Vu thanh tamMiru✮YoCiaoRyoran NhoHàn Khả Dii🍀🧡_Trang_🧡🍀Lâm Tư Thần

kethattinhtrongmuaShiroChibi NguyễnĐào Nguyệt Hằngtrần khánh linh

nhận thưởng

1

mình nghĩ bài này không thẻ làm được vì phải nhân từ 1 đến 99.Nên số này rất lớn ,không thể làm được.

=9.332621544394415e+115

mik ko biết nhưng mik tính máy tính nó ra như thế.

6 |x-3| - 9|x-3| = -21

( 6 - 9 ) |x-3| = - 21

-3 | x - 3| = - 21

| x -3 | =7

=>x- 3 = 7 hoặc x - 3 = -7

   x - 3 = 7

=> x = 10

x - 3 = -7

=> x =-4

=2041 nhé!

ngày sinh bạn là 28/8/2005

kq là 2041

ai đâu đâu thấy ai

Without his help, we would all have died.

=> Had it not been for his help, we would all have died.

Without his help, we would all have died

Had it not been his help, we would all have died
< day la dang dao ngu cua cau dieu kien nha loai 3 >

Bài 7:

a: ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)

\(\dfrac{x+5}{2x-1}-\dfrac{1-2x}{x+5}-2=0\)

=>\(\dfrac{x+5}{2x-1}+\dfrac{2x-1}{x+5}-2=0\)

=>\(\dfrac{\left(x+5\right)^2+\left(2x-1\right)^2}{\left(2x-1\right)\left(x+5\right)}=2\)

=>\(\left(x+5\right)^2+\left(2x-1\right)^2=2\left(2x-1\right)\left(x+5\right)\)

=>\(x^2+10x+25+4x^2-4x+1=2\left(2x^2+10x-x-5\right)\)

=>\(5x^2+6x+26-4x^2-18x+10=0\)

=>\(x^2-12x+36=0\)

=>\(\left(x-6\right)^2=0\)

=>x-6=0

=>x=6(nhận)

b: ĐKXĐ: \(x\notin\left\{3;-2;4\right\}\)

\(1-\dfrac{8}{x-4}=\dfrac{5}{3-x}-\dfrac{8-x}{x+2}\)

=>\(\dfrac{x-4-8}{x-4}=\dfrac{-5}{x-3}+\dfrac{x-8}{x+2}\)

=>\(\dfrac{x-12}{x-4}=\dfrac{-5\left(x+2\right)+\left(x-8\right)\left(x-3\right)}{\left(x-3\right)\left(x+2\right)}\)

=>\(\dfrac{x-12}{x-4}=\dfrac{-5x-10+x^2-11x+24}{\left(x-3\right)\left(x+2\right)}\)

=>\(\left(x-12\right)\left(x^2-x-6\right)=\left(x-4\right)\left(x^2-16x+14\right)\)

=>\(x^3-x^2-6x-12x^2+12x+72=x^3-16x^2+14x-4x^2+64x-56\)

=>\(-13x^2+6x+72=-20x^2+78x-56\)

=>\(7x^2-72x+128=0\)

=>\(\left[{}\begin{matrix}x=8\left(nhận\right)\\x=\dfrac{16}{7}\left(nhận\right)\end{matrix}\right.\)

c: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x-1}{x+2}+\dfrac{2}{x-2}=\dfrac{12}{x^2-4}\)

=>\(\dfrac{x-1}{x+2}+\dfrac{2}{x-2}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{\left(x-1\right)\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}\)

=>\(x^2-3x+2+2x+4=12\)

=>\(x^2-x-6=0\)

=>(x-3)(x+2)=0

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