\(\frac{1}{3}x-\frac{3}{5}=\frac{5}{6}x+2\)
Mn giải giúp e bài này vs ạ.Thanks mn trc nhé
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ĐKXĐ : X khác 1
pt <=> X^2+X+1/(X-1).(X^2+X+1) - 3X^2/(X-1).(X^2+X+1) = 2X.(X-1)/(X-1).(X^2+X+1)
<=> X^2+X+1/(X-1).(X^2+X+1) - 3X^2/(X-1).(X^2+X+1) - 2X^2-2X/(X-1).(X^2+X+1) = 0
<=> X^2+X+1-3X^2-2X^2+2X/(X-1).(X^2+X+1) = 0
<=> X^2+X+1-3X^2-2X^2+2X=0
<=> -4X^2+3X+1=0
<=> 4X^2-3X-1=0
<=> (X-1).(4X+1) = 0
<=> 4X+1=0 ( vì X khác 1 nên X-1 khác 0 )
<=> X = -1/4 (tm)
Vậy pt có tập nghiệm S = {-1/4}
Tk mk nha
\(\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x}{2021}\)
\(\Leftrightarrow\frac{x+2}{2019}+1+\frac{x+3}{2018}+1=\frac{x+4}{2017}+1+\frac{x}{2021}+1\)
\(\Leftrightarrow\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2021}\)
\(\Leftrightarrow x+2021=0\)
\(\Leftrightarrow x=-2021\)
CÁi này easy mà .-.
\(\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)
\(\Leftrightarrow\frac{\frac{\left(7-x\right)-\left(x-5\right)}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+\left(x-6\right)=0\)
\(\Leftrightarrow\frac{\frac{-2\left(x-6\right)}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{\frac{-2}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+1\right)=0\)
\(\Rightarrow x-6=0\Rightarrow x=6\)
a. (x-3)(x\(^2\)+6x+9)(x-1)(x\(^2\)+2x+1)(-x\(^2\)+2x+3)=0
\(\Leftrightarrow\)(x-3)(x\(^2\)+6x+9)(x-1)(x\(^2\)+2x+1)(x-3)(x+1)=0
a) \(3\left(x-4\right)+5=2\left(x+1\right)-8\)
\(\Leftrightarrow3x-12+5=2x+2-8\)
\(\Leftrightarrow x=1\)
Vậy : \(S=\left\{1\right\}\)
b) \(5\left(x+1\right)^2+2x=5x^2-3\)
\(\Leftrightarrow5x^2+10x+5+2x=5x^2-3\)
\(\Leftrightarrow12x=-8\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy : \(S=\left\{-\frac{2}{3}\right\}\)
c) \(\frac{4\left(x+2\right)}{15}=\frac{13x-9}{40}\)
\(\Leftrightarrow32\left(x+2\right)=3\left(13x-9\right)\)
\(\Leftrightarrow32x-39x=-27-64\)
\(\Leftrightarrow-7x=-91\)
\(\Leftrightarrow x=13\)
Vậy : \(S=\left\{13\right\}\)
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
\(\frac{1}{3}x-\frac{3}{5}=\frac{5}{6}x+2\)
\(\Leftrightarrow\frac{x}{3}-\frac{3}{5}=\frac{5x}{6}+2\)
\(\Leftrightarrow2x-\frac{18}{5}=5x+12\)
\(\Leftrightarrow2x-5x=\frac{18}{5}+12\)
\(\Leftrightarrow-3x=\frac{78}{5}\)
\(\Leftrightarrow3x=-\frac{78}{5}\)
\(\Leftrightarrow x=-\frac{26}{5}\)
Ps: đoạn nào không hiểu hỏi anh nhé. Nhớ k để tạo động lực cho anh nhé :33
# Aeri #