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Bác google được sinh ra để làm gì, đăng nhiều vc, google có hết mà ;v
Bài 1,2,3,4 đơn giản, tự làm :v
7) \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=\dfrac{abc}{c^3}+\dfrac{abc}{a^3}+\dfrac{abc}{b^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{1}{3abc}=\dfrac{1}{3}\)
P/S: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
5) ĐK: a>b>0
\(3a^2+3b^2=10ab\Leftrightarrow\left(a-3b\right)\left(3a-b\right)=0\)
Tự phân tích
Mà a>b>0=> Chọn a=3b
Thay vào
Bài 6 tương tự bài 5
Có bất mãn chỗ nào thì ib nha bạn :))
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b: \(=\dfrac{6x^2-x}{6x-1}=x\)
c: \(=\dfrac{2x+2y+3x-3y-4x}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x+y}\)
d: \(=\dfrac{x+1}{\left(x-1\right)^2}\cdot\dfrac{5\left(x-1\right)}{x+1}=\dfrac{5}{x-1}\)
f: \(=\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a: Thiếu vế phải rồi bạn
b: \(\Leftrightarrow\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)^2>=4xy\)
\(\Leftrightarrow\left(x-y\right)^2>=0\)(luôn đúng)
\(S=\frac{5\left(x-1\right)}{4}+\frac{5}{x-1}+\frac{9\left(y-1\right)}{4}+\frac{9}{y-1}+\frac{7}{4}\left(x+y\right)+\frac{7}{2}\)
\(S\ge2\sqrt{\frac{25\left(x-1\right)}{4\left(x-1\right)}}+2\sqrt{\frac{81\left(y-1\right)}{4}}+\frac{7}{4}.6+\frac{7}{2}=28\)
\(\Rightarrow S_{min}=28\) khi \(x=y=3\)
1)???
2) \(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=2+\dfrac{x^2-4x+4}{x^2-2x+1}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)
Vậy GTNN của A là 2 tại x=2.
3) \(\)Đặt \(a=\dfrac{1}{x+100}\Rightarrow x=\dfrac{1}{a}-100\)
\(D=\dfrac{x}{\left(x+100\right)^2}=a^2x=a^2\left(\dfrac{1}{a}-100\right)=a-100a^2=-100\left(a^2-\dfrac{a}{100}+\dfrac{1}{40000}-\dfrac{1}{40000}\right)=-100\left(a-\dfrac{1}{200}\right)^2+\dfrac{1}{400}\le\dfrac{1}{400}\)
Vậy GTLN của D là \(\dfrac{1}{400}\) tại \(a=\dfrac{1}{200}\Leftrightarrow x=100\)
áp dụng BĐT bunhiacopxki ta có
\(\left(1+2^2\right)\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
<=> \(5\left(x^2+4y^2\right)\ge1\)
<=> \(x^2+4y^2\ge\dfrac{1}{5}\) (đpcm)
S=5/4(x-1)+5/x-1+9/4(y-1)+9/(y-1)+7/4(x+y)+7/2
=5/4(x-1)+5/(x-1)+9/4(y-1)+9/y-1+14
=>S>=2*5/2+2*9/2+14=28
Dấu = xảy ra khi x=y=3