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2. b,
\(\sqrt{3x^2-2x}+3=2x\)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge\frac{2}{3}\\x\le0\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{3x^2-2x}=2x-3\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\3x^2-2x=\left(2x-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\3x^2-2x=4x^2-12x+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\4x^2-3x^2-12x+2x+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\x^2-10x+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\\left(x-1\right)\left(x-9\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow x=9\)
Vậy phương trình có 1 nghiệm duy nhất là x = 9.
2.a,
\(9\sqrt{\frac{4x-8}{9}}-5\sqrt{\frac{16x-32}{25}}+18\sqrt{\frac{25x^2-100}{81}}=15\sqrt{x^2-4}\)
ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow9\sqrt{\frac{4\left(x-2\right)}{9}}-5\sqrt{\frac{16\left(x-2\right)}{25}}+18\sqrt{\frac{25\left(x^2-4\right)}{81}}=15\sqrt{x^2-4}\)
\(\Leftrightarrow9.\frac{2}{3}\sqrt{\left(x-2\right)}-5.\frac{4}{5}\sqrt{\left(x-2\right)}+18.\frac{5}{9}\sqrt{\left(x-2\right)\left(x+2\right)}=15\sqrt{\left(x-2\right)\left(x+2\right)}\)\(\Leftrightarrow6\sqrt{\left(x-2\right)}-4\sqrt{\left(x-2\right)}+10\sqrt{\left(x-2\right)\left(x+2\right)}=15\sqrt{\left(x-2\right)\left(x+2\right)}\)\(\Leftrightarrow6\sqrt{\left(x-2\right)}-4\sqrt{\left(x-2\right)}+10\sqrt{\left(x-2\right)\left(x+2\right)}-15\sqrt{\left(x-2\right)\left(x+2\right)}=0\)\(\Leftrightarrow\sqrt{x-2}\left(6-4+10\sqrt{x+2}-15\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\sqrt{x-2}\left(2-5\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\2-5\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=\frac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=\frac{4}{25}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tmDKXD\right)\\x=-\frac{11}{6}\left(khongtmDKXD\right)\end{matrix}\right.\)
Vậy pt có 1 nghiệm duy nhất là x = 2.
5.
\(a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)^2=\frac{1}{2}\left(a^2+b^2\right)\left(a^2+b^2\right)\ge ab\left(a^2+b^2\right)\)
\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)
\(\Rightarrow VT\le\frac{ab}{ab\left(a^2+b^2\right)+ab}+\frac{bc}{bc\left(b^2+c^2\right)+bc}+\frac{ca}{ca\left(c^2+a^2\right)+ca}\)
\(VT\le\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\)
Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)
\(\Rightarrow VT\le\frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}\)
\(VT\le\frac{xyz}{xy\left(x+y\right)+xyz}+\frac{xyz}{yz\left(y+z\right)+xyz}+\frac{xyz}{zx\left(x+z\right)+xyz}\)
\(VT\le\frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{x+y+z}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
2. Đề bài bạn viết thiếu thì phải
3. a/
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2+5x+1}=a\\\sqrt{4x^2-4x+4}=b\end{matrix}\right.\)
\(\Rightarrow a-b=a^2-b^2\Leftrightarrow a-b=\left(a-b\right)\left(a+b\right)\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\)
- Với \(a=b\Rightarrow9x-3=0\Rightarrow x=...\)
- Với \(a+b=1\Rightarrow\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}=1\)
\(\Leftrightarrow\sqrt{4x^2+5x+1}+\sqrt{\left(2x-1\right)^2+3}=1\)
\(VT\ge\sqrt{3}>1\Rightarrow\) pt vô nghiệm
b/ ĐKXĐ: ...
\(2x+y+2\sqrt{2x+y}-3=0\)
\(\Leftrightarrow\left(\sqrt{2x+y}-1\right)\left(\sqrt{2x+y}+3\right)=0\)
\(\Leftrightarrow\sqrt{2x+y}=1\Rightarrow y=1-2x\)
Thay vào pt dưới:
\(x^2-2x\left(1-2x\right)=\left(1-2x\right)^2+2\)
\(\Leftrightarrow...\) bạn tự giải
3.Áp dụng BĐT \(\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)ta có
\(\frac{ab}{a+3b+2c}=ab.\frac{1}{\left(a+c\right)+2b+\left(b+c\right)}\le\frac{1}{9}ab.\left(\frac{1}{a+c}+\frac{1}{2b}+\frac{1}{b+c}\right)\)
TT \(\frac{bc}{b+3c+2a}\le\frac{bc}{9}.\left(\frac{1}{b+a}+\frac{1}{2c}+\frac{1}{c+a}\right)\)
\(\frac{ca}{c+3a+2b}\le\frac{ac}{9}.\left(\frac{1}{a+b}+\frac{1}{2a}+\frac{1}{b+c}\right)\)
=> \(VT\le\frac{1}{18}\left(a+b+c\right)+\Sigma.\frac{1}{9}.\left(\frac{bc}{a+c}+\frac{ba}{a+c}\right)=\frac{1}{18}\left(a+b+c\right)+\frac{1}{9}\left(a+b+c\right)=\frac{1}{6}\left(a+b+c\right)\)
Dấu bằng xảy ra khi a=b=c
cảm ơn bạn nhiều, bạn có thể giúp mình hai câu kia nữa được không
Đề: \(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\sqrt{3}\) ???
*Ta chứng minh : \(x^4-x^3+2\ge x+1\forall x>0\)
\(\Leftrightarrow x^4-x^3-x+1\ge0\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)\ge0\) ( đúng )
Do đó: \(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\) \(\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}=\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=1\)
3.
\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)
\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)