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29 tháng 7 2021
\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)
\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)
\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)
NH
11 tháng 3 2023
bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu
11 tháng 3 2023
a: =>\(4x-5=2x-2+x=3x-2\)
=>x=3
b: \(\Leftrightarrow7x-35=3x+6\)
=>4x=41
=>x=41/4
c: =>(2x+5)(x+5)-2x^2=0
=>2x^2+10x+5x+25-2x^2=0
=>15x=-25
=>x=-5/3
e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)
=>11(x^2-3x-4)=x(11x-34)
=>11x^2-33x-44=11x^2-34x
=>x=44
2 tháng 12 2021
\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)
\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)
11 tháng 3 2023
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
12 tháng 3 2023
d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>25(11x-4)=18(12x+1)
=>275x-100=216x+18
=>59x=118
=>x=2
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
10 tháng 4 2021
a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
10 tháng 4 2021
b)
ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)
Ta có: P=AB
\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)
\(=\dfrac{3x}{x+1}\)
Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)
\(\Leftrightarrow9\left(x+1\right)=6x\)
\(\Leftrightarrow9x-6x=-9\)
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)
16 tháng 6 2022
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
LD
18 tháng 1 2019
a.
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)=18\)
Đặt \(t=x^2+2x+1=\left(x+1\right)^2\left(t\ge0\right)\)
\(\Rightarrow\left(4t-1\right)\cdot t=18\)
\(\Leftrightarrow\left(2t\right)^2-2\cdot2t\cdot\dfrac{1}{4}+\dfrac{1}{16}=\dfrac{289}{16}\)
\(\Leftrightarrow\left(2t-\dfrac{1}{4}\right)^2=\dfrac{289}{16}\Leftrightarrow\left(t-\dfrac{1}{8}\right)^2=\dfrac{289}{64}\)
\(\Leftrightarrow\left[{}\begin{matrix}t-\dfrac{1}{8}=\dfrac{17}{8}\\t-\dfrac{1}{8}=-\dfrac{17}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\\t=-2\left(loai\right)\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2=\dfrac{9}{4}\Leftrightarrow\left[{}\begin{matrix}x+1=\dfrac{3}{2}\\x+1=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{5}{2};\dfrac{1}{2}\right\}\)
b.
Ta có:
- \(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
- \(x^2+11x+24=x^2+3x+8x+24=x\left(x+3\right)+8\left(x+3\right)=\left(x+3\right)\left(x+8\right)\)
- \(x^2+18x+80=x^2+8x+10x+80=x\left(x+8\right)+10\left(x+8\right)=\left(x+8\right)\left(x+10\right)\)
Thay vào phương trình, ta được:
\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{x+10-\left(x+1\right)}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{25}\Leftrightarrow\dfrac{9}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{25}\)
\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=25\)
\(\Leftrightarrow x^2+11x+\dfrac{121}{4}=\dfrac{181}{4}\)
\(\Leftrightarrow\left(x+\dfrac{11}{2}\right)^2=\dfrac{181}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{11}{2}=\dfrac{\sqrt{181}}{2}\\x+\dfrac{11}{2}=-\dfrac{\sqrt{181}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{181}}{2}\\x=\dfrac{-11-\sqrt{181}}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{-11+\sqrt{181}}{2};\dfrac{-11-\sqrt{181}}{2}\right\}\)
ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{x^2-9}\)
\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)+11x-3}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x-3}\)