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9 tháng 1 2021
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
PN
16 tháng 5 2021
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
DH
4 tháng 7 2017
a, \(-\left(x+3\right)\left(x-4\right)+\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow-\left(x^2-4x+3x-12\right)+x^2-1=10\)
\(\Rightarrow-x^2+x+12+x^2-1=10\)
\(\Rightarrow x=10+1-12\Rightarrow x=-1\)
b, \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)
\(\Rightarrow2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)
\(\Rightarrow2x^2-5x+2-2x^2+x+21=3\)
\(\Rightarrow-4x=3-21-2\Rightarrow-4x=-20\)
\(\Rightarrow x=5\)
Các câu còn lại làm tương tự! Phá ngoặc ra!
Chúc bạn học tốt!!!
NU
3
13 tháng 4 2021
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)
Vậy phương trình vô nghiệm
26 tháng 7 2021
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
TG
1
HV
14 tháng 4 2020
Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Lại Có: ĐKXĐ: x≠1,x≠2,x≠3,x≠4,x≠5,x≠6
\(\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{10}\)<=>\(\frac{1}{\left(x-6\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-1\right)}=\frac{1}{10}\)
<=>\(\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-1}=\frac{1}{10}\)
<=> \(\frac{1}{x-6}-\frac{1}{x-1}=\frac{1}{10}\)
<=> \(\frac{x-1-x+6}{\left(x-6\right)\left(x-1\right)}=\frac{1}{10}\)
<=> \(\frac{5}{\left(x-6\right)\left(x-1\right)}=\frac{1}{10}\)
<=>(x-6)(x-1)=50
<=>x2-7x+6-50=0
<=>x2+4x-11x-44=0
<=>x(x+4)-11(x+4)=0
<=>(x+4)(x-11)=0
<=>\(\left[{}\begin{matrix}x+4=0\\x-11=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=-4\\x=11\end{matrix}\right.\)(Thỏa mãn)
Vậy phương trình thuộc tập nghiệm S={-4;11}
\(\left(x^2+1\right)\left(x+3\right)+5=\left(x-3\right)\left(x+4\right)+10\)
\(\Leftrightarrow x^3+3x^2+x+3+5=x^2+x-12+10\)
\(\Leftrightarrow x^3+3x^2+x-x^2-x=-12+10-3-5\)
\(\Leftrightarrow x^2\left(x+2\right)=-10\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-10\\x+2=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-12\end{matrix}\right.\)
sau khi dùng phép thử ta nhận thấy \(x\ne-12\)
Vậy \(x\in\varnothing\)
gà thì đừng làm