Nhầm sorry mk tưởng cộng sory bạn nha Thái Viết Nam

Ta có : |17x - 5| - |17x + 5| = 0 

Mà |17x - 5| \(\ge\)0 ; |17x + 5| \(\ge\) 0

Nên \(\hept{\begin{cases}\left|17x-5\right|=0\\\left|17x+5\right|=0\end{cases}}\)

 <=>\(\hept{\begin{cases}17x-5=0\\17x+5=0\end{cases}}\)

  <=>  \(\hept{\begin{cases}17x=5\\17x=-5\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{5}{17}\\x=-\frac{5}{17}\end{cases}}\)

Mà x ko thể đồng thời bằng 2 giá trị 

Nên x thuộc rỗng 

lười vl , cần gấp k

<=> |17x - 5| = |17x + 5|

=> 17x - 5 = 17x + 5 hoặc 17x - 5 = -17x - 5

=> 0x = 10(loại) hoặc 34x = 0

<=> x = 0.

pt <=> | 17x - 5 | = | 17x + 5 |

\(\Leftrightarrow\orbr{\begin{cases}17x-5=17x+5\left(loai\right)\\17x-5=-17x-5\end{cases}}\)

\(\Leftrightarrow17x=-17x\)

\(\Leftrightarrow x=-x\)

\(\Leftrightarrow x=0\)

Lớp nguồn ra nhiều bài khó nhỉ

có sai đề ko bạn nếu ko sai đề thì mik nghĩ bài này có nhiều đáp án đấy

x= -2 hoặc x=0. Đảm bảo đúng!

\(\left|17x-5\right|-\left|17x+5\right|=0\)

\(\Rightarrow\left|17x-5\right|=\left|17x+5\right|\)

\(\Rightarrow\left(17x-5\right)^2=\left(17x+5\right)^2\)

\(\Rightarrow17^2.x^2-2\left(17x-5\right)+5^2=17^2.x^2+2\left(17x+5\right)+5^2\)

\(\Rightarrow-\left(34x-10\right)+25=34x+10+25\)

\(\Rightarrow-34x+10+25=34x+10+25\)

\(\Rightarrow68x=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

\(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)

=>\(17x^2-17x+8=x^2-17x+33\)

<=> \(16x^2-25=0\)

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

=> \(4x-5=0=>x=\dfrac{5}{4}\)

hoặc \(4x+5=0=>x=\dfrac{-5}{4}\)

(x+2)(x−3)(17x²−17x+8) - (x+2)(x−3)(x²−17x+33) = 0

\(\Leftrightarrow\)(x+2)(x−3).[(17x²−17x+8)-(x²−17x+33)] = 0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x+2 = 0}\\\text{x−3 = 0}\\\text{(17x^2−17x+8)-(x^2−17x+33) = 0}\end{matrix}\right.\)

\(\Leftrightarrow\)​\(\left[{}\begin{matrix}x=-2\\x=3\\17x^2-17x+8-x^2+17x-33=0\end{matrix}\right.\)​​

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\16x^2-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\\left(4x-5\right)\left(4x+5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\4x-5=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\4x=5\\4x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=\dfrac{5}{4}\\x=\dfrac{-5}{4}\end{matrix}\right.\)

Vậy S = \(\left\{-2;\dfrac{-5}{4};\dfrac{5}{4};3\right\}\)

a: =>|5x+4|=19

=>5x+4=19 hoặc 5x+4=-19

=>5x=15 hoặc 5x=-23

=>x=3 hoặc x=-23/5

b: =>3|2x-9|=33

=>|2x-9|=11

=>2x-9=11 hoặc 2x-9=-11

=>2x=20 hoặc 2x=-2

=>x=10 hoặc x=-1

d: =>|17x-5|=|17x+5|

=>17x-5=17x+5 hoặc 17x-5=-17x-5

=>34x=0

hay x=0

Bài 1:

\(A=a^2b^2\left(b-a\right)+b^2c^2\left(c-b\right)+c^2a^2\left(a-c\right)\)

\(=a^2b^2\left(b-c+c-a\right)+b^2c^2\left(c-a+a-b\right)+c^2a^2\left(a-c\right)\)

\(=a^2b^2\left(b-c\right)+a^2b^2\left(c-a\right)+b^2c^2\left(c-a\right)+b^2c^2\left(a-b\right)+c^2a^2\left(a-c\right)\)

\(=\left(c-a\right)\left(a^2b^2+b^2c^2-c^2a^2\right)+b^2\left[a^2\left(b-c\right)+c^2\left(a-b\right)\right]\)

\(=\left(c-a\right)\left(a^2b^2+b^2c^2-c^2a^2\right)+b^2\left(c-a\right)\left(ac-bc-ba\right)\)

\(=\left(c-a\right)\left[a^2b^2+b^2c^2-c^2a^2+b^2\left(ac-bc-ba\right)\right]\)

2/ \(\left(17x-5\right)^2+2\left(17x-5\right)\left(3x-2\right)+\left(3x-2\right)^2=0\)

\(\Leftrightarrow\left(17x-2+3x-2\right)^2=0\)

\(\Leftrightarrow20x-4=0\)

\(\Rightarrow x=\frac{1}{5}\)

Lời giải:

\(17x^2-2x^3-3x^4-4x-5=-3x^4-2x^3+17x^2-4x-5\)

\(=-3x^2(x^2+x-5)+x^3+2x^2-4x-5\)

\(=-3x^2(x^2+x-5)+x(x^2+x-5)+(x^2+x-5)\)

\(=(x^2+x-5)(-3x^2+x+1)\)

Do đó: $(17x^2-2x^3-3x^4-4x-5):(x^2+x-5)=(-3x^2+x+1)$