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\(pt\Leftrightarrow\sqrt{x}\left(\sqrt{x-1}+\sqrt{x+2}-2\sqrt{x}\right)=0\)
Đk: tự xác định
\(pt\Leftrightarrow\sqrt{x+3}-\left(\frac{1}{3}x+1\right)+\sqrt{6-x}-\left(-\frac{1}{3}x+2\right)-\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\frac{x+3-\left(\frac{1}{3}x+1\right)^2}{\sqrt{x+3}+\frac{1}{3}x+1}+\frac{6-x-\left(-\frac{1}{3}x+2\right)^2}{\sqrt{6-x}-\frac{1}{3}x+2}-\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\frac{-\frac{1}{9}\left(x+3\right)\left(x-6\right)}{\sqrt{x+3}+\frac{1}{3}x+1}+\frac{-\frac{1}{9}\left(x+3\right)\left(x-6\right)}{\sqrt{6-x}-\frac{1}{3}x+2}-\frac{\left(x+3\right)\left(x-6\right)}{\sqrt{-\left(x+3\right)\left(x-6\right)}}=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-6\right)\left(\frac{-\frac{1}{9}}{\sqrt{x+3}+\frac{1}{3}x+1}+\frac{-\frac{1}{9}}{\sqrt{6-x}-\frac{1}{3}x+2}-\frac{1}{\sqrt{-\left(x+3\right)\left(x-6\right)}}\right)=0\)
Dễ thấy:\(\frac{-\frac{1}{9}}{\sqrt{x+3}+\frac{1}{3}x+1}+\frac{-\frac{1}{9}}{\sqrt{6-x}-\frac{1}{3}x+2}-\frac{1}{\sqrt{-\left(x+3\right)\left(x-6\right)}}< 0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-6=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-3\\x=6\end{cases}}\)
\(2\left(x-4\right)\sqrt{x-2}+\left(x-2\right)\sqrt{x+1}+2\left(x-3\right)=0\)
ĐK:\(x\ge2\)
\(\Leftrightarrow2\left(x-4\right)\left(\sqrt{x-2}-1\right)+\left(x-2\right)\left(\sqrt{x+1}-2\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow2\left(x-4\right)\frac{x-2-1}{\sqrt{x-2}+1}+\left(x-2\right)\frac{x+1-4}{\sqrt{x+1}+2}-2\left(x-3\right)=0\)
\(\Leftrightarrow2\left(x-4\right)\frac{x-3}{\sqrt{x-2}+1}+\left(x-2\right)\frac{x-3}{\sqrt{x+1}+2}-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{2\left(x-4\right)}{\sqrt{x-2}+1}+\frac{x-2}{\sqrt{x+1}+2}-2\right)=0\)
Suy ra x=3
Bài 1:
Ta có:
\(A=(x^2-x)(x^2+3x+2)=x(x-1)(x+1)(x+2)\)
\(=[x(x+1)][(x-1)(x+2)]=(x^2+x)(x^2+x-2)\)
\(=(x^2+x)^2-2(x^2+x)=(x^2+x)^2-2(x^2+x)+1-1\)
\(=(x^2+x-1)^2-1\geq -1\)
Vậy GTNN của $A$ là $-1$. Dấu "=" xảy ra khi \((x^2+x-1)^2=0\Leftrightarrow x^2+x-1=0\Leftrightarrow x=\frac{-1\pm \sqrt{5}}{2}\)
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\(B=x^4+(x-2)^4+6x^2(x-2)^2=x^4+(x-2)^4-2x^2(x-2)^2+8x^2(x-2)^2\)
\(=[x^2-(x-2)^2]^2+8x^2(x-2)^2\)
\(=16(x-1)^2+8[x(x-2)]^2=16(x^2-2x+1)+8(x^2-2x)^2\)
\(=8[(x^2-2x)^2+2(x^2-2x+1)]=8[(x^2-2x)^2+2(x^2-2x)+1+1]\)
\(=8[(x^2-2x+1)^2+1]=8(x^2-2x+1)^2+8\geq 8\)
Vậy GTNN của biểu thức là $8$ khi \((x^2-2x+1)^2=0\Leftrightarrow (x-1)^4=0\Leftrightarrow x=1\)
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\(C=4x^2+4x-6|2x+1|+6=(4x^2+4x+1)-6|2x+1|+5\)
\(=|2x+1|^2-6|2x+1|+5\)
\(=|2x+1|^2-6|2x+1|+9-4=(|2x+1|-3)^2-4\geq -4\)
Vậy GTNN của biểu thức là $-4$ khi \(|2x+1|=3\Leftrightarrow x=1\) hoặc $x=-2$
Bài 2:
ĐKXĐ: \(x\geq 0\)
Áp dụng BĐT Cô-si cho các số không âm ta có:
\(x+1\geq 2\sqrt{x}\Rightarrow x+1+\sqrt{x}\geq 3\sqrt{x}\)
\(\Rightarrow E=\frac{\sqrt{x}}{x+1+\sqrt{x}}\leq \frac{\sqrt{x}}{3\sqrt{x}}=\frac{1}{3}\)
Vậy GTLN của $E$ là $\frac{1}{3}$ khi $x=1$
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\Leftrightarrow2x+1=2x^3+x^2+2x+1\)\(\Leftrightarrow2x^3+x^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\left(1\right)\)
\(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0!2x+1!=2x+1\)
\(\left(1\right)\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)
\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)
\(\hept{\begin{cases}2x+1=0\\-x^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}}\)
Chúc bạn học tốt !!!
Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left|2\left(1-x\right)\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2\left(1-x\right)=6\\2\left(1-x\right)=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)