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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
a)
\(A=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)
\(A-2=-\dfrac{3}{x^2-8x+22}=-\dfrac{3}{\left(x-4\right)^2+6}\ge-\dfrac{3}{6}=-\dfrac{1}{2}\)
\(A\ge\dfrac{3}{2}\) khi x =4
minh giai phan d, nha bn :
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
a)
-x2+x+1=-(x2-x-1)=\(-\left(x^2-2.\frac{1}{2}.x+\frac{1}{4}-\frac{5}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{5}{4}\right]=\frac{5}{4}-\left(x-\frac{1}{2}\right)^2\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\frac{5}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{5}{4}\Leftrightarrow-x^2+x+1\le\frac{5}{4}\)
Dấu "=" xảy ra khi (x-1/2)2=0 => x-1/2=0 => x=1/2
Vậy max của biểu thức -x2+x+1 là 5/4 khi x=1/2
b) câu này trình bày tương tự câu trên thôi
\(x^2+x+1=x^2+2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra khi x=-1/2
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
\(\dfrac{4a^2-9b^2}{a^2b^2}\div\dfrac{2ax+3bx}{2ab}\)
\(=\dfrac{\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2}\times\dfrac{2ab}{x\left(2a+3b\right)}\)
\(=\dfrac{2ab\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2x\left(2a+3b\right)}=\dfrac{4a-6b}{xab}\)
2 x25−4b2:15+2b
\(=\dfrac{2x}{\left(5-2b\right)\left(5+2b\right)}\times\dfrac{5+2b}{1}\)
\(=\dfrac{2x\left(5+2b\right)}{\left(5-2b\right)\left(5+2b\right)}=\dfrac{2x}{5-2b}\)
(2−a)22ab.b(2−a)+12
\(=\dfrac{\left(2-a\right)^2b}{2ab\left(2-a\right)}+\dfrac{1}{2}\)
\(=\dfrac{2b-ab}{2ab}+\dfrac{1}{2}\)
\(=\dfrac{2b-ab}{2ab}+\dfrac{ab}{2ab}=\dfrac{2b}{2ab}=\dfrac{1}{a}\)
2 b+22b−b2:b+1b+2b+23b−6
\(=\dfrac{2\left(b+1\right)}{b\left(2-b\right)}\times\dfrac{b}{b+1}+\dfrac{2b+2}{3b-6}\)
\(=\dfrac{2b\left(b+1\right)}{\left(2-b\right)b\left(b+1\right)}+\dfrac{2b+2}{3b-6}\)
\(=\dfrac{2}{2-b}-\dfrac{2\left(b+1\right)}{3\left(2-b\right)}\)
\(=\dfrac{6}{3\left(2-b\right)}-\dfrac{2\left(b+1\right)}{3\left(2-b\right)}\)
\(=\dfrac{6-2\left(b+1\right)}{3\left(2-b\right)}\)
\(=\dfrac{4-2b}{3\left(2-b\right)}=\dfrac{2\left(2-b\right)}{3\left(2-b\right)}=\dfrac{2}{3}\)
a.
\(F=\dfrac{a}{b+2}\Rightarrow F.b+2F=a\)
\(\Rightarrow2F=a-F.b\)
\(\Rightarrow4F^2=\left(a-F.b\right)^2\le\left(a^2+b^2\right)\left(1^2+F^2\right)=F^2+1\)
\(\Rightarrow3F^2\le1\)
\(\Rightarrow-\dfrac{1}{\sqrt{3}}\le F\le\dfrac{1}{\sqrt{3}}\)
Dấu "=" lần lượt xảy ra tại \(\left(a;b\right)=\left(-\dfrac{\sqrt{3}}{2};-\dfrac{1}{2}\right)\) và \(\left(\dfrac{\sqrt{3}}{2};-\dfrac{1}{2}\right)\)
b. Đặt \(\left\{{}\begin{matrix}a+b=x\\a-2b=y\end{matrix}\right.\) quay về câu a