Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Đặt \(A=x^2-2x+1\)
Ta có: \(A=x^2-2x+1=\left(x-1\right)^2\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow A_{min}=0\)
Dấu "=" xảy ra khi: \(x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(A_{min}=0\)\(\Leftrightarrow\)\(x=1\)
b) Ta có: \(M=x^2-3x+10\)
\(\Leftrightarrow M=\left(x^2-3x+\frac{9}{4}\right)+\frac{31}{4}\)
\(\Leftrightarrow M=\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\forall x\)
\(\Rightarrow\)\(M_{min}=\frac{31}{4}\)
Dấu "=" xảy ra khi: \(x-\frac{3}{2}=0\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy \(M_{min}=\frac{31}{4}\)\(\Leftrightarrow\)\(x=\frac{3}{2}\)
Bài 1:
\(x^2+5y^2+2x-4xy-10y+2014\)
\(=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-6y+9\right)+2004\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+2004\)
Do \(\left(x-2y+1\right)^2\ge0;\left(y-3\right)^2\ge0\Rightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2+2004>0\)
=> đpcm
Bài 2:
\(A=3.\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)...\left(2^{64}+1\right)+1\)
...
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=\left(2^{128}-1\right)+1=2^{128}\)
Vậy \(A=2^{128}\)
B1:
\(x^2+5y^2+2x-4xy-10y+2014\)
= \(\left(x^2+2x-4xy+1-4y+4y^2\right)+y^2-6y+9+2004\)
= \(\left[x^2+2x\left(1-2y\right)+\left(1-2y\right)^2\right]+\left(y-3\right)^2+2004\)
= \(\left(x-1+2y\right)^2+\left(y-3\right)^2+2004>0\) với mọi x
\(\Rightarrow\) Đpcm
B2:
\(A=3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^8-1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^{16}-1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^{32}-1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^{64}-1\right).\left(2^{64}+1\right)+1\)
\(A=2^{128}-1+1=2^{128}\)
Bài 1:
a) \(M=x^2-3x+10=\left(x^2-3x+\frac{9}{4}\right)+\frac{31}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
KL:...
2. a. \(A=12a-4a^2+3=-4\left(a-\frac{3}{2}\right)^2+12\)
Vì \(\left(a-\frac{3}{2}\right)^2\ge0\forall a\)\(\Rightarrow-4\left(a-\frac{3}{2}\right)^2+3\le3\)
Dấu "=" xảy ra \(\Leftrightarrow-4\left(a-\frac{3}{2}\right)^2=0\Leftrightarrow a-\frac{3}{2}=0\Leftrightarrow a=\frac{3}{2}\)
Vậy Amax = 3 <=> a = 3/2
b. \(B=4t-8v-v^2-t^2+2017=-\left(v^2+t^2-4t+8v+20\right)+2037\)
\(=-\left(t-2\right)^2-\left(v+4\right)^2+2037\)
Vì \(\left(t-2\right)^2\ge0;\left(v+4\right)^2\ge0\forall t;v\)
\(\Rightarrow-\left(t-2\right)^2-\left(v+4\right)^2+2037\le2037\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left(t-2\right)^2=0\\\left(v+4\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t-2=0\\v+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=2\\v=-4\end{cases}}\)
Vậy Bmax = 2037 <=> t = 2 ; v = - 4
c. \(C=m-\frac{m^2}{4}=-\frac{1}{4}\left(m-2\right)^2+1\)
Vì \(\left(m-2\right)^2\ge0\forall m\)\(\Rightarrow-\frac{1}{4}\left(m-2\right)^2+1\le1\)
Dấu "=" xảy ra \(\Leftrightarrow-\frac{1}{4}\left(m-2\right)^2=0\Leftrightarrow m-2=0\Leftrightarrow m=2\)
Vậy Cmax = 1 <=> m = 2
b1:
câu a,f áp dụng a2-b2=(a-b)(a+b)
câu b,c áp dụng a3-b3=(a-b)(a2+ab+b2)
câu d: \(x^2+2xy+x+2y=x\left(x+2y\right)+\left(x+2y\right)=\left(x+1\right)\left(x+2y\right)\)
câu e: \(7x^2-7xy-5x+5y=7x\left(x-y\right)-5\left(x-y\right)=\left(7x-5\right)\left(x-y\right)\)
câu g xem lại đề
\(P=y^2+8y+15\)
\(=\left(y^2+8y+16\right)-1\)
\(=\left(y+4\right)^2-1\ge-1\)
Dấu "=" xảy ra khi \(y=-4\)
\(2x^2+5y^2+4xy+8x-4y-100\)
\(=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)
\(=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)
\(\ge-120\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x+2y=0\\x+4=0\\y-2=0\end{cases}}\Rightarrow y=2;x=-4\)