Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2+y^2+z^2-2x+4y-6z=15\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=29\)
Đặt \(P=\left|2x-3y+4z-20\right|=\left|2\left(x-1\right)-3\left(y+2\right)+4\left(z-3\right)\right|\)
\(P^2=\left[2\left(x-1\right)-3\left(y+2\right)+4\left(z-3\right)\right]^2\)
\(P^2\le\left(2^2+3^2+4^2\right)\left[\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2\right]=29^2\)
\(\Rightarrow P\le29\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=29\\\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}\end{matrix}\right.\)
a: \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3-1-3\left(4x^2-4x+1\right)=15\)
=>\(12x^2+6x-12x^2+12x-3=15\)
=>18x=18
=>x=1
b: \(\Leftrightarrow y^3+6y^2+9y-y^3-8-6y^2+150=97\)
=>9y+142=97
=>9y=-45
=>y=-5
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
a: \(=4x^4y+6x^2y^2z-2x^5y\)
b: \(=\dfrac{24x^5}{6x^2}-\dfrac{12x^4}{6x^2}+\dfrac{6x^2}{6x^2}=4x^3-2x^2+1\)
c: \(=\dfrac{\left(2x-1\right)^2}{2x-1}=2x-1\)
d: \(=\dfrac{\left(x+5\right)\left(x^2-1\right)}{x+5}=x^2-1\)
\(\text{a) }P=x^3+6x^2y+12xy^2+8y^3\) tại \(x+2y=-5\) Chữa đề
\(\text{Ta có : }P=x^3+6x^2y+12xy^2+8y^3\\ P=x^3+3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2+y^3\\ P=\left(x+2y\right)^3\\ Thay\text{ }2y+x=-5\text{ vào biểu thức}\\ \text{Ta được: }P=\left(-5\right)^3\\ P=-125\\ \text{Vậy }P=-125\text{ }khi\text{ }2y+x=-5\)
\(\text{b) }Q=x^3-y^3\text{tại }x-y=20;xy=24\\ \text{Theo bài ra ta có: }x-y=10\\ \Rightarrow\left(x-y\right)^2=10^2\\ \Rightarrow x^2-2xy+y^2=100\\ \Rightarrow x^2+y^2=100+2xy\\ Thay\text{ }xy=24\text{ vào biểu thức ta được : }\\ x^2+y^2=100+2xy\\ \Rightarrow x^2+y^2=100+48\\ \Rightarrow x^2+y^2=148\\ \text{Ta lại có : }Q=x^3-y^3\\ Q=\left(x-y\right)\left(x^2+xy+y^2\right)\\ Thay\text{ }x-y=10;xy=24;x^2+y^2=148\text{ vào biểu thức }\\ \text{Ta được : }Q=10\left(148+24\right)\\ Q=1720\\ \text{Vậy }Q=1720\text{ }khi\text{ }x-y=20;xy=24\)
\(\)
\(x^6-1=\left(x^3-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\\ \RightarrowĐPCM\)
\(2005^3+125=\left(2005+5\right)\left(2005^2+2005\cdot5+5^2\right)=2010\left(2005^2+2005\cdot5+5^2\right)⋮2010\)\(x^2+y^2+z^2+3=2\left(x+y+z\right)\\ \Leftrightarrow x^2+y^2+x^2+3=2x+2y+2z\\ \Leftrightarrow x^2-2x+1+y^2-2y+1+z^2-2z+1=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\\ \left(x-1\right)^2\ge0;\left(y-1\right)^2\ge0;\left(z-1\right)^2\ge0\\ \Rightarrow\left(x-1\right)^2=\left(y-1\right)^2=\left(z-1\right)^2=0\\ \Rightarrow x-1=y-1=z-1=0\\ \Leftrightarrow x=y=z=1\)
b) \(2005^3+125\)
\(=2005^3+5^3\)
\(=\left(2005+5\right)\left(2005^2-2005.5+5^2\right)\)
\(=2010\left(2005^2-2005.5+5^2\right)\)\(⋮\) 2010
Vậy \(2005^3+125\) chia hết cho 2010