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9 tháng 3 2020
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
9 tháng 3 2020
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 <=> \(y\ge0\)
20 tháng 4 2018
a/ Đặt \(\hept{\begin{cases}\frac{x+1}{x-2}=a\\\frac{x+1}{x-4}=b\end{cases}}\) thì có
\(a^2+b-\frac{12b^2}{a^2}=0\)
\(\Leftrightarrow\left(a^2-3b\right)\left(a^2+4b\right)=0\)
b/ \(2x^2+3xy-2y^2=7\)
\(\Leftrightarrow\left(2x-y\right)\left(x+2y\right)=7\)
WR
11 tháng 6 2017
1)
a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)
(đk:x khác \(\frac{1}{2}\))
\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)
Vậy x=\(\frac{25}{7}\)
b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)
(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))
\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)
Vậy x=4
2)
\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)
\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)
\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 4 2019
a/
\(9x^2+25y^2+1+30xy-6x-10y+4y^2-20y+25=0\)
\(\Leftrightarrow\left(3x+5y-1\right)^2+\left(2y-5\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y-1=0\\2y-5=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{23}{6}\\y=\frac{5}{2}\end{matrix}\right.\)
b/
\(4x^2+4y^2+8xy+x^2-2x+1+y^2+2y+1=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
c/
\(y^2-2y+1+2=\frac{6}{x^2+2x+1+3}\)
\(\Leftrightarrow\left(y-1\right)^2+2=\frac{6}{\left(x+1\right)^2+3}\)
Ta có \(VT=\left(y-1\right)^2+2\ge2\)
\(\left(x+1\right)^2+3\ge3\Rightarrow VP=\frac{6}{\left(x+1\right)^2+3}\le\frac{6}{3}=2\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}y-1=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
d/
\(\frac{-9x^2+18x-9-8}{x^2-2x+1+2}=y^2+4y+4-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-8}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-18+10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow-9+\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2+5\)
Ta có \(\left(x-1\right)^2+2\ge2\Rightarrow\frac{10}{\left(x-1\right)^2+2}\le\frac{10}{2}=5\Rightarrow VT\le5\)
\(\left(y+2\right)^2+5\ge5\Rightarrow VP\ge5\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
20 tháng 7 2019
\(\text{a) }\frac{6}{x-4}-\frac{x}{x+2}=\frac{6}{x-4}.\frac{x}{x+2}\)
\(ĐKXĐ:x\ne-2;x\ne4\)
\(MTC:\left(x-4\right)\left(x+2\right)\)
\(\Leftrightarrow\frac{6\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}-\frac{x\left(x-4\right)}{\left(x-4\right)\left(x+2\right)}=\frac{6x}{\left(x-4\right)\left(x+2\right)}\)
\(\Rightarrow6\left(x+2\right)-x\left(x-4\right)=6x\)
\(\Leftrightarrow6x+12-x^2+4x=6x\)
\(\Leftrightarrow6x+12-x^2+4x-6x=0\)
\(\Leftrightarrow-x^2+4x+12=0\)
\(\Leftrightarrow-\left(x^2-4x-12\right)=0\)
\(\Leftrightarrow x^2-4x-12=0\)
\(\Leftrightarrow x^2+2x-6x-12=0\)
\(\Leftrightarrow x\left(x+2\right)-6\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)
\(\Leftrightarrow x=-2\left(\text{loại}\right)\text{ hoặc }x=6\left(\text{nhận}\right)\)
Vậy \(S=\left\{6\right\}\)
\(\text{b) }\frac{2x+3}{2x-1}=\frac{x-3}{x+5}\)
\(ĐKXĐ:x\ne\frac{1}{2};x\ne-5\)
\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\left[\text{Tỉ lệ thức}\right]\)
\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)
\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)
\(\Leftrightarrow2x^2+13x-2x^2+7x=3-15\)
\(\Leftrightarrow20x=-12\)
\(\Leftrightarrow x=\frac{-12}{20}=\frac{-3}{5}\)
Vậy \(S=\left\{\frac{-3}{5}\right\}\)
nhầm ở dòng thứ 2 từ dưới lên: \(\hept{\begin{cases}\left(x-1\right)^2+2=2\\\frac{6}{\left(x+1\right)^2+3}=2\end{cases}}\) ms đúng ,sau vẫn giải bth
\(y^2-2y+3=\frac{6}{x^2+2x+4}\)
\(y^2-2y+1+2=\frac{6}{x^2+2x+1+3}\)
\(\left(y-1\right)^2+2=\frac{6}{\left(x+1\right)^2+3}\)
Vì \(\left(y-1\right)^2+2\ge2\) với mọi y
\(\left(x+1\right)^2+3\ge3=>\frac{6}{\left(x+1\right)^2+3}\le\frac{6}{3}=2\) với mọi x
Ta có : \(VT\ge2;VP\le2\) nên để \(VT=VP\)
thì \(\hept{\begin{cases}\left(y-1\right)^2+2=0\\\frac{6}{\left(x+1\right)^2+3}=0\end{cases}}< =>\hept{\begin{cases}y=1\\x=-1\end{cases}}\)
Vậy x=-1;y=1 là nghiệm của pt........