Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow3x=-15\Rightarrow x=-5\)
b/ ĐKXĐ: \(x\ne\left\{-\frac{4}{3};1\right\}\)
\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2+37x+28=12x^2-7x-5\)
\(\Leftrightarrow44x=-33\Rightarrow x=-\frac{3}{4}\)
c/ ĐKXĐ: \(x\ne\left\{-\frac{1}{4};0\right\}\)
\(\Leftrightarrow\frac{3\left(x^2-1\right)}{4x+1}+\frac{2\left(1-x^2\right)}{x}-\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{3}{4x+1}-\frac{2}{x}-1\right)=0\)
TH1: \(x^2-1=0\Rightarrow x=\pm1\)
TH2: \(\frac{3}{4x+1}-\frac{2}{x}-1=0\Leftrightarrow3x-2\left(4x+1\right)-x\left(4x+1\right)=0\)
\(\Leftrightarrow4x^2+6x+2=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{2}\end{matrix}\right.\)
a/ ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)
\(\Leftrightarrow x^2+2x-15=x^2-9\)
\(\Leftrightarrow2x=6\Rightarrow x=3\) (ktm)
Vậy pt vô nghiệm
b/ ĐKXĐ: \(x\ne1\)
\(\Leftrightarrow\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2}{x^2+x+1}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow x^2+x+1+2\left(x-1\right)=3x^2\)
\(\Leftrightarrow2x^2-3x+1=0\Rightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=\frac{1}{2}\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne\pm4\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{\left(x-4\right)\left(x+4\right)}+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow5x^2-80+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2+16=5x^2+2x\)
\(\Rightarrow x=8\)
a/ ĐKXĐ: \(x\ne-1\)
\(\Leftrightarrow4\left(3-7x\right)=x+1\)
\(\Leftrightarrow12-28x=x+1\)
\(\Rightarrow29x=11\Rightarrow x=\frac{11}{29}\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(\Leftrightarrow1-\left(\sqrt{x}-2\right)=3-\sqrt{x}\)
\(\Leftrightarrow3=3\) (luôn đúng)
Vậy nghiệm của pt là \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne7\)
\(\Leftrightarrow8-x-8\left(x-7\right)=1\)
\(\Leftrightarrow8-x-8x+56=1\)
\(\Leftrightarrow-9x=-63\Rightarrow x=7\left(ktm\right)\)
Vậy pt vô nghiệm
d/ ĐKXĐ: \(x\ne4\)
\(\Leftrightarrow\frac{28}{6\left(x-4\right)}-\frac{6\left(x+2\right)}{6\left(x-4\right)}=\frac{-9}{6\left(x-4\right)}-\frac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
\(\Rightarrow x=5\)
e/ ĐKXĐ: \(x\ne\left\{-\frac{2}{3};\frac{1}{3}\right\}\)
\(\Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow3x=-15\Rightarrow x=-5\)
a/ ĐKXĐ: \(x\ne\pm5\)
\(\Leftrightarrow\left(x+5\right)^2-\left(x-5\right)^2=20\)
\(\Leftrightarrow\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow20x=20\Rightarrow x=1\)
b/ ĐKXĐ: \(x\ne\pm4\)
\(\Leftrightarrow2x\left(x-4\right)-4x=0\)
\(\Leftrightarrow2x^2-12x=0\)
\(\Leftrightarrow2x\left(x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne\pm\frac{2}{3}\)
\(\Leftrightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow6x=16\Rightarrow x=\frac{8}{3}\)
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\x\ne2\\x\ne\frac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)
Đặt \(x^2-x-1=a\) ta được:
\(\frac{4}{a-1}+\frac{2}{a}=5\Leftrightarrow4a+2\left(a-1\right)=5a\left(a-1\right)\)
\(\Leftrightarrow5a^2-11a+2=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x-1=2\\x^2-x-1=\frac{1}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3=0\\5x^2-5x-6=0\end{matrix}\right.\) (bấm máy)
b/ ĐKXĐ: \(x>2\)
Đặt \(\sqrt{x-2}=a>0\)
\(\frac{4}{a+1}-\frac{1}{a}=1\Leftrightarrow4a-\left(a+1\right)=a\left(a+1\right)\)
\(\Leftrightarrow a^2-2a+1=0\Rightarrow a=1\)
\(\Rightarrow\sqrt{x-2}=1\Rightarrow x=3\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\frac{4}{9}\end{matrix}\right.\)
\(\Leftrightarrow4\left(2-3\sqrt{x}\right)-\left(\sqrt{x}+1\right)=3\left(\sqrt{x}+1\right)\left(2-3\sqrt{x}\right)\)
\(\Leftrightarrow9x-10\sqrt{x}+1=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=\frac{1}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{81}\end{matrix}\right.\)
1.
\(f\left(x\right)=\frac{\left(x^2-3x\right)^2-2\left(x^2-3x\right)-8}{x^2-3x}=\frac{\left(x^2-3x-4\right)\left(x^2-3x+2\right)}{x^2-3x}\)
\(f\left(x\right)=\frac{\left(x+1\right)\left(x-1\right)\left(x-2\right)\left(x-4\right)}{x\left(x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{0;3\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{-1;1;2;4\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -1\\0< x< 1\\2< x< 3\\x>4\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}-1< x< 0\\1< x< 2\\3< x< 4\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{2x-2\left(x+1\right)-x\left(x+1\right)}{2x\left(x+1\right)}=\frac{-x^2-x-2}{2x\left(x+1\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{-1;0\right\}\)
\(f\left(x\right)>0\Rightarrow-1< x< 0\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -1\\x>0\end{matrix}\right.\)
3.
\(f\left(x\right)=\frac{x^2-4x+3+\left(x-1\right)\left(3-2x\right)}{3-2x}=\frac{-x^2+x}{3-2x}=\frac{x\left(1-x\right)}{3-2x}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\frac{3}{2}\)
\(f\left(x\right)=0\Rightarrow x=\left\{0;1\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}0< x< 1\\x>\frac{3}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< 0\\1< x< \frac{3}{2}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(2-x\right)\left(3x+4\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\pm\sqrt{3};-\frac{4}{3};2\right\}\)
\(f\left(x\right)=0\Rightarrow x=\pm1\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}-\sqrt{3}< x< -\frac{4}{3}\\-1< x< 1\\\sqrt{3}< x< 2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -\sqrt{3}\\-\frac{4}{3}< x< -1\\1< x< \sqrt{3}\\x>2\end{matrix}\right.\)