1, (x – 13)(2x +1 ) = 5 (x – 13)
2, x + 4 / x – 4 – x -4 / x+4 = x (x+16) / x’2 -16
1, (x – 13)(2x +1 ) = 5 (x – 13) 2, x + 4 / x – 4 – x -4 / x+4 = x (x+16) / x’2 -16
By Bella
By Bella
1, (x – 13)(2x +1 ) = 5 (x – 13)
2, x + 4 / x – 4 – x -4 / x+4 = x (x+16) / x’2 -16
Đáp án:
↓↓↓
Giải thích các bước giải:
`1,(x-13)(2x+1)=5(x-13)`
⇒`2x^2+x-26x-13=5x-65`
⇒`2x^2-25x-13-5x+65=0`
⇒`(x-2)(x-13)=0`
⇒`x={2;13}`
`2,(x+4)/(x-4)-(x-4)/(x+4)=(x(x+16))/(x^2-16)`
⇒`x`$\neq$ `±4`
⇒`(x+4)/(x-4)-(x-4)/(x+4)-(x^2+16x)/(x^2-16)=0`
⇒`((x+4)^2-(x-4)^2-(x^2+16x))/((x-4)(x+4))=0`
⇒`(x+4)^2-(x-4)^2-(x^2+16x)=0`
⇒`-x^2=0`
⇒`x=0`
`a)` `(x-13)(2x+1)=5(x-13)`
`<=>2x^2+x-26x-13=5x-65`
`<=>2x^2-25x-13=5x-65`
`<=>2x^2-25x-13-5x+65=0`
`<=>2x^2-30x+52=0`
`<=>2(x-15x+26)=0`
`<=>x^2-15+26=0`
`<=>x^2-2x-13x+26=0`
`<=>x(x-2)-13(x-2)=0`
`<=>(x-2)(x-13)=0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x-13=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=2\\x=13\end{array} \right.\)
Vậy phương trình có nghiệm `S={13;2}`
`b)` `frac{x+4}{x-4}-frac{x-4}{x+4}=frac{x(x+16)}{x^2-16}`
Điều kiện xác định: `x\ne4;x\ne-4`
`<=>frac{(x+4)(x+4)}{(x-4)(x+4)}-frac{(x-4)(x-4)}{(x+4)(x-4)}=frac{x(x+16)}{(x-4)(x+4)}`
`=>(x+4)(x+4)-(x-4)(x-4)=x(x+16)`
`<=>(x+4)^2-(x-4)^2=x^2+16x`
`<=>x^2+8x+16-(x^2-8x+16)=x^2+16x`
`<=>x^2+8x+16-x^2+8x-16=x^2+16x`
`<=>16x=x^2+16x`
`<=>x^2+16x-16x=0`
`<=>x^2=0`
`<=>x=0` (TMĐK)
Vậy phương trình trên có nghiệm `S={0}`