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a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{5x+2}{4-x^2}\left(x\ne\pm2\right)\)
\(=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x-8+3x+6-5x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x-2\right)\left(x+2\right)}\)
f) \(x^2+1-\frac{x^4-3x^2+2}{x^2-1}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x^2-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\left(x^2-2\right)\)
\(=x^2+1-x^2+2\)
\(=3\)
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)
\(\Leftrightarrow\)\(-4x=-1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy...
mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))
a) \(\left(\frac{1}{x}+2\right)=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(x^2+1\right)-\left(\frac{1}{x}+2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)x^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}+2=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\left(L\right)\end{cases}}\)
Vậy \(x=-\frac{1}{2}\)
s e thấy == câu này mọi ngừi ko tl vậy :v ( bài này cs cần đk ko -.- e chưa hc nên ko nắm chắc , kệ đi , cứ lm )
\(a,\left(\frac{1}{x}+2\right)=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(1+2x=x\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(1+2x=x^2+1+2x^3+2x\)
\(2x=x^2+2x^3+2x\)
\(0=x^2+2x^3\)
\(0=x^2\left(1+2x\right)\)
\(x=0;-\frac{1}{2}\)
\(\frac{1}{9}\left(x-3\right)^2-\frac{1}{25}\left(x+5\right)^2=0\)
\(\Leftrightarrow\left[\frac{1}{3}\left(x-3\right)\right]^2-\left[\frac{1}{5}\left(x+5\right)\right]^2=0\)
\(\Leftrightarrow\left(\frac{1}{3}x-1\right)^2-\left(\frac{1}{5}x+1\right)^2=0\)
\(\Leftrightarrow\left(\frac{2}{15}x-2\right)\left(\frac{8}{15}x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=15\\x=0\end{cases}}\)
\(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}\right)^2-\left(x-1-\frac{1}{x}\right)^2=0\)
\(\Leftrightarrow\left(2+\frac{2}{x}\right)\left(2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0L\end{cases}}\)
Vậy x= - 1
\(c,\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)
\(4x^2-3x-18=4x^2+3x\)
\(-3x-18=3x\)
\(-18=6x\)
\(x=-3\)
\(\left(\frac{2x}{3}+1\right)^2=\left(\frac{3x}{2}-1\right)^2\)
\(\Leftrightarrow\left(\frac{2x}{3}+1\right)^2-\left(\frac{3x}{2}-1\right)^2=0\)
\(\Leftrightarrow\left(\frac{13x}{6}\right)\left(\frac{-5x}{6}+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{12}{5}\end{cases}}\)
\(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\)
\(\left(x^2-17x+33\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)-\left(x+2\right)\left(x-3\right)\)
\(\left(x^2-17x+33\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(16x^2-25\right)=0\)
\(\Leftrightarrow x\in\left\{-2;3;\pm\frac{5}{4}\right\}\)
câu b e giải trc câu a nhưng dài lắm cj ơi , rút ngắn lại nghe
\(b,\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
\(17x^4-34x^3-77x^2+94x-48=x^4-18x^3+44x^2+69x-198\)
\(17x^4-34x^3-77x^2+94x-48-x^4+18x^3+44x^2-69x+198=0\)
\(\left(16x^3-48x^2-2x+75\right)\left(x+2\right)=0\)
\(\left(x-3\right)\left(16x^2-25\right)\left(x+2\right)=0\)
\(x=3;-2;\pm\frac{5}{4}\)
\(d,\frac{1}{9}\left(x-3\right)^2-\frac{1}{25}\left(x+5\right)^2=0\)
\(\frac{1\left(x-3\right)2}{9}-\frac{1}{25}\left(x+5\right)2=0\)
\(\frac{2\left(x-3\right)}{9}-\frac{1\left(x+5\right)2}{25}=0\)
\(\frac{2\left(x+3\right)}{9}-\frac{2\left(x+5\right)}{25}=0\)
\(2\left(\frac{x-3}{9}-\frac{x+5}{25}\right)=0\)
\(\frac{x-3}{9}-\frac{x+5}{25}=0\)
\(25\left(x-3\right)-9\left(x+5\right)=0\)
\(x=7,5\)
\(e,\left(\frac{2x}{3}+1\right)^2=\left(\frac{3x}{2}-1\right)^2\)
\(\frac{4x^2}{9}+\frac{4x}{3}+1=\frac{9x^2}{4}-3x+1\)
\(\frac{4x^2}{9}+\frac{4x}{3}=\frac{9x^2}{4}-3x\)
\(16x^2+48x=81x^2-108x\)
\(16x^2+48x-81x^2+108x=0\)
\(-65x^2+156x=0\)
\(-13x\left(5x-12\right)=0\)
\(x=0;\frac{12}{5}\)
\(f,\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)
\(\left(x+1+\frac{1}{x}\right)^2-\left(x-1-\frac{1}{x}\right)^2=0\)
\(x=-1\)