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d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
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a) \(\frac{x+3}{x-2}-\frac{2x+3}{x+2}=\frac{2x^2+5x+12}{x^2-4}\)
ĐKXĐ: \(\left\{\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)-\left(2x+3\right)\left(x-2\right)=2x^2+5x+12\)
\(\Leftrightarrow x^2+2x+3x+6-2x^2+4x-3x+6-2x^2-5x-12=0\)
\(\Leftrightarrow-3x^2+4x=0\)
\(\Leftrightarrow3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\3x=4\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\left(tmđk\right)\\x=\frac{4}{3}\left(tmđk\right)\end{matrix}\right.\)
Vậy: \(x=0;\frac{4}{3}\)
_Chúc bạn học tốt_
b) Ta có: \(\frac{2x+5}{x-3}+\frac{x-1}{x+3}=\frac{x^2+6x+18}{x^2-9}\)
ĐKXĐ: \(\left\{\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
\(\Leftrightarrow\frac{\left(2x+5\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{x^2+6x+18}{\left(x+3\right)\left(x-3\right)}\)
\(\Rightarrow\left(2x+5\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)=x^2+6x-18\)
\(\Leftrightarrow2x^2+6x+5x+15+x^2-3x-x+3-x^2-6x-18=0\)
\(\Leftrightarrow2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\2x=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x=0;-\frac{1}{2}\)
_Chúc bạn học tốt_
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\(\Leftrightarrow\frac{\frac{x-3+5}{5}}{4}=\frac{\frac{4x-3}{6}}{6}\Leftrightarrow\frac{x+2}{20}=\frac{4x-3}{36}\Leftrightarrow36x+72=80x-60\Leftrightarrow44x=132\Rightarrow x=2\)
\(\Leftrightarrow\frac{\frac{10x+x+2}{2}}{9}-\frac{\frac{x+3+75}{5}}{12}=x-2\)\(\Leftrightarrow\frac{11x+2}{18}-\frac{x+78}{60}=x-2\)\(\Leftrightarrow\left(\frac{11}{18}-\frac{1}{60}-1\right)x+\left(\frac{2}{18}-\frac{78}{60}+2\right)=0\).Giải típ nha, ko có Casio nên mk ko bấm
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a)\(\frac{\frac{x-3}{5}+1}{4}=\frac{\frac{2x}{3}-\frac{1}{2}}{6}\)
\(\Rightarrow\frac{\frac{x-3+5}{5}}{4}=\frac{\frac{4x}{6}-\frac{3}{6}}{6}\)
\(\Rightarrow\frac{\frac{x+2}{5}}{4}=\frac{\frac{4x-3}{6}}{6}\)
\(\Rightarrow\frac{x+2}{5}:4=\frac{4x-3}{6}:6\)
\(\Rightarrow\frac{x+2}{5}.\frac{1}{4}=\frac{4x-3}{6}.\frac{1}{6}\)
\(\Rightarrow\frac{x+2}{20}=\frac{4x-3}{36}\)
\(\Rightarrow36.\left(x+2\right)=20.\left(4x-3\right)\)
\(\Rightarrow36x+72=80x-60\)
\(\Rightarrow72+60=80x-36x\)
\(\Rightarrow132=44x\)
\(\Rightarrow x=\frac{132}{44}=3\)
Vậy x=3
b)Bn làm tương tự phần a nha
Chúc bn học tốt
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Bạn tham khảo link này nha:https://olm.vn/hoi-dap/detail/246637974159.html
Chúc bn học tốt
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\(\Leftrightarrow\frac{3x-3+5}{20}=|\frac{4x-3}{36}|\)
\(\Leftrightarrow\frac{3x+2}{5}=|\frac{4x-3}{9}|\)
\(\Leftrightarrow\left(3x+2\right)9=5|4x-3|\)
ĐK: \(3x+2\ge0\Rightarrow x\ge\frac{-2}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}5.\left(4x-3\right)=9.\left(3x+2\right)\\-5.\left(4x-3\right)=9.\left(3x+2\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}20x-15=27x+18\\-20x+15=27x+18\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}7x=-33\\-47x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-33}{7}\left(L\right)\\x=\frac{-3}{47}\end{cases}}}\)
Vậy x= - 3/47
b) \(\Leftrightarrow\frac{10x+x+2}{18}=\frac{x+3+75}{60}-2\)
\(\Leftrightarrow\frac{11x+2}{18}=\frac{x-42}{60}\)
\(\Leftrightarrow10.\left(11x+2\right)=3.\left(x-42\right)\)
\(\Leftrightarrow110x+20=3x-126\)
\(\Leftrightarrow107x=-146\Leftrightarrow x=\frac{-146}{107}\)
tìm x à
\(\frac{2x-3}{12}\)- \(\frac{x+2}{9}\)= 2
<=> \(\frac{6x-9}{36}\)- \(\frac{4x+8}{36}\)=2
<=> \(\frac{6x-9-4x-8}{36}\)=2
<=>\(\frac{2x-17}{36}\)=2
<=>2x-17=72
<=> 2x=72+12
<=>2x=84
<=> x=84:2
<=>x=41
Vậy x=41