a: TH1: x>=2

=>x-2=x+1

=>-2=1(loại)

TH2: x<2

=>2-x=x+1

=>-2x=-1

=>x=1/2(nhận)

b: Th1: x>=1

=>2x-2=x+2

=>x=4(nhận)

TH2: x<1

=>2-2x=x+2

=>-3x=0

=>x=0(nhận)

c: =>|2x-3|=3

=>2x-3=3 hoặc 2x-3=-3

=>x=0 hoặc x=3

d: =>3x+1=2x-3 hoặc 3x+1=3-2x

=>x=-4 hoặc x=2/5

e: =>2x+1=x+3 hoặc 2x+1=-x-3

=>x=2 hoặc x=-4/3

giúp mik với.Mình cảm ơn ạ

Dựa vào biểu đồ ven ta có: \(A=\left\{1;2;3;4;5\right\}\) ; \(B=\left\{1;2;3;6;9\right\}\)

`{([2x-3]/[y+2]-[3y]/[x+1]=5),([4x-6]/[y+2]+[5y]/[x+1]=3):}`  `ĐK: x \ne -1,y \ne -2`

`<=>{([2x-3]/[y+2]-3[y]/[x+1]=5),(2[2x-3]/[y+2]+5[y]/[x+1]=3):}`

Đặt `[2x-3]/[y+2]=a;y/[x+1]b` khi đó hệ ptr có dạng:

    `{(a-3b=5),(2a+5b=3):}`

`<=>{(a=34/11),(b=-7/11):}`

  `=>{([2x-3]/[y+2]=34/11),(y/[x+1]=-7/11):}`

`<=>{(22x-33=34y+68),(11y=-7x-7):}`

`<=>{(22x-34y=101),(7x+11y=-7):}`

`<=>{(x=291/160),(y=-287/160):}`   (t/m)

\(\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}-\dfrac{3y}{x+1}=5\\\dfrac{4x-6}{y+2}+\dfrac{5y}{x+1}=3\end{matrix}\right.\)

DKXD: \(x\ne1,y\ne-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}-\dfrac{3y}{x+1}=5\\\dfrac{2\left(2x-3\right)}{y+2}+\dfrac{5y}{x+1}=3\end{matrix}\right.\)

Đặt \(\dfrac{2x-3}{y+2}=a,\dfrac{y}{x+1}=b\)

Ta có hệ pt: \(\left\{{}\begin{matrix}a-3b=5\\2a+5b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{34}{11}\\b=-\dfrac{7}{11}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x-3}{y+2}=\dfrac{34}{11}\\\dfrac{y}{x+1}=-\dfrac{7}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11\left(2x-3\right)=34\left(y+2\right)\\-7\left(x+1\right)=11y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}22x-33=34y+68\\-7x-7=11y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}22x-34y=101\\-7x-11y=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{97}{42}\\y=-\dfrac{31}{21}\end{matrix}\right.\) (tmdk)

Thay `x=1` và `x=2` vào `x^3-mx^2+nx-2=0` có:

    `{(1-m+n-2=0),(8-4m+2n-2=0):}`

`<=>{(-m+n=1),(-2m+n=-3):}`

`<=>{(m=4),(-4+n=1):}`

`<=>{(m=4),(n=5):}`

Thế \(x=1\) và \(x=2\) vào \(x^3-mx^2+nx-2=0\) ta được:

\(\left\{{}\begin{matrix}1-m+n-2=0\\8-4m+2n-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-m+n=1\\-4m+2n=-6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m=4\\n=5\end{matrix}\right.\)

\(A\cap B=\varnothing\Leftrightarrow\left\{{}\begin{matrix}n\le3\\m\ge4\end{matrix}\right.\)

Đặt `{(sqrt (x-2) = a), (sqrt (x+2) = b):}`

`-> ab = sqrt(x^2 - 4)`.

Ptr trở thành:

`a - b = 2ab - (a^2 + b^2) + 2`

`a - b + a^2 + b^2 - 2ab - 2 =0`

`<=> (a-b)^2 + (a-b) - 2 = 0`

Đặt `a - b = t`.

`-> t^2 + t - 2 = 0`

`<=> (t-1)(t+2) = 0`

`->` \(\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)

`+, t = 1 -> a - b = 1 -> sqrt(x-2) - sqrt(x+2) = 1`

`-> sqrt(x-2) = 1 + sqrt(x+2)`

`-> x - 2 = 3 + x + 2sqrt(x+2)`

`-> sqrt(x+2) = -2,5`

`-> x = cancel O`.

`+, t = -2 -> sqrt(x-2) - sqrt(x+2) = -2`

`<=> sqrt(x-2) + 2 = sqrt(x+2)`

`<=> x + 2 + 4sqrt(x-2) = x + 2`

`<=> 4 sqrt(x-2) = 0`

`<=> sqrt(x-2) = 0`

`<=> x = 2`.

Vậy pt có một nghiệm duy nhất là `2`

a: \(=\dfrac{27}{7}:\dfrac{21}{5}=\dfrac{27}{7}\cdot\dfrac{5}{21}=\dfrac{5}{7}\cdot\dfrac{9}{7}=\dfrac{45}{49}\)

b: \(=\dfrac{13}{2}+\dfrac{13}{5}=\dfrac{65+26}{10}=\dfrac{91}{10}\)

\(a,A\cup B=\left\{1;2;3;4;5;6\right\}\)

\(b,A\cap B=\left\{3;4\right\}\)

c, A\B= \(\left\{1;2\right\}\)

d, B \ A =\(\left\{5;6\right\}\)

a) \(A\cup B=\left\{1;2;3;4;5;6\right\}\)

b) \(A\cap B=\left\{3;4\right\}\)

\(a,A\cup B=\left(-\infty;+\infty\right)\\ b,A\cap B=[3;6)\\ \)

c, \(A\) \ \(B\) = \((-\infty;3)\)

d, \(B\) \ \(A\) \(=[6;+\infty)\)