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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: $\sqrt{x+1}$ `+2(x+1) =x-1 +` $\sqrt{1-x}$ +$3\sqrt{1-x^2}$ help mé

Home/ Toán Lớp 9: $\sqrt{x+1}$ `+2(x+1) =x-1 +` $\sqrt{1-x}$ +$3\sqrt{1-x^2}$ help mé

Toán Lớp 9: $\sqrt{x+1}$ `+2(x+1) =x-1 +` $\sqrt{1-x}$ +$3\sqrt{1-x^2}$ help mé

Tháng Sáu 26, 2022 Toán học 1 Comment 0 views

Toán Lớp 9: $\sqrt{x+1}$ +2(x+1) =x-1 + $\sqrt{1-x}$ +$3\sqrt{1-x^2}$
help mé

Comments ( 1 )

  1. hothaibinh
    hothaibinh
    26 Tháng Sáu, 2022 at 12:46 chiều
    Reply
    Giải đáp và giải thích các bước giải:
    ĐK : -1≤x≤1
    \sqrt[x+1]+2(x+1)=x-1+\sqrt[1-x]+3\sqrt[1-x^2]
    =\sqrt[1+x]+2(x+1)=x-1+\sqrt[1-x]+3\sqrt[(1-x)(1+x)] (1)
    Đặt \sqrt[1+x]=u , \sqrt[1-x]=t (u,t>0)
    (1)⇔u+2u^2=-t^2+t+3ut
    ⇔ (u-t)^2+u(u-t)+(u-t)=0
    ⇔ (u-t)(u-t+u+1)=0
    ⇔ (u-t)(2u-t+1)=0
    ⇔ \(\left[ \begin{array}{l}u=t\\2u+1=t\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=0\\4(x+1)+4\sqrt{x+1}+1=1-x\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=0\\4\sqrt{x+1}=-(5x+4)(*)\end{array} \right.\) 
    (*)⇔4\sqrt[x+1]=-(5x+4) (-1≤x≤-5/4)
    ⇔ 25x^2+24x=0
    ⇔ x(25x+24)=0
    ⇔ \(\left[ \begin{array}{l}x=0\\x=\dfrac{-24}{25}\end{array} \right.\) 
    Vậy S={0,-{24}/{25}}

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222-9+11+12:2*14+14 = ? ( )

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