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Áp dụng bđt Cauchy-schwarz ta có:
\(\frac{4}{x+1}+\frac{9}{y+2}+\frac{25}{z+3}\ge\frac{\left(2+3+5\right)^2}{x+1+y+2+z+3}=\frac{10^2}{4+6}=10\)
Dấu "=" \(\Leftrightarrow\frac{2}{x+1}=\frac{3}{y+2}=\frac{5}{z+3}\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\\z=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Q=3x+9y+15z+x+x4+y+y9+z+z25
\ge 108+2.2+2.3+2.5=128≥108+2.2+2.3+2.5=128
Dấu "=" xảy ra khi x+3y+5z=36, x=\dfrac{4}x, y=\dfrac{9}y, z=\dfrac{25}z\Rightarrow x=2,y=3,z=5x+3y+5z=36,x=x4,y=y9,z=z25⇒x=2,y=3,z=5
bạn tham khảo nhé
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a,\(x^2-2x+1=25\)
\(\Rightarrow\left(x-1\right)^2=25\)
\(\Rightarrow x-1=\orbr{\begin{cases}-5\\5\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}-4\\6\end{cases}}\)
b,\(\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Rightarrow-\left(1+2x\right)\left(9-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1+2x=0\\9-2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{9}{2}\end{cases}}\)
Đặt là a, b, c... nhé
\(a)\) \(x^2-2x+1=25\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=5^2\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-4\end{cases}}}\)
Vậy \(x=-4\) hoặc \(x=6\)
\(b)\) \(\left(5-2x\right)^2-16=0\)
\(\Leftrightarrow\)\(\left(5-2x\right)^2-4^2=0\)
\(\Leftrightarrow\)\(\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Leftrightarrow\)\(\left(1-2x\right)\left(9-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}1-2x=0\\9-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}\)
Vậy \(x=\frac{1}{2}\) hoặc \(x=\frac{9}{2}\)
\(c)\) \(\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\)\(\left(x+2\right)^2-3^2=0\)
\(\Leftrightarrow\)\(\left(x+2-3\right)\left(x+2+3\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)
Vậy \(x=1\) hoặc \(x=-5\)
Chúc bạn học tốt ~
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\(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|< 0\)
Ta có: \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-y\right|+\left|y+\dfrac{9}{25}\right|\ge0\)
Mà \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|< 0\)
\(\Rightarrow\) không tìm được các giá trị x;y thỏa mãn đề bài
Vậy không tìm được các giá trị x;y thỏa mãn đề bài
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phân tích thành nhân tử:
\(x^2-9=x^2-3^2=\left(x+3\right)\left(x-3\right)\)
\(4x^2-25=\left(2x\right)^2-5^2=\left(2x+5\right)\left(2x-5\right)\)
\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
\(9x^2+6xy+y^2=\left(3x\right)^2+2\cdot3x\cdot1+y^2=\left(3x+y\right)^2\)
\(x^2+4y^2+4xy=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)
a. \(x^3-0.25x=0\Rightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\end{cases}}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\end{cases}}\)=> \(x\in\left\{0;\frac{1}{2};\frac{-1}{2}\right\}\)
b, \(x^2-10x=-25\)\(\Rightarrow x^2-10x+25=0\)
\(\Rightarrow\left(x-5\right)^2=0\Rightarrow x-5=0\Rightarrow x=5\)
a, \(x^2-9=x^2-3x+3x-9\)
\(=x\left(x-3\right)+3\left(x-3\right)=\left(x-3\right)\left(x+3\right)\)
b, \(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)
c, \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
d, \(9x^2+6xy+y^2=\left(3x\right)^2+2\left(3xy\right)+y^2\) \(=\left(3x+y\right)^2\)
e, \(6x-9-x^2=6x-18+9-x^2\) \(=6\left(x-3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x-3\right)\left(6-x-3\right)=\left(x-3\right)\left(3-x\right)\)
f, \(x^2+4y^2+4xy=x^2+2\left(2xy\right)+\left(2y\right)^2\)
\(\left(x+2y\right)^2\)
\(\)
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Lời giải:
Áp dụng BĐT Bunhiacopxky:
$[(x+\frac{1}{x})^2+(y+\frac{1}{y})^2](1+1)\geq (x+\frac{1}{x}+y+\frac{1}{y})^2$
$\Leftrightarrow (x+\frac{1}{x})^2+(y+\frac{1}{y})^2\geq \frac{1}{2}(x+y+\frac{1}{x}+\frac{1}{y})^2=\frac{1}{2}(1+\frac{1}{xy})^2$
Mà:
$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$ theo BĐT Cô-si
$\Rightarrow (x+\frac{1}{x})^2+(y+\frac{1}{y})^2\geq \frac{1}{2}(1+\frac{1}{\frac{1}{4}})^2=\frac{25}{2}$ (đpcm)
Dấu "=" xảy ra khi $x=y=\frac{1}{2}$
Ta có : \(\left|x-y\right|\ge0\)
\(\left|y+\frac{9}{25}\right|\ge0\)
Theo đề bài ra , ta có : \(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)
\(\Rightarrow\begin{cases}\left|x-y\right|=0\\\left|x+\frac{9}{25}\right|=0\end{cases}\Rightarrow\begin{cases}y=x\\x=-\frac{9}{25}\end{cases}\Rightarrow\begin{cases}y=-\frac{9}{25}\\x=-\frac{9}{25}\end{cases}\)
Vậy \(x=y=-\frac{9}{25}\)