\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)

\(10,\left(x+3\right)^2-x^2=45\)

\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)

Vậy \(S=\left\{6\right\}\)

\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)

\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)

\(49\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)

\(4x^2+8x+4+4x^2-4x+1-8x^2+8-11=0\)

\(4x+2=0\)

\(4x=2\)

\(x=-\frac{1}{2}\)

<=>4(x²+2x+1)+4x²-4x+1-8x²+8-11=0

<=>4x²+8x+4+4x²-4x+1-8x²+8-11=0

<=>4x+2=0

<=>2(2x+1)=0

<=>2x+1=0

<=>x=-1/2

XL mik nhầm

\(=\frac{1}{2}\)

~~~~~~~~~~

..........

////////////////

b) 3x - 6 - (8x + 4) - (10x + 15) = 50

=> 3x - 6 - 8x - 4 - 10x - 15  = 50

=> (3x - 8x - 10x)  =  6+ 4 + 15 + 50

=> -15x = 75 => x = 75 : (-15) = -5

c) => 2x - 3 = 2 - x hoặc 2x - 3 = - (2 - x) (Vì 2 số  có giá trị tuyệt đối bằng nhau thì chings bằng nhau hoặc đối nhau)

+) nếu 2x - 3 = 2 - x => 2x+ x = 2 + 3 => 3x = 5 => x = 5/3

+) nếu 2x - 3 = -(2 - x) => 2x - 3 = -2 + x => 2x - x = -2 + 3 => x = 1

Vậy x = 5/3 hoặc x = 1

a) (n-1)ⁿ⁺¹¹-(n-1)ⁿ=0

(n-1)ⁿ(n-1)¹¹-(n-1)ⁿ=0

(n-1)ⁿ[(n-1)¹¹-1]=0

(n-1)ⁿ=0 hoặc (n-1)¹¹-1=0

n-1=0   hoặc  (n-1)¹¹   =1

n=1      hoặc  n-1         =1

n=1      hoặc   n          =2

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

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

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(4(x-3)^2-(2x-1)(2x+1)=10\\\Rightarrow4(x^2-6x+9)-(4x^2-1)=10\\\Rightarrow4x^2-24x+36-4x^2+1=10\\\Rightarrow-24x+37=10\\\Rightarrow-24x=-27\\\Rightarrow x=\dfrac98\)

Bài 1:

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)

\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)

=>A<B

Bài 2:

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

=>4x+13=11

=>4x=-2

=>\(x=-\dfrac{1}{2}\)

Bài làm

a)dãy số U: \(2,7,12,...x\)

U là cấp số cộng\(\Rightarrow\left\{{}\begin{matrix}d=u_2-u_1=7-2=5\\u_1=2\end{matrix}\right.\)

\(U_n=U_1+\left(n-1\right)d\)

=> \(n=\dfrac{U_n-U_1}{d}+1=\dfrac{x-2}{5}+1=\dfrac{\left(x+3\right)}{5}\)

\(S_n=\dfrac{n\left(U_1+U_n\right)}{2}=\dfrac{\dfrac{\left(x+3\right)}{5}\left(2+x\right)}{2}=\dfrac{\left(x+3\right)\left(x+2\right)}{2.5}=245\)

\(x^2+5x+6=2450\)

\(x^2+5x-2444=0\)

\(\Delta=5^2-4.\left(-2444\right)=9801=\)99^2

\(\left\{{}\begin{matrix}x_1=\dfrac{-5-99}{2}< 0\left(loai\right)\\x_2=\dfrac{-5+99}{2}=47\end{matrix}\right.\)

Đáp số: x=47

b) Xét cấp số cộng 1, 6, 11, ..., 96. Ta có :

\(96=1+\left(n-1\right)5\Rightarrow n=20\)

Suy ra :

\(S_{20}=1+6+11+...+96=\dfrac{20\left(1+96\right)}{2}=970\)

và \(2x.20+970=1010\)

Từ đó : \(x=1\)