chiều rộng mảnh vườn là: 12 x 3/4 = 9

a) chu vi mảnh vườn là: (12 + 9) x 2 = 42

b) diện tích mảnh vườn là: 12 x 9 = 108

c) diện tích trồng rau là: 108 x 2/3 = 72

a) Chiều rộng mảnh vườn là:

`3/4 xx 12 = 9` 

Chu vi mảnh vườn là:

(12 + 9) x 2 = 42

b) Diện tích mành vườn là:

12 x 9 = 108 

c) Diện tích trồng rau là:

`2/3 xx 108 = 72` '

ĐS: ... 

\(\dfrac{3}{8}-\dfrac{1}{2}:\left(\dfrac{2x}{5}-\dfrac{2}{3}\right)=\dfrac{-8}{5}\\ \dfrac{3}{8}-\dfrac{1}{2}:\left(\dfrac{6x}{15}-\dfrac{10}{15}\right)=-\dfrac{8}{5}\\ \dfrac{3}{8}-\dfrac{1}{2}:\dfrac{6x-10}{15}=-\dfrac{8}{5}\\ \dfrac{1}{2}:\dfrac{6x-10}{15}=\dfrac{3}{8}+\dfrac{8}{5}\\ \dfrac{1}{2}\cdot\dfrac{15}{6x-10}=\dfrac{79}{40}\\ \dfrac{15}{6x-10}=\dfrac{79}{20}\\ 6x-10=15:\dfrac{79}{20}\\ 6x-10=\dfrac{300}{79}\\ 6x=\dfrac{300}{79}+10\\ 6x=\dfrac{1090}{79}\\ x=\dfrac{1090}{79}:6\\x =\dfrac{545}{237}\)

\(\dfrac{3}{8}-\dfrac{1}{2}:\left(\dfrac{2x}{5}-\dfrac{2}{3}\right)=-\dfrac{8}{5}\\ \dfrac{1}{2}:\left(\dfrac{2x}{5}-\dfrac{2}{3}\right)=\dfrac{79}{40}\\ \dfrac{2x}{5}-\dfrac{2}{3}=\dfrac{20}{79}\\ \dfrac{2x}{5}=\dfrac{218}{237}\\ 2x=\dfrac{1090}{237}\\ x=\dfrac{545}{237}\)

a) Với `m=-1` ta có:

\(\dfrac{2x-1}{2-x}+\dfrac{2x+1}{2+x}=\dfrac{4}{4-x^2}\left(x\ne\pm2\right)\\ \Leftrightarrow\dfrac{\left(2x-1\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\dfrac{\left(2x+1\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}=\dfrac{4}{\left(2-x\right)\left(2+x\right)}\\ \Leftrightarrow\left(2x-1\right)\left(2+x\right)+\left(2x+1\right)\left(2-x\right)=4\\ \Leftrightarrow\left(4x+2x^2-2-x\right)+\left(4x-2x^2+2-x\right)=4\\ \Leftrightarrow3x+2x^2-2+3x-2x^2+2=4\\ \Leftrightarrow6x=4\\ \Leftrightarrow x=\dfrac{4}{6}\\ \Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\) 

b) Vì pt có nghiệm `x=1` nên thay `x=1` vào pt ta có:

\(\dfrac{2\cdot1+m}{2-1}+\dfrac{2\cdot1-m}{2+1}=\dfrac{4}{4-1^2}\\ \Leftrightarrow2+m+\dfrac{2-m}{3}=\dfrac{4}{3}\\ \Leftrightarrow\dfrac{3\left(2+m\right)+2-m}{3}=\dfrac{4}{3}\\ \Leftrightarrow3\left(2+m\right)+2-m=4\\ \Leftrightarrow6+3m+2-m=4\\ \Leftrightarrow8-2m=4\\ \Leftrightarrow2m=6\\ \Leftrightarrow m=3\) 

\(\dfrac{-13}{15}=\dfrac{-13}{5}\cdot\dfrac{1}{3}=\dfrac{-13}{5}:3\)

\(\dfrac{-2}{11}=-2\cdot\dfrac{1}{11}=-2:11\)

A. không có x

B. 

\(\left(x-73\right)\times10^2-26=8^2+10\\ \left(x-73\right)\times100-26=64+10\\ \left(x-73\right)\times100-26=74\\ \left(x-73\right)\times100=26+74=100\\ x-73=1\\ x=1+73=74\)

C. 

\(\left(x+1\right)^3-2^5:2^3=60\\ \left(x+1\right)^3-2^2=60\\ \left(x+1\right)^3-4=60\\ \left(x+1\right)^3=60+4=64\\ \left(x+1\right)^3=4^3\\ x+1=4\\ x=4-1\\ x=3\)

D. 

\(\left(x+1\right)^4-1=80\\ \left(x+1\right)^4=80+1\\ \left(x+1\right)^4=3^4\)

TH1: 

\(x+1=3\\ x=3-1=2\)

TH2: 

\(x+1=-3\\ x=-3-1=-4\)

A sai đề ạ

a) ĐKXĐ: `x\ne3;x\ne-1`

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

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

`\Rightarrow x(x+1)+x(x-3)=4x`

`\Leftrightarrow 2x^2-2x=4x`

`\Leftrightarrow 2x^2-6x=0`

`\Leftrightarrow 2x(x-3)=0`

\(\Leftrightarrow \left[\begin{array}{} x=0(tm)\\x=3 (ktm)\end{array} \right.\)

b) ĐKXĐ: `x\ne-2`

`\frac{3x}{x^2-2x+4}=\frac{3}{x+2}+\frac{72}{x^3+8}`

`\Leftrightarrow \frac{3x(x+2)}{(x+2)(x^2-2x+4)}=\frac{3(x^2-2x+4)}{(x+2)(x^2-2x+4)}+\frac{72}{(x+2)(x^2-2x+4)}`

`\Rightarrow 3x^2+6x=3x^2-6x+12+72`

`\Leftrightarrow 12x=84`

`\Leftrightarrow x=7(tm)`

$\mathtt{Toru}$

\(a.\dfrac{3}{4}:x=-\dfrac{1}{2}\\ x=\dfrac{3}{4}:\left(-\dfrac{1}{2}\right)\\ x=-\dfrac{3}{2}\\ b.x:\left(-\dfrac{2}{3}\right)=-\dfrac{5}{6}\\ x=\left(-\dfrac{5}{6}\right)\cdot\left(-\dfrac{2}{3}\right)\\ x=\dfrac{5}{9}\\ c.\dfrac{1}{2}:x-\dfrac{1}{2}=-\dfrac{3}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}\\ x=\dfrac{1}{2}:\left(-\dfrac{1}{4}\right)\\ x=-2\\ d.\dfrac{1}{2}\cdot x+\dfrac{3}{4}=-\dfrac{5}{6}\\ \dfrac{1}{2}\cdot x=-\dfrac{19}{12}\\ x=\dfrac{-19}{12}:\dfrac{1}{2}=-\dfrac{19}{12}\cdot2\\ x=-\dfrac{19}{6}\)

\(e.x:\dfrac{1}{2}+\dfrac{1}{2}=-\dfrac{5}{6}\\ x:\dfrac{1}{2}=-\dfrac{4}{3}\\ x=-\dfrac{4}{3}\cdot\dfrac{1}{2}\\ x=-\dfrac{2}{3}\\ g.\dfrac{4}{3}-\dfrac{1}{2}\cdot x=\dfrac{1}{2}\\ \dfrac{1}{2}\cdot x=\dfrac{5}{6}\\ x=\dfrac{5}{6}:\dfrac{1}{2}=\dfrac{5}{6}\cdot2\\ x=\dfrac{5}{3}\\ h.\dfrac{3}{8}\cdot x-\dfrac{1}{2}=\dfrac{3}{4}\\ \dfrac{3}{8}\cdot x=\dfrac{1}{4}\\ x=\dfrac{1}{4}:\dfrac{3}{8}=\dfrac{1}{4}\cdot\dfrac{8}{3}\\ x=\dfrac{2}{3}\)

mình làm lại câu h xíu

\(\dfrac{3}{8}\cdot x-\dfrac{1}{2}=\dfrac{3}{4}\\ \dfrac{3}{8}\cdot x=\dfrac{5}{4}\\ x=\dfrac{5}{4}:\dfrac{3}{8}=\dfrac{5}{4}\cdot\dfrac{8}{3}\\ x=\dfrac{10}{3}\)

a) 

\(\dfrac{4}{x+3}-\dfrac{3}{x-5}=0\left(x\ne-3;x\ne5\right)\\ \Leftrightarrow\dfrac{4\left(x-5\right)}{\left(x+3\right)\left(x-5\right)}-\dfrac{3\left(x+3\right)}{\left(x+3\right)\left(x-5\right)}=0\\ \Leftrightarrow4\left(x-5\right)-3\left(x+3\right)=0\\ \Leftrightarrow4x-20-3x-9=0\\ \Leftrightarrow x-29=0\\ \Leftrightarrow x=29\left(tm\right)\)

b) 

\(\dfrac{1}{x+2}-\dfrac{1}{x-2}=\dfrac{3x-12}{x^2-4}\left(x\ne\pm2\right)\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x-2}=\dfrac{3x-12}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x-12}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow x-2-x-2=3x-12\\ \Leftrightarrow-4=3x-12\\ \Leftrightarrow3x=-4+12\\ \Leftrightarrow3x=8\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)

c) 

\(\dfrac{1}{x+1}-\dfrac{4}{x^2-x+1}=\dfrac{2x^2+1}{x^3+1}\left(x\ne-1\right)\\ \Leftrightarrow\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x^2+1}{\left(x+1\right)\left(x^2-x+1\right)}\\ \Leftrightarrow x^2-x+1-4x-4=2x^2+1\\ \Leftrightarrow x^2-5x-3=2x^2+1\\ \Leftrightarrow2x^2-x^2+5x+1+3=0\\ \Leftrightarrow x^2+5x+4=0\\ \Leftrightarrow\left(x+1\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)

a) Đặt: `4x-5=t`

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

\(\Rightarrow4x-5=-1\\ \Leftrightarrow4x=-1+5\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\) 

b) Đặt: \(x^2-x=t\)

\(t\left(t+1\right)=6\\ \Leftrightarrow t^2+t-6=0\\ \Leftrightarrow t^2-2t+3t-6=0\\ \Leftrightarrow t\left(t-2\right)+3\left(t-2\right)=0\\ \Leftrightarrow\left(t-2\right)\left(t+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\)

Với: 

\(t=2\Rightarrow x^2-x=2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Với: 

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

Mà: `x^2-x+3>0` nên vô lý