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a) \(B=\frac{3x^2+6x+10}{x^2+2x+5}\)
\(\Leftrightarrow B=3-\frac{5}{x^2+2x+5}\)
\(\Leftrightarrow B=3-\frac{5}{5\left(\frac{x^2}{5}+\frac{2x}{5}+\frac{5}{5}\right)}\Leftrightarrow B=3-\frac{1}{\frac{\left(x^2+2x+1\right)}{5}+\frac{4}{5}}\)( cho \(\left(x+1\right)^2=0\))
\(\Leftrightarrow maxB=3-\frac{1}{\frac{4}{5}}=\frac{7}{4}\) KHI X= -1
c) \(D=x^2-2x+y^2+4y+7\)
\(\Leftrightarrow D=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2\)
\(\Leftrightarrow D=\left(x-1\right)^2+\left(y+2\right)^2+2\)
\(\Leftrightarrow minD=2\)KHI X= 1 và Y= -2
e) Câu này đề có vẻ sai bạn kiểm tra lại giúp mk ! mk làm theo đề đúng nka !
\(E=\frac{x^2-4x+1}{x^2}\)
\(\Leftrightarrow E=\frac{x^2\left(1-\frac{4}{x}+\frac{1}{x^2}\right)}{x^2}=1-\frac{4}{x}+\frac{1}{x^2}\)
ĐẶT \(y=\frac{1}{x}\)\(\Leftrightarrow minE=-3\)KHI X = 1/2
Hai câu còn lại tối mk giải tiếp mk bận đi học rùi bạn thông cảm
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) \(x^2+1-\frac{x^4-3x^2+2}{x^2-1}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x^2-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\left(x^2-2\right)\)
\(=x^2+1-x^2+2\)
\(=3\)
ko bt có sai ko nữa mà mình tìm ra câu a hai nghiệm:\(\frac{-11+\sqrt{69}}{26}\)
và \(\frac{-11-\sqrt{69}}{29}\)
d) \(\frac{1}{2x-3}-\frac{3}{x.\left(2x-3\right)}=\frac{5}{x}\)
\(\Leftrightarrow\frac{x}{x.\left(2x-3\right)}-\frac{3}{x.\left(2x-3\right)}=\frac{5.\left(2x-3\right)}{x.\left(2x-3\right)}\)
\(\Leftrightarrow x-3=5.\left(2x-3\right)\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=\left(-15\right)+3\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow9x=12\)
\(\Leftrightarrow x=12:9\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{4}{3}\right\}.\)
Chúc bạn học tốt!
c, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
- Ta có : \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
=> \(\frac{12\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}=\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}\)
=> \(12\left(x-3\right)-8\left(x-1\right)=8\left(x-1\right)\)
=> \(12x-36-8x+8-8x+8=0\)
=> \(-4x-20=0\)
=> \(x=-5\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-5\right\}\)
b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\2x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
Ta có : \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
=> \(\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
=> \(x-3=5\left(2x-3\right)\)
=> \(x-3-10x+15=0\)
=> \(-9x=-12\)
=> \(x=\frac{4}{3}\) ( TM )
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{4}{3}\right\}\)
\(a,\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\frac{2-x}{\left(x+1\right)\left(2-x\right)}+\frac{5x+5}{\left(2-x\right)\left(x+1\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow7+4x=15\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
\(\Leftrightarrow Ptvn\)
\(b,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{10x-15}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-3-10x+15=0\)
\(\Leftrightarrow-9x+12=0\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow\frac{4}{3}\)
\(c,\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{6x-18}{\left(x-1\right)\left(x-3\right)}-\frac{4x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4x-4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow6x-18-4x+4=4x-4\)
\(\Leftrightarrow2x-14=4x-4\)
\(\Leftrightarrow-2x=10\)
\(\Leftrightarrow x=-5\)
\(d,\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{3x-9}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow3x-9+2x-4=x-1\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow Ptvn\)
Vậy .................................
Ta có : \(P=2x^2-8x+1=2\left(x^2-4x\right)+1=2\left(x^2-4x+4-4\right)+1=2\left(x-2\right)^2-7\)
Vì \(2\left(x-2\right)^2\ge0\forall x\)
Nên : \(P=2\left(x-2\right)^2-7\ge-7\forall x\in R\)
Vậy \(P_{min}=-7\) khi x = 2
tách thành hàng đẳng thức
nhớ tickkkkk cho mình nha
a/ B=\(\frac{2}{-x^2+6x-12}=\frac{2}{-\left(x-3\right)^2-3}\ge\frac{-2}{3}\) dau bang khi x =0