Ko hiểu đề!

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

<=>\(\left(\frac{7}{2}+2x\right).\frac{8}{3}=\frac{16}{3}\)

<=>\(\frac{28}{3}+\frac{16x}{3}=\frac{16}{3}\)

<=>\(\frac{16x}{3}=\frac{-2}{3}\)

<=>\(16x=-2\)

<=>\(x=\frac{-1}{8}\)

vậy \(x=\frac{-1}{8}\)

b,\(\left|2x+3\right|=5\)

xét x<0,ta co: \(\left|2x+3\right|=5\)<=> \(-2x+3=5\)<=>\(-2x=2\)<=>\(x=-1\)(loại)

xét x>0,ta co:\(\left|2x+3\right|=5\)<=>\(2x+3=5\)<=>\(2x=2\)<=>\(x=1\)

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

<=>\(\frac{3x-6}{12}=\frac{20+4x}{12}\)

=>\(3x-6=20+4x\)

<=>\(3x-6-20-4x=0\)

<=>\(-x-26=0\)

<=>\(-x=26\)

<=>\(x=-26\)

kl:.......

Dài quá thì sao mik học được mấy ý chính đây ?

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

\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

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

__

`(x+2)(5-3x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

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

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

__

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

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

\(\left(-24\right):\left(2-x\right)=6\)

\(\left(2-x\right)=\left(-24\right):6\)

\(\left(2-x\right)=-4\)

\(x=2-\left(-4\right)\)

\(x=6\)

(-24) :(2-x) =6

2-x = 6 x (-24)

2-x= -146

x = 2- (-146)

Vậy x = 146

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

      a) \(\frac{x}{3}\)\(-\)\(\frac{1}{8}\)\(=\)\(\frac{5}{8}\)

\(\Rightarrow\) \(\frac{x}{3}\)=\(\frac{5}{8}+\frac{1}{8}\)

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

\(\Rightarrow\)  x=3.3:4

\(\Rightarrow\) x=2,25

Vậy x=2,25

     b) \(3\frac{1}{3}\).x-\(6\frac{3}{4}=3\frac{1}{4}\)

\(\Rightarrow\)\(\frac{10}{3}\).x-\(\frac{27}{4}=\frac{13}{4}\)

\(\Rightarrow\) \(\frac{10}{3}\).x=\(\frac{13}{4}+\frac{27}{4}\)

\(\Rightarrow\) \(\frac{10}{3}\).x=10

\(\Rightarrow\) x=\(10\div\frac{10}{3}\)

\(\Rightarrow\) x=3

Vậy x=3

Ủng hộ mik nha mọi người !

Ta có : \(\frac{x}{3}-\frac{1}{8}=\frac{5}{8}\)

\(\Leftrightarrow\frac{x}{3}=\frac{5}{8}-\frac{1}{8}\)

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

\(\Leftrightarrow x=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

bai 1

\(\frac{7}{4}\)+ \(\frac{5}{6}\):5 - 0,375.2.\(^{\left(-2\right)^2}\)= \(\frac{7}{4}\)+ \(\frac{5}{6}\)x\(\frac{1}{5}\)- \(\frac{15}{4}\). 2.4=\(\frac{7}{4}\)+\(\frac{1}{6}\)-\(\frac{15}{4}\).8=\(\frac{42}{24}\)+\(\frac{4}{24}\)-30=\(\frac{11}{6}\)-30=-169/6

\(\frac{1}{4}\)+\(\frac{3}{4}\). \(\left(\frac{-1}{2}+\frac{2}{3}\right)\)=\(\frac{1}{4}\)+ \(\frac{3}{4}\).\(\left(\frac{-3}{6}+\frac{4}{6}\right)\)= \(\frac{1}{4}+\frac{3}{4}.\frac{1}{6}=\frac{1}{4}+\frac{3}{8}\)= \(\frac{5}{8}\)

tick di minh noi

mấy bạn ơi địa lý nha