A = x - x ^ 2 - 1 <0. với mọi x
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1 tháng 7 2023
`+)axx2+bxx1=cxx2+axx1<=>2a+b=2c+a<=>2c-a=b`
`+)cxx3+axx1=bxx2+axx1<=>3c+a=2b+a<=>3c=2b<=>c=2/3b`
mà `2c-a=b` nên `a=2c-b=4/3b-b=1/3b`
Khi đó: `cxx2+axx2=2(a+c)=2(1/3b+2/3b)=2b`
Vậy dấu hỏi chấm cần điền là `2`
29 tháng 8 2021
e: \(\left(a^2-1\right)\left(a^2+a+1\right)\left(a^2-a+1\right)\)
\(=\left(a^3-1\right)\left(a^3+1\right)\)
\(=a^6-1\)
19 tháng 5 2022
a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x-3|=2
=>x-3=2 hoặc x-3=-2
=>x=5(nhận) hoặc x=1(loại)
Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)
c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow2x^2-x+1=0\)
hay \(x\in\varnothing\)
19 tháng 5 2022
f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)
-Vậy \(A_{min}=4\)
21 tháng 6 2023
1: =(8+a^3)(8-a^3)=64-a^6
2: =x^3-6x^2+12x-8-x(x^2-1)+6x^2-18x
=x^3-6x-8-x^3+x
=-5x-8
3: =x^3+3x^2+3x+1-x^3+1-3x^2-3x
=2
T
28 tháng 8 2023
1) \(3x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+5\right)\)
2) \(4x(x-2y)-8y(2y-x)\)
\(=4x\left(x-2y\right)+8y\left(x-2y\right)\)
\(=\left(4x+8y\right)\left(x-2y\right)\)
\(=4\left(x+2y\right)\left(x-2y\right)\)
3) \(a^2\left(x-1\right)+b^2\left(1-x\right)\)
\(=a^2\left(x-1\right)-b^2\left(x-1\right)\)
\(=\left(a^2-b^2\right)\left(x-1\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(x-1\right)\)
4) \(3x\left(x-a\right)+4a\left(a-x\right)\)
\(=3x\left(x-a\right)-4a\left(x-a\right)\)
\(=\left(x-a\right)\left(3x-4a\right)\)
5) \(5x\left(x-y\right)^2+10y^2\left(y-x\right)^2\)
\(=5x\left(x-y\right)^2+10y^2\left(x-y\right)^2\)
\(=\left(5x+10y^2\right)\left(x-y\right)^2\)
\(=5\left(x+2y^2\right)\left(x-y\right)^2\)
6) \(3x\left(x-3\right)^2+9\left(3-x\right)^2\)
\(=3x\left(x-3\right)^2+9\left(x-3\right)^2\)
\(=\left(3x+9\right)\left(x-3\right)^2\)
\(=3\left(x+3\right)\left(x-3\right)^2\)
7) \(x\left(m-a\right)^2-y\left(a-m\right)^2\)
\(=x\left(a-m\right)^2-y\left(a-m\right)^2\)
\(=\left(x-y\right)\left(a-m\right)^2\)
8) \(6y^2\left(x-1\right)^2+9y\left(1-x\right)^2\)
\(=6y^2\left(x-1\right)^2+9y\left(x-1\right)^2\)
\(=\left(6y^2+9x\right)\left(x-1\right)^2\)
\(=3\left(2y^2+3x\right)\left(x-1\right)^2\)
#Ayumu
DK
5 tháng 6 2021
Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)
\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)
Dấu "=" \(x=y=\dfrac{1}{2}\)
\(A=x-x^2-1\)
\(A=-\left(x^2-x+1\right)\)
\(A=-\left(x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(A=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\)
Mà: \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\)
Và: \(-\dfrac{3}{4}< 0\)
\(\Rightarrow A=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< 0\forall x\)