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\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}+\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow x+15=0\)
\(\Leftrightarrow x=-15\)
Vậy x = -15.
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\\ \Leftrightarrow\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\\ \Leftrightarrow\dfrac{x+15}{13}+\dfrac{2x+30}{15}=\dfrac{3x+45}{37}+\dfrac{4x+60}{9}\)
\(\Leftrightarrow\left(x+15\right)\dfrac{1}{13}+\left(x+15\right)\dfrac{2}{15}=\left(x+15\right)\dfrac{3}{37}+\left(x+15\right)\dfrac{4}{9}\\ \Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
vì:\(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\ne0\) nên:
x+15=0 =>x=-15
vậy phương trình có tập nghiệm là S={-15}
a. \(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)
\(\Leftrightarrow2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
\(\Leftrightarrow12x+10-10x-3=8x+4x+2\)
\(\Leftrightarrow12x-10x-8x-4x=2-10+3\)
\(\Leftrightarrow-10x=-5\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\)
\(\Leftrightarrow6x^2+2=6x^2+6x+6\)
\(\Leftrightarrow6x^2-6x^2-6x=6-2\Leftrightarrow-6x=4\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
c. \(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\left(\dfrac{x+2}{13}+1\right)+\left(\dfrac{2x+45}{15}-1\right)=\left(\dfrac{3x+8}{37}+1\right)+\left(\dfrac{4x+69}{9}-1\right)\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
Vì \(\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)>0\)
\(\Leftrightarrow x+15=0\Leftrightarrow x=-15\)
<=>(x+2/13)+1+(2x+45/15)-1 = (3x+8/37)+1+(4x+69/9)-1
<=> x+15/13 + 2x+30/15 = 3x+45/37 + 4x+60/9
<=>(x+15)(1/13+2/15-3/37-4/9) = 0
<=> x+15=0( vì 1/13+2/15-3/37-4/9=0)
<=>x=-15......
\(\text{a) }\left|2-5x\right|=\left|3x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}2-5x=3x+1\\2-5x=-3x-1\end{matrix}\right. \Leftrightarrow\left[{}\begin{matrix}-5x-3x=1-2\\-5x+3x=-1-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-1\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{1}{8};\dfrac{3}{2}\right\}\)
\(\text{b) }\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)
ĐXKĐ của phương trình \(:x\ne\pm5\)
\(\text{Ta có }:\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{2\left(25-x^2\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x+5\right)\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{9\left(x+5\right)}{12\left(x+5\right)\left(x-5\right)}-\dfrac{90}{12\left(x+5\right)\left(x-5\right)}+\dfrac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\\ \Rightarrow9x+45-90+14x-70=0\\ \Leftrightarrow23x=115\\ \Leftrightarrow x=5\left(KTM\right)\)
Vậy phương trình vô nghiệm
\(\text{c) }\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\\ \Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\\ \Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}=0\\ \Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\\ \Leftrightarrow x+60=0\left(\text{Vì }\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\\ \Leftrightarrow x=-60\)
Vậy \(x=-60\) là nghiệm của phương trình
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=-10\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=-10\)
\(.....................\)
đến đây thì dễ rồi :)
Mấy này bạn quy đồng lên cùng mẫu xong khử mẫu rồi giải. Dễ mà.
Giải phương trình
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\)\(\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)
\(\Leftrightarrow\)\(\dfrac{x+2}{13}+\dfrac{13}{13}+\dfrac{2x+45}{15}-\dfrac{15}{15}=\dfrac{3x+8}{37}+\dfrac{37}{37}+\dfrac{4x+69}{9}-\dfrac{9}{9}\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2x+30}{15}=\dfrac{3x+45}{37}+\dfrac{4x+60}{9}\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)=\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)-\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=0\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-15\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\ne0\end{matrix}\right.\)
Do đó: \(x=-15\)
Vậy \(S=\left\{-15\right\}\)
Hai ý gần cuối dùng dấu ngoặc nhọn {} nhé bạn!