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Tính và thu gọn biểu thức.
Các bạn ơi giúp mk với. Mk cần gấp lắm ạ.Các bạn giải chi tiết giúp mk nhé . Mk hứa sẽ vote 5* cho . Mk cảm ơn nhiều.
a, \(2\sqrt{3}-\sqrt{4+x^2}=0\Leftrightarrow\sqrt{4+x^2}=2\sqrt{3}\)
\(\Leftrightarrow x^2+4=12\Leftrightarrow x^2=8\Leftrightarrow x=\pm2\sqrt{2}\)
b, \(\sqrt{16x+16}-\sqrt{9x+9}=0\)ĐK : x >= -1
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=0\Leftrightarrow\sqrt{x+1}=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
c, \(\sqrt{4\left(x+2\right)^2}=8\Leftrightarrow2\left|x+2\right|=8\Leftrightarrow\left|x+2\right|=4\)
TH1 : \(x+2=4\Leftrightarrow x=2\)
TH2 : \(x+2=-4\Leftrightarrow x=-6\)
c: Ta có: \(\sqrt{4\left(x+2\right)^2}=8\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
\(\sqrt{x^2-6x+9}\) \(-\frac{\sqrt{3}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=0\)
\(\Leftrightarrow\left|x-3\right|-\sqrt{3}=0\)
\(\Leftrightarrow\left|x-3\right|=\sqrt{3}\)
th1 \(x\ge3\Rightarrow x-3=\sqrt{3}\Rightarrow x=3+\sqrt{3}\)
th2 \(x< 3\Rightarrow3-x=\sqrt{3}\Rightarrow x=3-\sqrt{3}\)
Áp dụng bđt Holder ta được:
\(9\left(a^3+b^3+c^3\right)=3.3.\left(a^3+b^3+c^3\right)=\left(1+1+1\right)\left(1+1+1\right)\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3=1\Rightarrow A\ge\frac{1}{9}\)
Dấu bằng xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
c/m bất đẳng thức Holder:
Cho a,b,c,x,y,z,m,n,p là các số thực dương. Khi đó ta có:
\(\left(a^3+b^3+c^3\right)\left(x^3+y^3+z^3\right)\left(m^3+n^3+p^3\right)\ge\left(axm+byn+czp\right)^3\)
Sử dụng bất đẳng thức AM-GM (Cô-si) ta có:
\(\frac{a^3}{a^3+b^3+c^3}+\frac{x^3}{x^3+y^3+z^3}+\frac{m^3}{m^3+n^3+p^3}\ge\frac{3axm}{\sqrt[3]{\left(a^3+b^3+c^3\right)\left(x^3+y^3+z^3\right)\left(m^3+n^3+p^3\right)}}\)
Tương tự:
\(\frac{b^3}{a^3+b^3+c^3}+\frac{y^3}{x^3+y^3+z^3}+\frac{n^3}{m^3+n^3+p^3}\ge\frac{3byn}{\sqrt[3]{\left(a^3+b^3+c^3\right)\left(x^3+y^3+z^3\right)\left(m^3+n^3+p^3\right)}}\)
\(\frac{c^3}{a^3+b^3+c^3}+\frac{z^3}{x^3+y^3+z^3}+\frac{p^3}{m^3+n^3+p^3}\ge\frac{3czp}{\sqrt[3]{\left(a^3+b^3+c^3\right)\left(x^3+y^3+z^3\right)\left(m^3+n^3+p^3\right)}}\)
\(\Rightarrow3\ge\frac{3axm+3byn+3czp}{\sqrt[3]{\left(a^3+b^3+c^3\right)\left(x^3+y^3+z^3\right)\left(m^3+n^3+p^3\right)}}\)
\(\Leftrightarrow\sqrt[3]{\left(a^3+b^3+c^3\right)\left(x^3+y^3+z^3\right)\left(m^3+n^3+p^3\right)}\ge axm+byn+czp\)
\(\Leftrightarrow\left(a^3+b^3+c^3\right)\left(x^3+y^3+z^3\right)\left(m^3+n^3+p^3\right)\ge\left(axm+byn+czp\right)^3\)
Đẳng thức xảy ra khi các biến bằng nhau
\(=\sqrt{\left(x+2y^2\right)^2}-\sqrt{\left(2x-3y^2\right)^2}\)\(=x+2y^2-3y^2+2x=3x-y^2=3\cdot\sqrt{2}-1\)
\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)
\(=\)\(2\sqrt{3}.\sqrt{3}+\sqrt{5}.\sqrt{3}-\sqrt{60}\)
\(=\)\(6+\sqrt{15}-\sqrt{60}\)
\(=\)\(6-\sqrt{15}\)
\(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)
=\(6+\sqrt{15}-\sqrt{2^2.15}\)
=\(6+\sqrt{15}-2\sqrt{15}\)
=\(6-\sqrt{15}\)
k mk nha
b: Tọa độ giao điểm là:
\(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=-\dfrac{3}{2}x-1\\y=\dfrac{1}{2}x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\y=\dfrac{1}{2}x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-4\end{matrix}\right.\)
a.
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x+3}=a>0\\\sqrt{x^2+x+2}=b>0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=x+1\)
Pt trở thành:
\(a+b=2\left(a^2-b^2\right)\)
\(\Leftrightarrow a+b=\left(2a-2b\right)\left(a+b\right)\)
\(\Leftrightarrow2a-2b=1\) (do \(a+b>0\))
\(\Leftrightarrow2a=2b+1\)
\(\Leftrightarrow2\sqrt{x^2+2x+3}=2\sqrt{x^2+x+2}+1\)
\(\Leftrightarrow4\left(x^2+2x+3\right)=4\left(x^2+x+2\right)+1+4\sqrt{x^2+x+2}\)
\(\Leftrightarrow4x+3=4\sqrt{x^2+x+2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{3}{4}\\16\left(x^2+x+2\right)=\left(4x+3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{3}{4}\\8x=23\end{matrix}\right.\) \(\Rightarrow x=\dfrac{23}{8}\)
b.
ĐKXĐ: \(x\ge3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-3}=a\ge0\\\sqrt{x+2}=b>0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=-5\)
Phương trình trở thành:
\(\left(a-b\right)\left(ab+1\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(vô-nghiệm\right)\\ab+1=a+b\end{matrix}\right.\)
\(\Rightarrow ab-a-b+1=0\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=1\\\sqrt{x+2}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\left(ktm\right)\end{matrix}\right.\)
a) \(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) \(\sqrt{\left(1-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+3-\sqrt{5}\)
=2
c) \(\sqrt{21-12\sqrt{3}}-\sqrt{13-4\sqrt{3}}\)
\(=2\sqrt{3}-3-2\sqrt{3}+1\)
=-2