4/1 x 13/15

= 52/15

52/15

mình có câu trl rồi nhé:

4 x 13/15 = 4 x 13/15 = 52/15 nhé!

a) x + 7 = -12

x = (-12) - 7

x = 19

b) x - 15 = -21

x = (-21) + 15

x = -6

c) 13 - x = 20

x = 13 - 20

x = -7

a, x+7=-12

\(\Leftrightarrow\) x= -19

b, x-15=-21

\(\Leftrightarrow\) x= -6

c, 13-x=20

\(\Leftrightarrow\) x=-7

a) x+7=-12\(\Leftrightarrow\) x=-12-7\(\Leftrightarrow\) x=-19

b) x-15=-21\(\Leftrightarrow\) x=-21+15\(\Leftrightarrow\) x=-6

c)13-x=20\(\Leftrightarrow\) -x=20-13\(\Leftrightarrow\)-x=7\(\Leftrightarrow\) x=-7

cái này cấp 1 cũng biết ..........

\(a.\)

\(x+7=-12\)

\(\Rightarrow x=-12-7=-19\)

Vậy \(x=-19\)

\(b.\)

\(x-15=-21\)

\(\Rightarrow x=-21+15=-6\)

Vậy \(x=-6\)

\(c.\)

\(13-x=20\)

\(\Rightarrow x=-13-20=-7\)

Vậy \(x-=-7\)

để nhận được câu trả lời nhanh và chi tiết thì bạn vui lòng chia nhỏ ra để đăng nhé! Mỗi lần chỉ nên đăng 1 - 2 câu thôi!

Bạn cho đề yêu cầu gì đã bạn

( x-1)(y+5)=28

\(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{8}{x}\)

\(\Rightarrow x^2=8\cdot2\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

2, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)

\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)

\(\Leftrightarrow12x+8-18x+12=45\)

\(\Leftrightarrow12x-18x=45-12-8\)

\(\Leftrightarrow-6x=25\)

\(\Leftrightarrow x=\dfrac{-25}{6}\)

Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)

\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(3-x\right)-x\left(3+x\right)=8x-6\)

\(\Leftrightarrow3x-x^2+6-2x-3x-x^2=8x-6\)

\(\Leftrightarrow-x^2-x^2+3x-2x-3x-8x=-6+6\)

\(\Leftrightarrow-2x^2-10x=0\)

\(\Leftrightarrow-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;5\right\}\)