(4x+1)(12x-1)(3x+2)(x+1)-4=0 mọi người giải giúp em với ạ 03/07/2021 Bởi Adalynn (4x+1)(12x-1)(3x+2)(x+1)-4=0 mọi người giải giúp em với ạ
Đáp án + Giải thích các bước giải: `(4x+1)(12x-1)(3x+2)(x+1)-4=0` `<=> [(4x+1)(3x+2)][(12x-1)(x+1)]-4=0` `<=> (12x^2+11x+2)(12x^2+11x-1)-4=0` Đặt `a=12x^2+11x+2` (với `a>=0`) phương trình trở thành : `a.(a+3)-4=0` `<=> a^2+3a-4=0` `<=> a^2-a+4a-4=0` `<=> a.(a-1)+4.(a-1)=0` `<=> (a+4).(a-1)=0` `<=>` \(\left[ \begin{array}{l}a=-4 \ \ \rm (ktm)\\a=1 \ \ \rm ™\end{array} \right.\) `=> 12x^2+11x+2=1` `<=> 12x^2+11x+1=0` `<=> 12.(x^2+11/12x+1/12)=0` `<=> x^2+11/12x+1/12=0` `<=> x^2 + 2 . x . 11/24 + (11/24)^2+1/22-(11/24)^2=0` `<=> (x+11/24)^2-73/576=0` `<=> (x+11/24)^2=73/576` `<=>` \(\left[ \begin{array}{l}x+\dfrac{11}{24}=\dfrac{\sqrt{73}}{24}\\x+\dfrac{11}{24}=-\dfrac{\sqrt{73}}{24}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{73}-11}{24}\\x=\dfrac{-\sqrt{73}-11}{24}\end{array} \right.\) Vậy `S={(sqrt(73)-11)/24;(-sqrt(73)-11)/24}` Bình luận
Đáp án + Giải thích các bước giải: `(4x+1)(12x-1)(3x+2)(x+1)-4=0` `<=>(4x+1)(3x+2)(12x-1)(x+1)-4=0` `<=>(12x^2+11x+2)(12x^2+11x-1)-4=0` Đặt `12x^2+11x+2=t(t>=0)` `<=>t(t+3)-4=0` `<=>t^2+3t-4=0` `<=>t^2-t+4t-4=0` `<=>t(t-1)+4(t-1)=0` `<=>(t-1)(t+4)=0` `<=>`\(\left[ \begin{array}{l}t=1(tm)\\t=-4(ktm)\end{array} \right.\) Do đó : `12x^2+11x+2=1<=>12x^2+11x+1=0` `<=>12(x^2+11/12x+1/12)=0` `<=>12[x^2+2*x*11/24+(11/24)^2]-73/48=0` `<=>12(x+11/24)^2=73/48` `<=>(x+11/24)^2=73/576` `<=>` `(x+11/24)^2=pm(sqrt73)/24` TH1 : `x + 11/24=(sqrt73)/24=>x=(sqrt73)/24-11/24=(sqrt73-11)/24` TH2 : `x + 11/24=-(sqrt73)/24<=>x=-(sqrt73)/24-11/24=-(sqrt73-11)/24` Vậy `S={(sqrt73-11)/24,-(sqrt73-11)/24}` Bình luận
Đáp án + Giải thích các bước giải:
`(4x+1)(12x-1)(3x+2)(x+1)-4=0`
`<=> [(4x+1)(3x+2)][(12x-1)(x+1)]-4=0`
`<=> (12x^2+11x+2)(12x^2+11x-1)-4=0`
Đặt `a=12x^2+11x+2` (với `a>=0`) phương trình trở thành :
`a.(a+3)-4=0`
`<=> a^2+3a-4=0`
`<=> a^2-a+4a-4=0`
`<=> a.(a-1)+4.(a-1)=0`
`<=> (a+4).(a-1)=0`
`<=>` \(\left[ \begin{array}{l}a=-4 \ \ \rm (ktm)\\a=1 \ \ \rm ™\end{array} \right.\)
`=> 12x^2+11x+2=1`
`<=> 12x^2+11x+1=0`
`<=> 12.(x^2+11/12x+1/12)=0`
`<=> x^2+11/12x+1/12=0`
`<=> x^2 + 2 . x . 11/24 + (11/24)^2+1/22-(11/24)^2=0`
`<=> (x+11/24)^2-73/576=0`
`<=> (x+11/24)^2=73/576`
`<=>` \(\left[ \begin{array}{l}x+\dfrac{11}{24}=\dfrac{\sqrt{73}}{24}\\x+\dfrac{11}{24}=-\dfrac{\sqrt{73}}{24}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{73}-11}{24}\\x=\dfrac{-\sqrt{73}-11}{24}\end{array} \right.\)
Vậy `S={(sqrt(73)-11)/24;(-sqrt(73)-11)/24}`
Đáp án + Giải thích các bước giải:
`(4x+1)(12x-1)(3x+2)(x+1)-4=0`
`<=>(4x+1)(3x+2)(12x-1)(x+1)-4=0`
`<=>(12x^2+11x+2)(12x^2+11x-1)-4=0`
Đặt `12x^2+11x+2=t(t>=0)`
`<=>t(t+3)-4=0`
`<=>t^2+3t-4=0`
`<=>t^2-t+4t-4=0`
`<=>t(t-1)+4(t-1)=0`
`<=>(t-1)(t+4)=0`
`<=>`\(\left[ \begin{array}{l}t=1(tm)\\t=-4(ktm)\end{array} \right.\)
Do đó : `12x^2+11x+2=1<=>12x^2+11x+1=0`
`<=>12(x^2+11/12x+1/12)=0`
`<=>12[x^2+2*x*11/24+(11/24)^2]-73/48=0`
`<=>12(x+11/24)^2=73/48`
`<=>(x+11/24)^2=73/576`
`<=>` `(x+11/24)^2=pm(sqrt73)/24`
TH1 : `x + 11/24=(sqrt73)/24=>x=(sqrt73)/24-11/24=(sqrt73-11)/24`
TH2 : `x + 11/24=-(sqrt73)/24<=>x=-(sqrt73)/24-11/24=-(sqrt73-11)/24`
Vậy `S={(sqrt73-11)/24,-(sqrt73-11)/24}`