Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, Đặt \(x^2-4x+8=a\left(a>0\right)\)
\(\Rightarrow a-2=\frac{21}{a+2}\)
\(\Leftrightarrow a^2-4=21\Rightarrow a^2=25\Rightarrow a=5\)
Thay vào là ra
b) ĐK: \(y\ne1\)
bpt <=> \(\frac{4\left(1-y\right)}{1-y^3}+\frac{1+y+y^2}{1-y^3}+\frac{2y^2-5}{1-y^3}\le0\)
<=> \(\frac{3y^2-3y}{1-y^3}\le0\)
\(\Leftrightarrow\frac{y\left(y-1\right)}{\left(y-1\right)\left(y^2+y+1\right)}\ge0\)
\(\Leftrightarrow\frac{y}{y^2+y+1}\ge0\)
vì \(y^2+y+1=\left(y+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
nên bpt <=> \(y\ge0\)
a) \(7x-8=4x+7\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
b) \(\frac{5x-4}{12}=\frac{16x+1}{7}\)
\(\Leftrightarrow35x-28=192x+12\)
\(\Leftrightarrow157x=-40\Leftrightarrow x=\frac{-40}{157}\)
c)\(ĐKXĐ:x\ne\pm2\)
\(\frac{y+1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)
\(\Rightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+y^2-4}{y^2-4}\)
\(\Rightarrow\frac{y^2+3y+2-5y+10}{y^2-4}=\frac{12+y^2-4}{y^2-4}\)
\(\Rightarrow y^2-2y+12=12+y^2-4\)
\(\Rightarrow-2y=-4\Leftrightarrow y=2\left(ktm\right)\)
Vậy pt vô nghiệm
\(1,\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}=\frac{x^2+y^2+z^2}{5}=\frac{x^2}{5}+\frac{y^2}{5}+\frac{z^2}{5}\)
\(=>\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}-\left(\frac{x^2}{5}+\frac{y^2}{5}+\frac{z^2}{5}\right)=0\)
\(=>\left(\frac{x^2}{2}-\frac{x^2}{5}\right)+\left(\frac{y^2}{3}-\frac{y^2}{5}\right)+\left(\frac{z^2}{4}-\frac{z^2}{5}\right)=0\)
\(=>\left(\frac{5x^2}{10}-\frac{2x^2}{10}\right)+\left(\frac{5y^2}{15}-\frac{3y^2}{15}\right)+\left(\frac{5z^2}{20}-\frac{4z^2}{20}\right)=0\)
\(=>\frac{3}{10}x^2+\frac{2}{15}y^2+\frac{1}{20}z^2=0\)
Tổng 3 số không âm=0 <=> chúng đều=0
\(< =>\frac{3}{10}x^2=\frac{2}{15}y^2=\frac{1}{20}z^2=0< =>x=y=z=0\)
Vậy x=y=z=0
\(2,x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)
\(=>x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}-4=0\)
\(=>\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)=0\)
\(=>\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)
\(=>\left(x^2-2.x.\frac{1}{x}+\frac{1}{x^2}\right)+\left(y^2-2.y.\frac{1}{y}+\frac{1}{y^2}\right)=0\)
\(=>\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
Tổng 2 số không âm=0 <=> chúng đều=0
\(< =>\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}< =>\hept{\begin{cases}x=\frac{1}{x}\\y=\frac{1}{y}\end{cases}< =>\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}}}\)\(< =>\hept{\begin{cases}x\in\left\{-1;1\right\}\\y\in\left\{-1;1\right\}\end{cases}}\)
Vậy có 4 cặp (x;y) cần tìm là (1;1) ;(1;-1);(-1;1);(-1;-1)
ĐKXĐ : \(x\ne\pm1\)
Pt \(\Leftrightarrow\frac{2\cdot3\cdot\left(y+1\right)+\left(y-1\right)}{3\left(y-1\right)\left(y+1\right)}-\frac{1}{6}=0\)
\(\Leftrightarrow\frac{7y+5}{3\left(y-1\right)\left(y+1\right)}-\frac{3\left(y-1\right)\left(y+1\right)}{6\left(y-1\right)\left(y+1\right)}=0\)
\(\Leftrightarrow14y+10-3y^2+3=0\)
\(\Leftrightarrow3y^2+14y+13=0\)
\(\Leftrightarrow y=-\frac{7}{3}\pm\frac{\sqrt{10}}{3}\)
Từ phương trình, ta suy ra:
\(\frac{y^2-2y}{y^2-4}-\frac{3y+6}{y^2-4}=\frac{y^2+8}{y^2-4}\)
\(\Leftrightarrow\frac{y^2-5y-6}{y^2-4}=\frac{y^2+8}{y^2-4}\)
\(\Leftrightarrow\frac{-5y-14}{y^2-4}=0\)
\(\Leftrightarrow-5y-14=0\)(với ĐKXĐ \(x\ne\pm2\))
\(\Leftrightarrow-5y=14\)
\(\Leftrightarrow y=\frac{-14}{5}\)(phù hợp với ĐKXĐ)
Vậy phương trình có nghiệm duy nhất y=-14/5
\(\frac{y}{y+2}-\frac{3}{y-2}=\frac{y^2+8}{y^2-4}\left(y\ne\pm2\right)\)
\(\Leftrightarrow\frac{y\left(y-2\right)}{\left(y-2\right)\left(y+3\right)}-\frac{3\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}-\frac{y^2+8}{\left(y-2\right)\left(y+2\right)}=0\)
\(\Leftrightarrow\frac{y^2-2y}{\left(y-2\right)\left(y+2\right)}-\frac{3y+6}{\left(y-2\right)\left(y+2\right)}-\frac{y^2+8}{\left(y-2\right)\left(y+2\right)}=0\)
\(\Rightarrow y^2-2y-3y-6-y^2-8=0\)
\(\Leftrightarrow-5y-14=0\)
<=> -5y=14
<=> x=-14/5