \(\frac{6}{11}\)

Vậy \(x=-\frac{6}{11}\)

Tìm \(x\) câu b:

\(x\) x 25% = 0,5

\(x\times0,25\) = 0,5

\(x=0,5:0,25\)

\(x=2\)

Vậy \(x=2\)

Câu 1a:

1/3x + 2/5(x + 1) = 0

1/3x + 2/5x + 2/5 = 0

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

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

x.(5/15 + 6/15) = -2/5

x.11/15 = - 2/5

x = - 2/5 : 11/15

x = - 6/11

Vậy x = -6/11

Câu b:

x . 25%. x = 0,5

x.x = 0,5 : 25%

x^2 = 2

x = - \(\sqrt2\); x = \(\sqrt2\)

Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )

Câu 1a:

1/3x + 2/5(x + 1) = 0

1/3x + 2/5x + 2/5 = 0

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

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

x.(5/15 + 6/15) = -2/5

x.11/15 = - 2/5

x = - 2/5 : 11/15

x = - 6/11

Vậy x = -6/11

Câu b:

x . 25%. x = 0,5

x.x = 0,5 : 25%

x^2 = 2

x = - \(\sqrt2\); x = \(\sqrt2\)

Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )

bài 1:

a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)

\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)

\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)

\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)

c) Cho mình hỏi x ở đâu vậy ???

d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)

\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)

\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)

\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)

\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)

f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)

\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)

\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)

\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)

\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)

\(x=\dfrac{-5}{7}\)

bài 2:

Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)

\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)

\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)

Vậy \(a=-15;b=81\)

Tìm x:
b) 1/3.x+2/5.(x-1)=0

\(<=> \dfrac{1}{3} .x +\dfrac{2}{5}x - \dfrac{2}{5} =0\)

\(<=> \dfrac{11}{15}x = \dfrac{2}{5}\)

\(<=> x= \dfrac{6}{11}\)

Vậy \( x= \dfrac{6}{11}\)
c) (2x-3).(6-2x)=0
\(<=> \begin{cases} 2x-3=0 \\ 6-2x=0 \end{cases}\) \(<=> \begin{cases} 2x=3 \\ -2x=-6 \end{cases}\) \(<=>\begin{cases} x=\dfrac{3}{2} \\ x=3 \end{cases}\)

Vậy \(x=( \dfrac{3}{2} ; 3)\)
d) -2/3-1/3.(2x-5)= 3/2

\(<=> 2x-5= \dfrac{5}{2}\)

\(<=> 2x= \dfrac{15}{2}\)

\(<=> x= \dfrac{15}{4}\)

Vậy \(x= \dfrac{15}{4}\)
f) 1/3.x-1/2=4 và 1/2 (Hỗn số ý '^')

\(<=> \dfrac{1}{3} x -\dfrac{1}{2} = \dfrac{9}{2}\)

\(<=> \dfrac{1}{3}x =5\)

\(<=> x= 15\)

Vậy \(x= 15\)

Cảm ơn cậu nhiều nhé!

a) Ta có: \(\frac{x+1}{3}=\frac{2}{6}\)

⇔\(x=\frac{2\cdot3}{6}-1=\frac{6}{6}-1=1-1=0\)

Vậy: x=0

b) Ta có: \(\frac{x-1}{4}=\frac{1}{-2}\)

⇔\(x=\frac{1\cdot4}{-2}+1=\frac{4}{-2}+1=-1\)

Vậy: x=-1

c) Ta có: \(\frac{-1}{6}=\frac{3}{2x}\)

⇔\(2x=\frac{3\cdot6}{-1}=-18\)

hay x=-9

Vậy: x=-9

d) Ta có: \(\frac{x+1}{3}=\frac{3}{x+1}\)

⇔\(\left(x+1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: x∈{2;-4}

e) Ta có: \(\frac{4}{5}=\frac{-12}{9-x}\)

⇔\(9-x=\frac{-12\cdot5}{4}=-15\)

hay x=24

Vậy: x=24

f) Ta có: \(\frac{x-1}{-4}=\frac{-4}{x-1}\)

⇔\(\left(x-1\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy: x∈{5;-3}

g) Ta có: \(\frac{5-x}{2}=\frac{2}{5-x}\)

⇔\(\left(5-x\right)^2=4\)

⇔\(\left[{}\begin{matrix}5-x=2\\5-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

Vậy: x∈{3;7}

h) Ta có: \(\frac{4-x}{-5}=\frac{-5}{4-x}\)

⇔\(\left(4-x\right)^2=25\)

⇔\(\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

Vậy: x∈{-1;9}

Cảm ơn bạn

\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(x-7\right)}{\left(x-4\right)\left(x-7\right)}=\frac{\left(x+5\right)\left(x-4\right)}{\left(x-4\right)\left(x-7\right)}\)

\(\Rightarrow\)\(^{x^2-7x+x-7=x^2-4x+5x-20}\)

\(\Leftrightarrow\)\(x^2-x^2-6x-x-7+20=0\)

\(\Leftrightarrow-7x+13=0\)

\(\Leftrightarrow-7x=-13\)

\(\Rightarrow x=\frac{13}{7}\)

\(pt\Leftrightarrow\left(x+1\right)\left(x-7\right)=\left(x-4\right)\left(x+5\right)\)

\(\Leftrightarrow x^2-7x+x-7=x^2+5x-4x-20\)

\(\Leftrightarrow-7x=-13\Rightarrow x=\frac{13}{7}\)