x. ( 4 + 40 ) = 101 . 11

x. 44           = 1111

x                 = 1111 : 44

x                 = 25,25

x04 + 40x = 101.11

=> 100x + 4 + 400 + x = 1111

=> 101x + 404 = 1111

=> x = 7

x04+40x = 100xX +4 + 400 +X = 101xX +404 = 101xX +101x4 =101x (X+4) = 101x11 => X+4 =11 => X =7

a/ (x+1)+(x+2)+(x+3)+....+(x+50)=2275

<=> (x+x+...+x)+(1+2+3+...+50)=2275

<=> \(50x+\frac{50\left(50+1\right)}{2}=2275\)

<=> 50.x+1275=2275

<=> 50x=1000

=> x=1000:50  => x=20

b/ x04+40x=101x11

<=> 100x+4+400+x=1111

<=> 101x=707

=> x=707:101  => x=7

a, ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 50 ) = 2275

x x 50 + ( 50 + 1 ) x ( 50 : 2 ) = 2257

x x 50 + 51 x 25 = 2257

x x 50 + 1275 = 2257

x x 50 = 2257 - 1275

x x 50 = 982

x = 982 : 50

x = 19,64

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=2275\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+50\right)=2275\)

\(50x+\frac{50\left(50+1\right)}{2}=2275\)

\(50x+1275=2275\)

50x = 2275 - 1275

50x = 1000

x = 20

B x04 + 40x = 101 . 11

100 .x + 4 + 400 + x = 101 . 11

101 . x = 707

=> x = 7

Dau . la dau nhan

a) \(\Rightarrow\left(x-1\right)^3=0\Rightarrow x=1\)

b) \(\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)(do \(\left\{{}\begin{matrix}x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\))

c) \(\Rightarrow4x\left(x^2-9\right)=0\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

d) \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)

a) \(x^3-3x^2+3x-1=0\Rightarrow\left(x-1\right)^3=0\Rightarrow x-1=0\)

      \(\Rightarrow x=1\)

b) \(x^6-1=0\Rightarrow\left(x^3\right)^2-1=0\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) \(4x^3-36x=0\Rightarrow4x\left(x^2-36\right)=0\Rightarrow4x\left(x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-6=0\\x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)

d) \(x^3-6x^2+12x-8=0\) (đề bài như vậy mới làm đc, nếu là +8 thì mình xin bó tay nhé)

     \(\Rightarrow x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3=0\)

     \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)

cdbdb

a: ta có: \(x^2+3x-\left(2x+6\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

b: Ta có: \(5x+20-x^2-4x=0\)

\(\Leftrightarrow\left(x+4\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\)

a: P(x) chia hết cho x-2

=>x^4-2x^3+3x^3-6x^2+12x^2-24x-16x+32+m-2017 chia hết cho x-2

=>m-2017=0

=>m=2017

b: P(x)=x^4+x^3+6x^2-40x+32

P(x)=0

=>x^4-2x^3+3x^3-6x^2+12x^2-24x-16x+32=0

=>(x-2)(x^3+3x^2+12x-16)=0

=>x^3+3x^2+12x-16=0 hoặc x-2=0

=>x^3-x^2+4x^2-4x+16x-16=0 hoặc x-2=0

=>x-1=0 hoặc x=2

=>x=1 hoặc x=2

Chọn B.