\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

a) \(\dfrac{x^3-1}{x^2+x+1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)

b) \(\dfrac{x^2+2xy+y^2}{2x^2+xy-y^2}\)

\(=\dfrac{\left(x+y\right)^2}{x^2+xy+x^2-y^2}=\dfrac{\left(x+y\right)^2}{x\left(x+y\right)+\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{\left(2x-y\right)\left(x+y\right)}=\dfrac{x+y}{\left(2x-y\right)}\)

c) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x^3-a^3\right)}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x-a\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}\)

\(=ax\left(x-a\right)\)

h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)

Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)

\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)

Thay a,b:

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)

a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

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

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

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)

a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)

b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)

`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`

`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`

`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`

`<=> -55x +20 = 24x-138`

`<=> -55x -24x=-138-20`

`<=>-79x=-158`

`<=> x=2`

Vậy pt có nghiệm `x=2`

`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)

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

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

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

`<=> x^2 + x =0`

`<=>x(x+1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-1\end{matrix}\right.\)

Vậy pt có nghiệm `x=-1`

`c,2x^3 + 6x^2 =x^2 +3x`

`<=> 2x^3 + 6x^2 -x^2 -3x=0`

`<=> 2x^3 + 5x^2 -3x=0`

`->` Đề có sai ko ạ ?

`d,` \(\left|x-4\right|+3x=5\) `(1)`

Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :

`x-4 = 5-3x`

`<=> x+3x=5+4`

`<=> 4x=9`

`<=> x= 9/4 (t//m)`

Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :

`-(x-4) =5-3x`

`<=> -x +4=5-3x`

`<=> -x+3x=5-4`

`<=> 2x =1`

`<=>x=1/2 ( kt//m)`

Vậy phương trình có nghiệm `x=9/4`

đây là phương trình mà đâu phải bất phương trình đâu