x^2 -7x +7
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\(A=\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
\(A=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
\(A=\dfrac{x^2-\left(x-7\right)^2}{x\left(x-7\right)\left(x+7\right)}\times\dfrac{x^2+7x}{2x-7}+\dfrac{x}{7-x}\)
\(A=\dfrac{x^2-x^2+14x-49}{x\left(x-7\right)\left(x+7\right)}\times\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)
\(A=\dfrac{7\left(2x-7\right)}{x\left(x-7\right)\left(x+7\right)}\times\dfrac{x\left(x+7\right)}{2x-7}+\dfrac{x}{7-x}\)
\(A=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)
\(A=\dfrac{7-x}{x-7}\)
\(A=-\dfrac{7-x}{7-x}\)
\(A=-1\)
\(A=\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
\(A=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right).\dfrac{x^2+7x}{2x-7}-\dfrac{x}{x-7}\)
\(A=\dfrac{x^2-\left(x-7\right)^2}{x\left(x-7\right)\left(x+7\right)}.\dfrac{x\left(x+7\right)}{2x-7}-\dfrac{x}{x-7}\)
\(A=\dfrac{\left(x-x+7\right)\left(x+x-7\right)}{x-7}.\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)
\(A=\dfrac{\left(x-x+7\right)\left(2x-7\right)}{x-7}.\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)
\(A=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)
\(A=\dfrac{7-x}{x-7}=\dfrac{-\left(x-7\right)}{x-7}=-1\)

1: =>|x-5|=5-7x+7x+28=33
=>x-5=33 hoặc x-5=-33
=>x=38 hoặc x=-28
3: 2|x-6|+7x-2=|x-6|+7x
=>|x-6|=2
=>x-6=2 hoặc x-6=-2
=>x=8 hoặc x=4

Mấy bài thực hiện phép tính bn tự làm nha
3.|x+4|-2.(x-1)=7-2x
3.|x-4|-2x-2=7-2x
=>2.|x-4| =7-2x+2x+2
=>2.|x-4| =9-4x
=>|x-4| =4,5-2x
=>x-4=4,5-2x hoặc x-4=-4,5+2x
2+2x=4,5+4 4,5-4=2x-2
3x =8,5 0,5 =x
x =8,5:3
x =2,833
Vậy x\(\in\){2,833;0,5}
Phần còn lại bn làm tương tự nha
Chúc bn học tốt

`4)x^2-5x+6`
`=x^2-2x-3x+6`
`=x(x-2)-3(x-2)=(x-2)(x-3)`
`5)x^2+7x+10`
`=x^2+5x+2x+10`
`=x(x+5)+2(x+5)=(x+5)(x+2)`
`6)x+7\sqrt{x}+10` `ĐK: x >= 0`
`=(\sqrt{x})^2+5\sqrt{x}+2\sqrt{x}+10`
`=\sqrt{x}(\sqrt{x}+5)+2(\sqrt{x}+5)=(\sqrt{x}+5)(\sqrt{x}+2)`
`7)3x^4+7x^2+4`
`=3x^4+3x^2+4x^2+4`
`=3x^2(x^2+1)+4(x^2+1)=(x^2+1)(3x^2+4)`
`8)x^2-x-2`
`=x^2-2x+x-2`
`=x(x-2)+(x-2)=(x-2)(x+1)`
`9)x^6-x^3-2`
`=x^6+x^3-2x^3-2`
`=x^3(x^3+1)-2(x^3+1)`
`=(x^3+1)(x^3-2)`.

1.
2|x-6|+7x-2=|x-6|+7x
2|x-6| - |x-6|=7x-(7x-2)
|x-6| = 2
=>x-6 = +2
*x-6=2 *x-6 = -2
x =2+6 x = (-2)+6
x =8 x = 4
2.
|x-5|-7(x+4)=5-7x
|x-5|-7x-28 =5-7x
|x-5|-28 =5-7x+7x
|x-5|-28 = 5
|x-5| = 5+28
|x-5| = 33
=>x-5 = +33
*x-5=33 *x-5=-33
x =38 x = -28
3.
3|x+4|-2(x-1)=7-2x
3|x+4|-2x+2 =7-2x
3|x+4|-2 =7-2x+2x
3|x+4|-2 =7
3|x+4| =7+2
3|x+4| = 9
|x+4| =9:3
|x+4| = 3
=>x+4 = +3
*x+4=3 *x+4=-3
x =-1 x = -7

\(1,\\ a,=7x^3-49x^2+21x\\ b,=x^2-x-42\\ c,=x^2-16x+64\\ d,=9x^2+12x+4\\ e,=x^2-16-25+10x-x^2=10x-41\\ 2,\\ a,\Rightarrow2\left(x-7\right)=19\\ \Rightarrow x-7=\dfrac{19}{2}\Rightarrow x=\dfrac{33}{2}\\ b,\Rightarrow4x^2-20x+25-4x^2+3x-2x=50\\ \Rightarrow-19x=25\Rightarrow x=-\dfrac{25}{19}\)

1, \(A=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(A=5x^3-15x+7x^2-5x^3-7x^2\)
\(A=\left(5x^3-5x^3\right)+\left(7x^2-7x^2\right)-15x\)
\(A=-15x\)
Thay \(x=-5\) vào A ta được:
\(-15\cdot-5=75\)
Vậy: ....
2. \(B=x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(B=x^3-3x+7x^2-5x^3-7x^2\)
\(B=\left(x^3-5x^3\right)+\left(7x^2-7x^2\right)-3x\)
\(B=-4x^3-3x\)
Thay \(x=10,y=-1\) vào B ta được:
\(-4\cdot10^3-3\cdot10=-4\cdot1000-3\cdot10=-4000-30=-4030\)
Vậy: ....

a) ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-2x=-2+9\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\dfrac{7}{19}\)
Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)

7 \(\times\) ( 2\(x\) - 5) - 5 \(\times\) (7\(x\) - 2) + 2 \(\times\) (5\(x\) - 7) = (\(x\) - 2) - (\(x\) +4)
14\(x\) - 35 - 35\(x\) + 10 + 10\(x\) - 14 = \(x\) - 2 - \(x\) - 4
(14\(x\) - 35\(x\) + 10\(x\)) - (35 - 10+ 14) = -6
(- 21 \(x\) + 10\(x\)) - (25 + 14) = - 6
-11\(x\) - 39 = - 6
-11\(x\) = - 6 + 39
- 11\(x\) = 33
\(x\) = 33 : (-11)
\(x\) = - 3
14x - 35 -35x + 10 + 10x - 14 = x-2-x-4
-11x -39 = -6
11x = -33
x= -3

a)\(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow2\cdot\left(x^3+1\right)+7x\cdot\left(x+1\right)=0\)
\(\Leftrightarrow2\cdot\left(x+1\right)\cdot\left(x^2+x+1\right)+7x\cdot\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left[2\cdot\left(x^2+x+1\right)+7x\right]=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(2x^2-2x+2+7x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(2x^2+5x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(2x+1\right)\cdot\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{-1}{2}\\x=-2\end{matrix}\right.\)
b)\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)
\(\Leftrightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\cdot\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Rightarrow x+66=0\)
\(\Rightarrow x=-66\)
Yêu cầu là gì thế bn
chắc là tìm x đấy ah.