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a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
a: \(=\dfrac{\sqrt{m}\left(m+4n-4\sqrt{mn}\right)}{\sqrt{mn}\left(\sqrt{m}-2\sqrt{n}\right)}\)
\(=\dfrac{1}{\sqrt{n}}\cdot\left(\sqrt{m}-2\sqrt{n}\right)\)
b: \(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
c: \(=\sqrt{5^2\cdot2\cdot x^2y^4\cdot xy}-\dfrac{2y^2}{x^2}\cdot4\sqrt{2}\cdot x^3\sqrt{xy}+\dfrac{3}{2}xy\cdot\sqrt{2}\cdot y\cdot\sqrt{xy}\)
\(=5xy^2\sqrt{2xy}-8\sqrt{2xy}xy^2+\dfrac{3}{2}xy^2\cdot\sqrt{2xy}\)
\(=-\dfrac{3}{2}\sqrt{2xy}\)
d: \(=\left(x+2\right)\cdot\dfrac{\sqrt{2x-3}}{\sqrt{x+2}}=\sqrt{\left(2x-3\right)\left(x+2\right)}\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)-xy=100\\xy-\left(x-2\right)\left(y-2\right)=64\end{matrix}\right.\)
=>xy+3x+2y+6-xy=100 và xy-xy+2x+2y-4=64
=>3x+2y=94 và 2x+2y=68
=>x=26 và x+y=34
=>x=26 và y=8
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3+2}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5y+20-11}{y+4}=9\end{matrix}\right.\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+1}-\dfrac{2}{y+4}=4-3=1\\\dfrac{-2}{x+1}+\dfrac{11}{y+4}=9+5-2=12\end{matrix}\right.\)
=>x+1=18/35; y+4=9/13
=>x=-17/35; y=-43/18
Chắc đề là \(x+y+z=3\)
Ta có:
\(\left(2x+y+z\right)^2=\left(x+y+x+z\right)^2\ge4\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow P\le\dfrac{x}{4\left(x+y\right)\left(x+z\right)}+\dfrac{y}{4\left(x+y\right)\left(y+z\right)}+\dfrac{z}{4\left(x+z\right)\left(y+z\right)}\)
\(\Rightarrow P\le\dfrac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{4\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\dfrac{xy+yz+zx}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Mặt khác:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(zy+yz+zx\right)=\dfrac{8}{3}\left(xy+yz+zx\right)\)
\(\Rightarrow P\le\dfrac{xy+yz+zx}{2.\dfrac{8}{3}\left(xy+yz+zx\right)}=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Bài 1:
\(P=(x+1)\left(1+\frac{1}{y}\right)+(y+1)\left(1+\frac{1}{x}\right)\)
\(=2+x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}\)
Áp dụng BĐT Cô-si:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(x+\frac{1}{2x}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
\(y+\frac{1}{2y}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
Áp dụng BĐT SVac-xơ kết hợp với Cô-si:
\(\frac{1}{2x}+\frac{1}{2y}\geq \frac{4}{2x+2y}=\frac{2}{x+y}\geq \frac{2}{\sqrt{2(x^2+y^2)}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Cộng các BĐT trên :
\(\Rightarrow P\geq 2+2+\sqrt{2}+\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
Vậy \(P_{\min}=4+3\sqrt{2}\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)
Bài 2:
Áp dụng BĐT Svac-xơ:
\(\frac{1}{a+3b}+\frac{1}{b+a+2c}\geq \frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
\(\frac{1}{b+3c}+\frac{1}{b+c+2a}\geq \frac{4}{2b+4c+2a}=\frac{2}{b+2c+a}\)
\(\frac{1}{c+3a}+\frac{1}{c+a+2b}\geq \frac{4}{2c+4a+2b}=\frac{2}{c+2a+b}\)
Cộng theo vế và rút gọn :
\(\Rightarrow \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\geq \frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
