Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)
2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)
\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)
\(=25\left(a-b\right)^2=25\cdot100=2500\)
c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)
\(\Leftrightarrow V\ge-1\forall x\)
Dấu '=' xảy ra khi x=1
\(M=x^4-x^3-x^3+x^2+x^2-2x+1\)
\(=x^3\left(x-1\right)-x^2\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x-1\right)\left(x^3-x^2\right)+\left(x-1\right)^2\)
\(=\left(x-1\right)^2\cdot x^2+\left(x-1\right)^2=\left(x-1\right)^2\left(x^2+1\right)\)
\(\left(x-1\right)^2\ge0\)\(\forall x\)
\(x^2+1\ge1\)\(\forall x\)
Do đó: \(M>=1\)
Dấu = xảy ra khi x=0
\(M=x^4-x^3-x^3+x^2+x^2-2x+1\)
\(=x^3\left(x-1\right)-x^2\left(x-1\right)+\left(x-1\right)^2\)
\(=x^2\left(x-1\right)^2+\left(x-1\right)^2\)
\(=\left(x^2+1\right)\left(x-1\right)^2\)
\(\left(x-1\right)^2>=0\forall x\)
\(x^2+1>=1\forall x\)
Do đó: \(\left(x-1\right)^2\cdot\left(x^2+1\right)>=0\forall x\)
Dấu = xảy ra khi x=1
\(A=x^2-2x+5=x^2-2.x.1+1^2+4\)
=\(\left(x+1\right)^2+4\)
Vì \(\left(x+1\right)^2\ge0\)với mọi x nên \(\left(x+1\right)^2+4\ge4\)
Vậy Min A =4. Dấu bằng xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Các câu khác tương tự
\(A=x^2-2x+5\)
\(A=x^2-2x+1+4\)
\(A=\left(x-1\right)^2+4>0\forall x\)
vậy ko tìm được \(MIN\) \(A\)
\(M=\left(x+1\right)\left(2x-1\right)\)
\(M=2x^2-x+2x-1\)
\(M=2x^2+x-1\)
\(M=2\left(x^2+\frac{1}{2}x-\frac{1}{2}\right)\)
\(M=2\left(x^2+2.\frac{1}{4}x+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}\right)\)
\(M=2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\)
\(M=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)
dấu \("="\) xảy ra \(\Leftrightarrow x+\frac{1}{4}=0\Leftrightarrow x=-\frac{1}{4}\)
vậy \(MIN\) \(M=\frac{-9}{8}\Leftrightarrow x=-\frac{1}{4}\)