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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
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}-\frac{1}{x+1}-\frac{1}{x+3}-\frac{1}{x+4}-\frac{1}{x+6}=0\)
\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{\left(x+4\right)\left(x+5\right)}-\frac{1}{\left(x+6\right)\left(x+7\right)}=0\)
\(\Leftrightarrow\frac{8x+20}{x\left(x+1\right)\left(x+4\right)\left(x+5\right)}+\frac{8x+36}{\left(x+2\right)\left(x+3\right)\left(x+6\right)\left(x+7\right)}=0\).Đến đây mk chịu
a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)
=> \(3x-3-2x-6=-15\)
=> \(3x-3-2x-6+15=0\)
=> \(x=-6\)
Vậy phương trình có nghiệm là x = -6 .
b, Ta có : \(3\left(x-1\right)+2=3x-1\)
=> \(3x-3+2=3x-1\)
=> \(3x-3+2-3x+1=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)
=> \(14-35x-5=16-24x\)
=> \(14-35x-5-16+24x=0\)
=> \(-35x+24x=7\)
=> \(x=\frac{-7}{11}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .
Bài 2 :
a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)
=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)
=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)
=> \(x+15x-3=2x-16-10x-15\)
=> \(x+15x-3-2x+16+10x+15=0\)
=> \(24x+28=0\)
=> \(x=\frac{-28}{24}=\frac{-7}{6}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .
b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)
=> \(6x+24-30x+120=10x-15x+30\)
=> \(6x+24-30x+120-10x+15x-30=0\)
=> \(-19x+114=0\)
=> \(x=\frac{-114}{-19}=6\)
Vậy phương trình có nghiệm là x = 6 .
bai 1
1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0
<=>(2x)^2-5^2=0
<=>(2x+5)*(2x-5)=0
<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự
2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)
\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)
Đặt \(x^2+5x+3=t\)
\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)
\(\Leftrightarrow t^2-9=280\)
\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)
\(\Leftrightarrow x^2-2x+7x-14=0\)
\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0
\(\Leftrightarrow\) x = 2 hoặc x = - 7
Vậy x = 2 hoặc x = -7.
3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)
\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)
\(\Leftrightarrow x^3+12x^2+46x+60=0\)
\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)
\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)
\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)
\(\Leftrightarrow x=-6\)
Vậy x = -6.
Nhiều vậy ai làm hết được :P
1) \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)
\(\Leftrightarrow\frac{3x-8}{3}=\frac{4x-1}{4}\)
\(\Leftrightarrow4\left(3x-8\right)=3\left(4x-1\right)\)
\(\Leftrightarrow12x-32=12x-3\)(vô lí)
Vậy pt vô nghiệm
P/s: mấy câu sau tương tự thôi mà :)))
nhăm nhe 1 câu thôi
\(10,\frac{3+5x}{5}-3=\frac{9x-3}{4}\)
\(\Leftrightarrow\frac{3+5x-15}{5}=\frac{9x-3}{4}\)
\(\Leftrightarrow\frac{-12+5x}{5}=\frac{9x-3}{4}\)
\(\Leftrightarrow\left(-12+5x\right)5=\left(9x-3\right)4\)
\(\Leftrightarrow-60+25x=36x-12\)
\(\Leftrightarrow26x-36x=-12+60\)
\(\Leftrightarrow-10x=48\)
\(\Leftrightarrow x=-4,8\)
a)
\(\left(5x+3\right)\cdot\left(x^2+4\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=4\end{matrix}\right.\)
b)
\(\left(4x-1\right)\cdot\left(x-3\right)-\left(x-2\right)\cdot\left(5x+2\right)=0\\ \Leftrightarrow4x^2-12x-x+3-5x^2-2x+10x+4=0\\ \Leftrightarrow-x^2-5x+7=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{5+\sqrt{53}}{2}\\-\frac{5-\sqrt{53}}{2}\end{matrix}\right.\)
c)
\(\left(x+3\right)\cdot\left(x-5\right)+\left(x+3\right)\cdot\left(3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(x-5+3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(4x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)
d)
\(\left(x+6\right)\cdot\left(3x-1\right)+x^2-36=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x^2-36\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x+6\right)\cdot\left(x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1+x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(4x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
e)
\(0.75x\cdot\left(x+5\right)=\left(x+5\right)\cdot\left(3-1.25x\right)\\ \Leftrightarrow0.75x\cdot\left(x+5\right)-\left(x+5\right)\cdot\left(3-1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(0.75x-3+1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)
Vậy phương trình vô nghiệm
oke cảm ơn bn nhìu :)))