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7 tháng 4 2020
Áp dụng BĐT Cauchy ta có: \(\frac{1}{a^2+1}=\frac{\left(a^2+1\right)-a^2}{a^2+1}=1-\frac{a^2}{a^2+1}\ge1-\frac{a^2}{2a}=1-\frac{a}{2}\)
Hoàn toàn tương tự ta được
\(\frac{1}{b^2+1}\ge1-\frac{b}{2};\frac{1}{c^2+1}\ge1-\frac{c}{2};\frac{1}{d^2+1}\ge1-\frac{d}{2}\)
Cộng theo vế của từng BĐT trên ta được
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}+\frac{1}{d^2+1\ge2}\)
Dấu "=" xảy ra khi a=b=c=d=1
Nguồn: Nguyễn Thị Thúy
21 tháng 7 2020
1.a) (2 + 1)(22 + 1)((24 + 1)(28 + 1) = (22 - 1)(22 + 1)(24 + 1)(28 + 1) = (24 - 1)(24 + 1)(28 + 1)
= (28 - 1)(28 + 1) = 216 - 1
b) 7(23 + 1)(26 + 1)(212 + 1)(224 + 1) = (23 - 1)(23 + 1)(26 + 1)(212 + 1)(224 + 1)
= (26 - 1)(26 + 1)(212 + 1)(224 + 1) = (212 - 1)(212 + 1)(224 + 1) = (224 - 1)(224 + 1) = 248 - 1
c) (x2 - x + 1)(x2 + x + 1)(x2 - 1) = [(x2 - x + 1)(x + 1)][(x2 + x + 1)(x - 1)] = (x3 + 1)(x3 - 1) = x6 - 1
21 tháng 7 2020
2. Đặt A = 4x - x2 - 1 = -(x^2 - 4x + 4) + 3 = -(x - 2)2 + 3 \(\le\)3 \(\forall\)x
Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2
Vậy MaxA = 3 khi x = 2
16 tháng 7 2020
Bài 1:
a) Ta có: \(\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=4x^2-4x+1+4\left(x^2+2x-3\right)-2\left(25-30x+9x^2\right)\)
\(=4x^2-4x+1+4x^2+8x-12-50+60x-18x^2\)
\(=-10x^2+64x-61\)
b) Ta có: \(\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1\right)^2-\left(2a\right)^2-\left(2a^2+1\right)^2\)
\(=-4a^2\)
c) Ta có: \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=\left(9x-1+1-5x\right)^2\)
\(=\left(4x\right)^2=16x^2\)
d)
Sửa đề: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)
Ta có: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)
\(=\left(x^2+5x-1+5x-1\right)^2\)
\(=\left(x^2+10x-2\right)^2\)
\(=x^4+100x^2+4+20x^3-40x-4x^2\)
\(=x^4+20x^3+96x^2-40x+4\)
e) Ta có: \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x\left(x^2-1\right)-\left(x^3+1\right)\)
\(=x^3-x-x^3-1\)
=-x-1
f) Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
NM
Nguyễn Minh Quang
Giáo viên
5 tháng 3 2022
ta có :
\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)
\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)
\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)
\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)
PT
17 tháng 10 2017
1. \(\dfrac{1}{x-1}-\dfrac{1}{x+1}\)
\(=\dfrac{1.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1+\left(-x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1-x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{1}{x^2-1}\)
2. \(\dfrac{x}{x^2-1}-\dfrac{1}{x-1}\)
\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}+\dfrac{-\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+\left(-x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{-1}{x^2-1}\)
3. \(\dfrac{1}{x\left(x-y\right)}-\dfrac{1}{x\left(x-y\right)}\)
\(=\dfrac{1}{y\left(x-y\right)}+\dfrac{-1}{x\left(x-y\right)}\)
\(=\dfrac{1x}{y\left(x-y\right)x}+\dfrac{-1y}{x\left(x-y\right)y}\)
\(=\dfrac{x}{xy\left(x-y\right)}+\dfrac{-y}{xy\left(x-y\right)}\)
\(=\dfrac{x-y}{xy\left(x-y\right)}=\dfrac{1}{xy}\)
4. \(\dfrac{1}{x}-\dfrac{1}{x-1}\)
\(=\dfrac{1\left(x-1\right)}{x\left(x-1\right)}-\dfrac{1x}{\left(x-1\right)x}\)
\(=\dfrac{x-1}{x\left(x-1\right)}+\dfrac{-x}{x\left(x-1\right)}\)
\(=\dfrac{\left(x-1\right)-x}{x\left(x-1\right)}\)
\(=\dfrac{-1}{x\left(x-1\right)}\)
5. \(\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(=\dfrac{1\left(x+1\right)}{x\left(x+1\right)}-\dfrac{1x}{\left(x+1\right)x}\)
\(=\dfrac{x+1}{x\left(x+1\right)}+\dfrac{-x}{x\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}\)
6. \(\dfrac{1}{2x^2-10x}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1.2x}{2x\left(x-5\right)}\)
\(=\dfrac{1}{2x\left(x-5\right)}+\dfrac{-2x}{2x\left(x-5\right)}\)
\(=\dfrac{1-2x}{2x\left(x-5\right)}\)
7. \(\dfrac{x-1}{x^2-1}.\dfrac{x+1}{x+3}\)
\(=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x^2-1\right)\left(x+3\right)}\)
\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x+3\right)}\)
8. \(\dfrac{2}{2x^2+10x}.\dfrac{x+5}{3x}\)
\(=\dfrac{2x\left(x+5\right)}{2x^2+10x.3x}\)
\(=\dfrac{2\left(x+5\right)}{2x\left(x+5\right)3x}\)
\(=\dfrac{2}{6x^2}=\dfrac{1}{3x^2}\)
S
1 tháng 7 2018
a,\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
b,\(=\left(2^3-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
tiếp tục giống bài a
c, \(=\left[x^2-\left(x-1\right)\right]\left[x^2+\left(x+1\right)\right]\left(x^2-1\right)=\left(x^2-x^2+1\right)\left(x^2-1\right)=x^2-1\)
15 nha bạn ơi
15 nha bạn