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a: \(\left\{{}\begin{matrix}3x-2y=1\\2x+4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-4y=2\\2x+4y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x=5\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\2y=3x-1=\dfrac{15}{8}-1=\dfrac{7}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=\dfrac{7}{16}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}4x-3y=1\\-x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-3y=1\\-4x+8y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1+2y=-1+2=1\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\dfrac{2}{3}x+\dfrac{4}{3}y=1\\\dfrac{1}{2}x-\dfrac{3}{4}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=3\\2x-3y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{41}{14}\\y=-\dfrac{5}{7}\end{matrix}\right.\)
a) đkxđ x≥0 , x ≠1
\(K=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)b)
\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2-1}{\sqrt{x}-2}=1-\dfrac{1}{\sqrt{x}-2}\)
để K ∈ z thì \(\dfrac{-1}{\sqrt{x}-2}\) nguyên
=> √x -2 ∈ Ư(-1)={-1;1}
=> x ∈ {1; 9}
vậy ...
a: \(=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{x-2\sqrt{x}}\)
\(=\dfrac{x-3\sqrt{x}}{x-2\sqrt{x}}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
b: Để K là số nguyên thì \(\sqrt{x}-2-1⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1\right\}\)
hay x=9
c: Để K là số âm thì \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}< 0\)
=>4<x<9
ta có:\(A=\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}+\dfrac{2007}{ab+bc+ca}\)
Áp dụng BĐT cauchy-schwarz:
\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(1+2\right)^2}{\left(a+b+c\right)^2}\ge\dfrac{9}{9}=1\)
mà \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le3\)
do đó \(A\ge1+\dfrac{2007}{3}=670\)
dấu = xảy ra khi và chỉ khi a=b=c=1(làm tắt)
a) \(\sqrt{\left(\sqrt{7-2}\right)^2}=\sqrt{5}\)
b)\(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-3\sqrt{2}\right)^2}\)
=\(\sqrt{2}-1-2+3\sqrt{2}=4\sqrt{2}-3\)
c)\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=2\sqrt{3}\)
d) hình như bn ghi sai
e)\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}\right):\sqrt{2}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{3}+1}\right):\sqrt{2}\)
=\(\dfrac{\sqrt{2+\sqrt{3}}\left(\sqrt{3}+1\right)+\sqrt{2-\sqrt{3}}\left(\sqrt{3}-1\right)}{2\sqrt{2}}\)
=\(\dfrac{\sqrt{6+3}+\sqrt{2+\sqrt{3}}+\sqrt{6-3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{2+\sqrt{3}}+\sqrt{3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{3}}{2\sqrt{2}}\)
f) \(\sqrt{9a^2}+3a-7=-3a+3a-7=-7\)
g)\(\dfrac{\sqrt{4x^2-4x+1}}{4x-2}+3x+2\)
=\(\dfrac{\sqrt{\left(2x-1\right)^2}}{4x-2}+3x+2=\dfrac{2x-1}{2\left(2x-1\right)}+3x+2\)
=\(\dfrac{1}{2}+3x+2=\dfrac{5}{2}+3x\)
h)\(\sqrt{\left(5a-1\right)^2}+2a-3\)
nếu a<0 :\(-5a+1+2a-3=-3a-2\)
nếu a>0 : \(5a-1+2a-3=7a-4\)
i)\(\sqrt{\dfrac{2a}{5}}.\sqrt{\dfrac{5a}{18}}+2\left(a-1\right)\)
=\(\sqrt{\dfrac{10a^2}{90}}+2a-2=\sqrt{\dfrac{a^2}{9}}+2a-2\)
=\(\dfrac{a}{3}+2a-2=\dfrac{7a}{3}-2\)
Vừa lúc đang kiếm bđt :))
Đặt \(A=\dfrac{2}{ab}+\dfrac{3}{a^2+b^2}\)
\(A=\dfrac{4}{2ab}+\dfrac{4}{a^2+b^2}-\dfrac{1}{a^2+b^2}\)
Áp dụng hệ quả của bđt AM-GM ta có:
\(A=4\left(\dfrac{1}{2ab}+\dfrac{1}{a^2+b^2}\right)-\dfrac{1}{a^2+b^2}\ge4\cdot\dfrac{4}{2ab+a^2+b^2}-\dfrac{1}{\dfrac{\left(a+b\right)^2}{2}}=16-2=14\)
"="<=>a=b=0,5
=>đpcm
Đặt \(\left(a^{\dfrac{1}{3}};b^{\dfrac{1}{3}};c^{\dfrac{1}{3}}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\\\left(a^3;b^3;c^3\right)\rightarrow\left(x^9;y^9;z^9\right)\end{matrix}\right.\)
\(BDT\Leftrightarrow\dfrac{1}{2x^9+3x^3+2}+\dfrac{1}{2y^9+3y^3+2}+\dfrac{1}{2z^9+3z^3+2}\ge\dfrac{3}{7}\)
Ta có BĐT: \(\dfrac{1}{2x^9+3x^3+2}\ge\dfrac{3}{7\left(x^{12}+x^6+1\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(7x^9+x^6+8x^3-1\right)}{7\left(x^6-x^3+1\right)\left(x^6+x^3+1\right)\left(2x^9+3x^3+2\right)}\ge0\) *Đúng*
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\ge\dfrac{3}{7}\left(\dfrac{1}{x^{12}+x^6+1}+\dfrac{1}{y^{12}+y^6+1}+\dfrac{1}{z^{12}+z^6+1}\right)\)
Cần chứng minh \(\dfrac{1}{x^{12}+x^6+1}+\dfrac{1}{y^{12}+y^6+1}+\dfrac{1}{z^{12}+z^6+1}\ge1\)
Đặt tiếp \(\left(x^6;y^6;z^6\right)\rightarrow\left(n;h;t\right)\) thì có:
\(\dfrac{1}{n^2+n+1}+\dfrac{1}{h^2+h+1}+\dfrac{1}{t^2+t+1}\ge1\forall nht=1;n,h,t>0\)
Cái này đã làm rồi Here - còn tại sao lại đặt và có BĐT phụ như vậy thì ko nói nhé :)
Sửa đề \("="\rightarrow"+"\)
Áp dụng BĐT cauchy, ta có:\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
\(\Leftrightarrow\sum\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\right)\\ \Leftrightarrow\sum\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{ab^2c+abc+ab}+\dfrac{b}{abc+ab+b}\right)=\dfrac{1}{2}\cdot1=\dfrac{1}{2}\)
Dấu \("="\Leftrightarrow a=b=c=1\)
Ghi Cô Si cho dễ hiểu chí