ta có P=\(\frac{x^2}{\sqrt{xy+3x}}+...\ge\frac{\left(x+y+z\right)^2}{\sqrt{xy+3x}+...}=\frac{9}{\sqrt{xy+3x}+...}\)
mà \(\left(\sqrt{xy+3x}+...\right)^2\le3\left(xy+...+3x+...\right)\le3\left(3+9\right)=36\Rightarrow\sqrt{xy+3x}+...\le6\)
=>\(P\ge\frac{3}{2}\)
ta có P=\(\frac{x^2}{x\sqrt{y+3}}+\frac{y^2}{y\sqrt{z+3}}+\frac{z^2}{z\sqrt{x+3}}\ge\frac{\left(x+y+z\right)^2}{x\sqrt{y+3}+y\sqrt{z+3}+z\sqrt{x+3}}\)
mà \(\left(x\sqrt{y+3}+...\right)^2\le\left(x+y+z\right)\left(xy+yz+zx+3x+3y+3z\right)\le3\left(9+3\right)=36\) ( vì xy+yz+zx<=3)
=>\(x\sqrt{y+3}+...\le6\Rightarrow P\ge\frac{9}{6}=\frac{3}{2}\)
dấu = xảy ra <=> x=y=z=1
giải \(\sqrt{x^2-3x+2}+1=x+\frac{1}{\sqrt{x}}\)
b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)
\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
tính hộ chúa con cuối với " ko dùng coccoc math " 100% sai " bạn nào có máy tính casio bấm hộ "
\(x^2+3=x+8+2x-x^2+2x\sqrt{8+2x-x^2}.\)
\(2x^2-3x-5=2x\sqrt{8+2x-x^2}\)
\(4x^4-12x^3-11x^2+30x+25=-4x^4+8x^3+32x^2\)
\(\left(X+1\right)^2\left(2x-5\right)^2+4x^4-8x^3-32x^2=0\)
\(\left(X-1\right)\left(8x^3-12x^2-55x-25\right)=0\)
\(8x^3-12x^2-55x-25=0\)
\(\Delta=144+1320=1464>0\)
\(k=\frac{47520+3456+43200}{2\sqrt{1464^3}}=\frac{94176}{2\sqrt{1464^3}}=\frac{47088}{\sqrt{1464^3}}< 1\)
\(x1=\frac{2\sqrt{1464}cos\left(arccos\left(\frac{47088}{\sqrt{1464^3}}\right)-\frac{2pi}{3}\right)+12}{24}=?\)
x2=...
x3=......
\(B=\frac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3-2\sqrt{2}}-\sqrt{2}\right)}{\sqrt{3}}=\sqrt{3-2\sqrt{2}}-\sqrt{2}=\sqrt{2-2\sqrt{2}+1}-\sqrt{2}\)
\(=\sqrt{2}-1-\sqrt{2}=-1\)
ĐKXD: tự làm nhé =='
từ đề
\(\Rightarrow A=\frac{2\sqrt{y}-9}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}-\frac{\left(\sqrt{y}+3\right)\left(\sqrt{y}-3\right)}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}+\frac{\left(2\sqrt{y}+1\right)\left(\sqrt{y}-2\right)}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}\)
\(\Rightarrow A=\frac{2\sqrt{y}-9-y+9+2y-3\sqrt{y}-2}{MC}\)
\(\Rightarrow A=\frac{y-\sqrt{y}-2}{MC}=\frac{\left(\sqrt{y}-2\right)\left(\sqrt{y}+1\right)}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}=\frac{\sqrt{y}+1}{\sqrt{y}-3}\)
Đc nhé bác :D
Sensodai: ĐỀ NGHỊ CÁC THÀNH PHẦN TAY NHANH HƠN NÃO K CMT NHÉ =='
CHỨNG MINH RẰNG VỚI MỌI SỐ DƯƠNG N THÌ
GIÚP MÌNH VỚI
1+\(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
\(ab+bc+ca=1\)\(\Rightarrow\)\(\hept{\begin{cases}a+b+c\ge\sqrt{3}\\a^2+b^2+c^2\ge1\end{cases}}\)
\(\left(a-\frac{1}{\sqrt{3}}\right)^2\ge0\)\(\Leftrightarrow\)\(a\le\frac{\sqrt{3}}{2}a^2+\frac{\sqrt{3}}{6}\)
\(P=\Sigma\frac{a^2\left(1-2b\right)^2}{b\left(1-2b\right)}\ge\frac{\left(a+b+c-2\right)^2}{\left(a+b+c\right)-2\left(a^2+b^2+c^2\right)}\ge\frac{\left(a+b+c-2\right)^2}{\frac{\sqrt{3}-4}{2}\Sigma a^2+\frac{\sqrt{3}}{2}}\ge\sqrt{3}-2\)
\(VT=a+b+\frac{1}{a}+\frac{1}{b}=\left(a+\frac{1}{2a}\right)+\left(b+\frac{1}{2b}\right)+\frac{1}{2a}+\frac{1}{2b}\)
để ý \(1=a^2+b^2\ge2ab\Leftrightarrow ab\le\frac{1}{2}\)
\(\frac{1}{2a}+\frac{1}{2b}\ge2\sqrt{\frac{1}{4ab}}\ge2\sqrt{\frac{1}{2}}\)
\(a+\frac{1}{2a}\ge2\sqrt{\frac{1}{2}}\)
\(b+\frac{1}{2b}\ge2\sqrt{\frac{1}{2}}\)
+ 3 vế thì ta được \(VT\ge6\sqrt{\frac{1}{2}}\) dấu = khi \(\frac{1}{2a}=\frac{1}{2b}....a=\frac{1}{2a}....b=\frac{1}{2b}\)
