giai PT
\(\dfrac{x}{60}+ \dfrac{x}{45}+1,5=\dfrac{32}{5}\)
Những câu hỏi liên quan
tim xdfrac{5}{6}x-dfrac{3}{4}dfrac{-1}{4}+dfrac{2}{3}65x−434−1+32-1dfrac{1}{2}-dfrac{2}{3}xdfrac{5}{6}-left(dfrac{-2}{5}right)−121−32x65−(5−2)left(dfrac{4}{5}:x+1,5right):dfrac{2}{3}-1,5(54:x+1,5):32−1,5dfrac{4}{3}x-dfrac{2}{3}dfrac{1}{4}-x34x−3241−xgiup minh nhe minh dang can gap
Đọc tiếp
tim x
\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}65x−43=4−1+32
-1\dfrac{1}{2}-\dfrac{2}{3}x=\dfrac{5}{6}-\left(\dfrac{-2}{5}\right)−121−32x=65−(5−2)
\left(\dfrac{4}{5}:x+1,5\right):\dfrac{2}{3}=-1,5(54:x+1,5):32=−1,5
\dfrac{4}{3}x-\dfrac{2}{3}=\dfrac{1}{4}-x34x−32=41−x
giup minh nhe minh dang can gap
giai pt
\(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25-125+6}\)
đề tào lao, nhìn VP biết có vấn đề
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Xét thấy : \(\sqrt{25-125+6}\)<0
mà: \(\sqrt{a}\)\(\ge\)0
\(\Rightarrow\)Vế phải ko hợp lí\(\Rightarrow\)X vô nghiệm
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giải pt sau
a)\(\dfrac{60}{x}=\dfrac{4}{3}+\dfrac{60-x}{x+4}\)
b)\(\dfrac{100}{x}-\dfrac{100}{x+20}=\dfrac{5}{6}\)
c)\(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
Helppppp
b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)
=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)
=>x*(x+20)=400*6=2400
=>x^2+20x-2400=0
=>(x+60)(x-40)=0
=>x=-60 hoặc x=40
c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
=>(2x+1)^2-(2x-1)^2=8
=>4x^2+4x+1-4x^2+4x-1=8
=>8x=8
=>x=1(nhận)
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giải pt
\(\dfrac{3x-0,4}{2}+\dfrac{1,5-2x}{3}=\dfrac{x+0,5}{5}\)
<=>\(\dfrac{\left(3x-0,4\right)15}{30}+\dfrac{\left(1,5-2x\right)10}{30}=\dfrac{\left(x+0,5\right)6}{30}\)
=>\(\left(3x-0,4\right)15+\left(1,5-2x\right)10=\left(x+0,5\right)6\)
<=>\(45x-6+15-20x=6x+3\)
<=>\(45x-20x-6x=6+3-15\)
<=>\(19x=-6\)
<=>\(x=\dfrac{-6}{19}\)
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Giai PT
\(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{-1}{x\left(x+1\right)}\)
\(\Leftrightarrow x^2-x-x-1=-1\)
=>x(x-2)=0
=>x=2
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ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.\)
\(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow\dfrac{x\left(x-1\right)}{x\left(x+1\right)}-\dfrac{\left(x+1\right)}{x\left(x+1\right)}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2-x-x-1}{x\left(x+1\right)}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow x^2-2x-1=-1\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
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Giai PT : \(\dfrac{x}{40}+\dfrac{x}{30}=0,75\)
Giai pt
\(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\)
\(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\left(ĐKXĐ:x\ne\pm1\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow2x^2+2x-x^2+x=x^2-1\)
\(\Leftrightarrow2x^2+2x-x^2+x-x^2+1=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow x=\dfrac{-1}{3}\left(nhận\right)\)
-Vậy \(S=\left\{\dfrac{-1}{3}\right\}\)
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Giai hệ phương trình :
/x+3/ + \(\dfrac{2}{y}\) = 3
2/x+2/ - \(\dfrac{5}{y}\) = -3
mk cần gấp ạ
1.giai pt
\(\dfrac{2x-1}{x+3}=\dfrac{2x+1}{x-3}\)
\(\dfrac{x^2+3}{x-2}\) =x+5
ĐKXĐ x≠3 ; x≠-3
\(\dfrac{2x-1}{x+3}=\dfrac{2x+1}{x-3}\)
=> (2x-1)(x-3)=(2x+1)(x+3)
⇔2x2-6x-x+3=2x2+6x+x+3
⇔2x2-2x2-7x-6x=3-3
⇔ -13x=0
⇔x=0 (tm)
vậy phương trình trên có tập no S={0}
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