PP chung ở cả 3 câu,nói ngắn gọn nhé:

Chứng mình x khác 0,hay nói cách khác x=0 không là nghiệm của phương trình.

Chia cả tử và mẫu cho x ,rồi giải bình thường bằng cách đặt ẩn phụ.

Vd ở câu a>>>4/(4x-8+7/x)+3/(4x-10+7/x)=1.Sau đó đặt 4x+7/x=a>>>4/(a-8)+3/(a-10)=1>>>giải bình thường,các câu sau tương tự

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\left(DK:x\ne-\frac{2}{5};x\ne-\frac{1}{5}\right)\)

\(\Rightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\Rightarrow20x^2+34x+6=20x^2+33x+10\Rightarrow x=4\)(thoả mãn)

Vậy x = 4

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(\Leftrightarrow2x\left(10x+2\right)+3\left(10x+2\right)=5x\left(4x+5\right)\)

\(\Leftrightarrow20x^2+4x+20x+6=20x^2+25x+9x+10\)

\(\Leftrightarrow20x^2+4x+20x+6-\left(20x^2+25x+9x+10\right)=0\)\(\Rightarrow20x^2+24x+6-\left(20x^2+34x+10\right)=0\)

\(\Leftrightarrow-10x-4=0\)

\(\Leftrightarrow-10x=4\)

\(\Leftrightarrow x=-\frac{4}{10}\)

Mình nhầm 

2x+3/5x+2=4x+5/10x+2

=> (2x+3)(10x+2)=(5x+2)(4x+5)

=> 20x^2+4x+30x+6=10x^2+25x+8x+10 ( Vì cả hai vế đều có 10x^2 nên ta xóa đi )

=> 34x+6=33x+10

=> 34x-33x=-6+10

=> x=4

\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)

\(a,5x\left(x-2020\right)+x=2020\)

\(< =>5x\left(x-2020\right)+x-2020=0\)

\(< =>\left(5x+1\right)\left(x-2020\right)=0\)

\(< =>\orbr{\begin{cases}5x+1=0\\x-2020=0\end{cases}}\)

\(< =>\orbr{\begin{cases}5x=-1\\x=2020\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=2020\end{cases}}}\)

\(b,4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(< =>4\left(x^2-20x+25\right)-\left(4x^2+4x+1\right)=0\)

\(< =>4x^2-80x+100-4x^2-4x-1=0\)

\(< =>-84x+99=0< =>84x=99< =>x=\frac{99}{84}\)

2x+35x+2=2(2x+3)2(5x+2)=4x+610x+4" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">2x+35x+2=2(2x+3)2(5x+2)=4x+610x+4

4x+610x+4=4x+510x+2=4x+6&#x2212;4x&#x2212;510x+4&#x2212;10x&#x2212;2=12" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-table; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">4x+610x+4=4x+510x+2=4x+6−4x−510x+4−10x−2=12

2x+35x+2=12&#x21D4;5x+2=4x+6&#x21D4;5x=4x+4&#x21D4;x=4" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">2x+35x+2=12⇔5x+2=4x+6⇔5x=4x+4⇔x=4

a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\left(ĐK:x\ne0;x\ne-4\right)\)

\(=\frac{6}{x\left(x+4\right)}+\frac{3}{2\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(4+x\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)

b) \(\frac{4xy-5}{140x^3y}-\frac{6y^2-5}{10x^3y}\left(ĐK:x,y\ne0\right)\)

\(=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{2y\left(2x-3y\right)}{10x^3y}=\frac{2x-3y}{5x^3}\)

\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)

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

\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )

\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)

\(c,\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\) (x khác 1 ; khác -1)

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

(2x + 3)(10x +2) = (5x +2 )(4x + 5) 

20x^2 + 30x +4x +6 = 20x^2 +8x +25x +10

20x^2 + 34x +6 = 20x^2 + 33x +10

x= 4

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(\Leftrightarrow4x+30x-25x-8x=10-6\)

\(\Leftrightarrow x=4\)

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+8x+25x+10\)

\(\Leftrightarrow20x^2-20x^2+4x+30x-8x-25x=10-6\)

\(\Leftrightarrow x=4\)