bài 1: tìm x a, (x – 4) ² – 25 = 0 b, ( 3x – 7) ² – 4(x+1) ²=0 c, x ³ + x ² + x +1 = 0 d, x ³+x ² – x – 1= 0 e, (2x – 1) ³ – 4(x ²+1) (2x -1) = 0 f,

bài 1: tìm x
a, (x – 4) ² – 25 = 0
b, ( 3x – 7) ² – 4(x+1) ²=0
c, x ³ + x ² + x +1 = 0
d, x ³+x ² – x – 1= 0
e, (2x – 1) ³ – 4(x ²+1) (2x -1) = 0
f, (x+4) (5x+9) – x ² + 16 = 0
giúp mình với. :))

0 bình luận về “bài 1: tìm x a, (x – 4) ² – 25 = 0 b, ( 3x – 7) ² – 4(x+1) ²=0 c, x ³ + x ² + x +1 = 0 d, x ³+x ² – x – 1= 0 e, (2x – 1) ³ – 4(x ²+1) (2x -1) = 0 f,”

  1. Đáp án: `↓`

     

    Giải thích các bước giải:

    `a, (x – 4) ² – 25 = 0`

    `(x-4)²= 25`

    `(x-4)²= (±5)²`

    \(\left[ \begin{array}{l}x-4=5\\x-4=-5\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=9\\x=-1\end{array} \right.\)

    ___________

    `b, ( 3x – 7) ² – 4(x+1) ²=0`

    `9x²-42x+49-4(x²+2x+1)=0`

    `9x²-42x+49-4x²-8x-4=0`

    `5x²-50x + 45 = 0 `

    `5(x²-10x+9)=0`

    `5(x²-x-9x+9)=0`

    `5[(x²-x)-(9x-9)]=0`

    `5[x(x-1)-9(x-1)]=0`

    `5(x-1)(x-9)=0`

    \(\left[ \begin{array}{l}x-1=0\\x-9=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1\\x=9\end{array} \right.\)

    _____________

    `c, x ³ + x ² + x +1 = 0`

    `(x³+x)+(x²+1)=0`

    `x(x²+1)+(x²+1)=0`

    `(x²+1)(x+1)=0`

    \(\left[ \begin{array}{l}x^2+1=0\\x+1=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x^2=-1(loại)\\x=-1\end{array} \right.\)

    _______________

    `d, x ³+x ² – x – 1= 0`

    `(x³-x)+(x²-1)= 0`

    `x(x²-1)+(x²-1)=0`

    `(x²-1)(x+1)=0`

    `(x+1)(x-1)(x+1)=0`

    `(x+1)²(x-1)=0`\(\left[ \begin{array}{l}(x+1)^2=0\\x-1=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=-1\\x=1\end{array} \right.\) 

    ________________

    `e, (2x – 1) ³ – 4(x ²+1) (2x -1) = 0`

    `(2x-1){(2x-1)²-[4(x²+1)]}=0`

    `(2x-1)[4x²-4x+1-(4x²+4)]=0`

    `(2x-1)(4x²-4x+1-4x²-4)=0`

    `(2x-1)(-4x-3)=0`

    `\(\left[ \begin{array}{l}2x-1=0\\-4x-3=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1/2\\x=-3/4\end{array} \right.\) 

    ___________________

    `f, (x+4) (5x+9) – x ² + 16 = 0`

    `(x+4) (5x+9)-(x²-16)=0`

    `(x+4) (5x+9)-(x-4)(x+4)=0`

    `(x+4)[(5x+9)-(x-4)]=0`

    `(x-4)(5x+9-x+4)=0`

    `(x-4)(4x+16)=0`

    \(\left[ \begin{array}{l}x-4=0\\4x+16=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)

    Chúc bn học tốt !!!!

    ~Giang~

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