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a: Để A là số nguyên thì 3x+9-11 chia hết cho x+3

\(\Leftrightarrow x+3\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{-2;-4;8;-14\right\}\)

b: Để B là số nguyên thì -x+4 chia hết cho x+6

=>\(x+6\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

hay \(x\in\left\{-5;-7;-4;-8;-1;-11;4;-16\right\}\)

c: Để C là số nguyên thì x-8 chia hết cho 2x-1

=>2x-16 chia hết cho 2x-1

=>\(2x-1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

hay \(x\in\left\{1;0;2;-1;3;-2;8;-7\right\}\)

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

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

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

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

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)

1 tháng 9 2023

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

1 tháng 9 2023

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

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

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

NV
18 tháng 3 2021

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

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

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

NV
18 tháng 3 2021

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

2 tháng 12 2021

\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)

\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)

 

AH
Akai Haruma
Giáo viên
13 tháng 12 2021

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

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

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

 

9 tháng 5 2018

Mấy này bạn quy đồng lên cùng mẫu xong khử mẫu rồi giải. Dễ mà.

26 tháng 1 2022

a) \(\dfrac{3x-4}{x+1}=2\)

⇔ \(2x+2=3x-4\)

⇒ \(x=6\)

b: \(\Leftrightarrow4\left(2x^2-3\right)=8x^2+3x\)

=>3x=-12

hay x=-4(nhận)

c: \(\Leftrightarrow6x^2+9x+4x+6-6=0\)

=>x(6x+13)=0

=>x=0 hoặc x=-13/6

a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8-12+20x=0\)

\(\Leftrightarrow21x-4=0\)

\(\Leftrightarrow21x=4\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)

AH
Akai Haruma
Giáo viên
5 tháng 3 2021

Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!

5 tháng 2 2022

TK

https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5

a: \(\Leftrightarrow4x-5=2x-2+x\)

=>4x-5=3x-2

=>x=3(nhận)

b: =>7x-35=3x+6

=>4x=41

hay x=41/4(nhận)

c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

=>-6x+16=-5x+11

=>-x=-5

hay x=5(nhận)

d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)

\(\Leftrightarrow4x=16\)

hay x=4(nhận)