\(x+\left(13-15\right)=5+\left(10-7\right)\)

\(< =>x-2=8=>x=10\)

Ta có: \(x+\left(13-15\right)=5+10-7\)

\(\Leftrightarrow x-2=15-7=8\)

hay x=10

x + (13 - 15) = 5 + (10 - 7)

x + (13 - 15) = 5 + 3

x + (13 - 15) = 8

x + (-2) = 8

x = 8 - (-2)

x = 10

Chúc bạn học tốt!

\(\dfrac{x}{10}=\dfrac{-3}{5}\\ \Rightarrow x=\dfrac{10.\left(-3\right)}{5}=-6\\ ------------\\ x+\dfrac{5}{12}=\dfrac{-2}{3}\\ x=\dfrac{-2}{3}-\dfrac{5}{12}\\ x=\dfrac{-8}{12}-\dfrac{5}{12}\\ x=\dfrac{-13}{12}\)

`x`=`-6`

`x` = `-13/12`

x/10=-3/5

x/10=-3.2/5.2

x/10=-6/10

x=-6

x+5/12=-2/3

x=-2/3-5/12

x=8/12-5/12

x=3/12

x=1/4

\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

Câu 3                           

 a, x= 53/5                                                x= 0

\(a,\Rightarrow x=\dfrac{7}{3}+\dfrac{6}{7}=\dfrac{49+18}{21}=\dfrac{67}{21}\\ b,\Rightarrow x=\dfrac{15}{7}-\dfrac{13}{14}=\dfrac{30-13}{14}=\dfrac{17}{14}\\ c,\Rightarrow3x=\dfrac{9}{4}-\dfrac{5}{6}=\dfrac{27-10}{12}=\dfrac{17}{12}\\ \Rightarrow x=\dfrac{17}{12}\cdot\dfrac{1}{3}=\dfrac{17}{36}\)

\(x=\dfrac{7}{3}+\dfrac{6}{7}=\dfrac{67}{21}\)

\(x=\dfrac{15}{7}-\dfrac{13}{14}=\dfrac{17}{14}\)

\(x=\left(\dfrac{9}{4}-\dfrac{5}{6}\right):3=\dfrac{17}{36}\)

\(x-\dfrac{6}{7}=\dfrac{7}{3}\\ \Rightarrow x=\dfrac{7}{3}+\dfrac{6}{7}\\ \Rightarrow x=\dfrac{67}{21}\)

\(x+\dfrac{13}{14}=\dfrac{15}{7}\\ \Rightarrow x=\dfrac{15}{7}-\dfrac{13}{14}\\ \Rightarrow x=\dfrac{17}{14}\)

\(\dfrac{9}{4}-3\times x=\dfrac{5}{6}\\ \Rightarrow3\times x=\dfrac{9}{4}-\dfrac{5}{6}\\ \Rightarrow3\times x=\dfrac{17}{12}\\ \Rightarrow x=\dfrac{17}{12}:3\\ \Rightarrow x=\dfrac{17}{36}\)

xx : 5 = 83

xx = 83 x 5 = 415

xx:5=83

xx   = 83x5

xx   = 415

a. (x80x - 801).12 = 0

⇔ x80 x (- 801) = 0

⇔ -64080x = 801

⇔ x = 0

(mấy câu tiếp mik ko hiểu lắm bn viết lại rõ đề rồi mik giải tiếp)