gọi A = {1;3;5;...;2x+1 / x thuộc N}

=> Số số hạng của A là:

(2x + 1 - 1) : 2 = x.

=> (1 + 2x + 1)x : 2 = 224

<=> 2(x+1)x:2=224

<=> x(x+1)=224

Mặt khác: x và x+1 là 2 số tự nhiên liên tiếp và tích của chúng chỉ tận cùng = 0;2;6

=> o tồn tại stn x thỏa mãn đề bài.

Vậy x thuộc tập rỗng

2 mũ x nhân 7=224        (3x+5) mũ 2=289                   phần c mình chịu T-T

2 mũ x=224:7                 (3x+5) mũ 2=17 mũ 2

2 mũ x=32                       3x+5=17

2 mũ 5=32                      3x=17-2

=>x=5                               3x=15

                                         x=15:3

                                         x=5

a )  2^x.7=224

     2^x=224:7

     2^x=32=2⁵

       ^{vậy x= 5}

b)(3X+5)²=289

    (3X+5)²=17²

      ^{=> 3X+5=17}

           ^3X=17-5

           ^3X=12

            ^X=12:3=4

C)3^2x+1.11=2673

     3^2x+1=2673:11

    3^2x+1=243

   3^2x+1=3⁵

  =>2x+1=5

      2x=5-1

    2x= 4

  x=4:2

x=2

\(2x-124=x+224\\ \Rightarrow2x+x=224+124\\ \Rightarrow3x=348\\ \Rightarrow x=348:3\\ \Rightarrow x=116\)

2\(x\) - 124 = \(x\) + 224

2\(x\) - \(x\)    = 224 + 124

\(x\)           = 348

(x - 5)² = 16

=> (x - 5)² = 4²

=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

(2x - 1)³ = -64

=> (2x - 1)³ = -4³

=> 2x - 1 = -4

=> 2x = -4 + 1

=> 2x = -3

=> x = -3/2

( x - 5)² = 16

=> (x - 5)² = 4²

=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

a) \(\left(x-1\right)^3=125\)

\(\Leftrightarrow\left(x-1\right)^3=5^3\)

\(\Leftrightarrow x-1=5\)

\(\Leftrightarrow x=5+1\)

\(\Leftrightarrow x=6\)

Vậy \(x=6\)

b) \(2^{x+2}-2^x=96\)

\(\Leftrightarrow\left(2^2-1\right)\cdot2^x=96\)

\(\Leftrightarrow\left(4-1\right)\cdot2^x=96\)

\(\Leftrightarrow3\cdot2^x=96\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

c) \(\left(2x+1\right)^3=343\)

\(\Leftrightarrow\left(2x+1\right)^3=7^3\)

\(\Leftrightarrow2x+1=7\)

\(\Leftrightarrow2x=7-1\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

d) \(720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\)

\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\left(đk:x\ne23\right)\)

\(\Leftrightarrow720:\left(41-2x+5\right)=8\cdot5\)

\(\Leftrightarrow720:\left(46-2x\right)=40\)

\(\Leftrightarrow\dfrac{720}{46-2x}=40\)

\(\Leftrightarrow\dfrac{720}{2\left(23-x\right)}=40\)

\(\Leftrightarrow\dfrac{360}{23-x}=40\)

\(\Leftrightarrow360=40\left(23-x\right)\)

\(\Leftrightarrow9=23-x\)

\(\Leftrightarrow x=23-9\)

\(\Leftrightarrow x=14\left(đk:x\ne23\right)\)

\(\Leftrightarrow x=14\)

Vậy \(x=14\)

e) \(2^x\cdot7=224\)

\(\Leftrightarrow7\cdot2^x=224\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

f) \(\left(3x+5\right)^2=289\)

\(\Leftrightarrow3x+5=\pm17\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=17\\3x+5=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{22}{3}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{22}{3};x_2=4\)

a)\(\left(x-1\right)^3=125\Leftrightarrow\left(x-1\right)^3=5^3\Leftrightarrow x-1=5\Leftrightarrow x=6\)b)\(2^{x+2}-2^x=96\Leftrightarrow2^x.2^2-2^x=96\Leftrightarrow2^x\left(2^2-1\right)=96\Leftrightarrow2^x.3=96\Leftrightarrow2^x=32\Leftrightarrow x=5\)c)\(\left(2x-1\right)^3=343\Leftrightarrow\left(2x-1\right)^3=7^3\Leftrightarrow2x-1=7\Rightarrow2x=8\Rightarrow x=4\)d)\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)

\(720:\left[41-\left(2x-5\right)\right]=40\Leftrightarrow\left[41-\left(2x-5\right)\right]=720:40=18\)

\(\Leftrightarrow41-2x+5=18\Leftrightarrow36-2x=18\Leftrightarrow2x=18\Leftrightarrow x=9\)

e)\(2^x.7=224\Leftrightarrow2^x=224:7=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)

f) \(\left(3x+5\right)^2=289\Leftrightarrow\left(3x+5\right)=17^2\Leftrightarrow3x+5=17\Leftrightarrow3x=12\Leftrightarrow x=4\)

`1)4x^2+1=4x^2+4x+1-4x=(2x+1)^2-4x=(2x-2\sqrt{x}+1)(2x+2\sqrt{x}+1)` (với `x >= 0`)

  `->` Ko có đ/á

(Câu này mình nghĩ là `4x^4+1` chứ nhỉ?)

`2)4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2`

        `=(2x^2+y^2)-(2xy)^2`

        `=(2x^2-2xy+y^2)(2x^2+2xy+y^2)`

   `->bb C`

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

1) `2x(3x-1)-(2x+1)(x-3)`

`=6x^2-2x-2x^2+6x-x+3`

`=4x^2+3x+3`

2) `3(x^2-3x)-(4x+2)(x-1)`

`=3x^2-9x-4x^2+4x-2x+2`

`=-x^2-7x+2`

3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`

`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`

`=3x^2-15x-x^2+4x-4-4x^2+9`

`=-2x^2-11x+5`

4) `(2x-3)^2+(2x-1)(x+4)`

`=4x^2-12x+9+2x^2+8x-x-4`

`=6x^2-5x+5`

\(\left(2x+1\right)^2+\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)\)

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

\(=4x^2+4x+1+4x^2-4x+1-8x^2+2\)

\(=4\)