1/2(x+1)=1/2
Những câu hỏi liên quan
$$ \frac{x^4-(x-1)^2}{(x^2+1)^2-x^2}+\frac{x^2-(x^2-1)^2}{x^2*(x+1)^2-1}+\frac{x^2*(x-1)^2-1}{x^4-(x+1)^2} $$
Chứng minh đẳng thức:
a, (x^2-2x/2x^2+8-2x^2/8-4x+2x^2-x^3)(1-1/x-2/x^2)=x+1/2x
b, [2/3x-2/x+1(x+1/3x-x-1)]:x-1/x=2x/x-1
c, [2/(x+1)^3(1/x+1)+1/x^2+2x+1(1/x^2+1)]:x-1/x^3=x/x-1
1. x+1/x-1 - x-1/x+1 = 4/x2-1
2. x+2/x-2 + x/x+2 = 2
3. 2x/x-1 - 1/x+2 = 2
4. x/x2-25 - 1-x/x-5 = 1/x+5
5. x-1/x+2 - 9/x2-4 = -3/x-2
6. 3x-3/x2-9 - 1/x-3 = x+1/x+3
Xem chi tiết
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
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a, (2+x/2-x - 4x^2/x^2-4 - 2-x/2+x):x^2-3x/2x^2-x^3
a, 2x-1/2x+1 :(2x-1+2-4x/2x+1)
b,(1/1-x -1) : (x- 1-2x/1-x +1)
c,(1/x + x-2/x^2 -4 -2+x/x^2+2x):(x^2+2x+1)/x^2
d,{1/x^2 + 1/y^2 + 2/ x+y.(1/x + 1/y )} : x^3+y^3/x^2+y^2
đề là thực hiện phép tính
a: \(=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{2x^2-x^3}{x^2-3x}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: \(=\dfrac{2x-1}{2x+1}:\left(2x-1+\dfrac{2-4x}{2x+1}\right)\)
\(=\dfrac{2x-1}{2x+1}:\dfrac{4x^2-1+2-4x}{2x+1}\)
\(=\dfrac{2x-1}{4x^2-4x+1}=\dfrac{1}{2x-1}\)
c: \(=\left(\dfrac{1}{1-x}-1\right):\left(x+1-\dfrac{2x-1}{x-1}\right)\)
\(=\dfrac{1-1+x}{1-x}:\dfrac{x^2-1-2x+1}{x-1}\)
\(=\dfrac{-x}{x-1}\cdot\dfrac{x-1}{x\left(x-2\right)}=\dfrac{-1}{x-2}\)
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5 tháng 6 2021 lúc 15:55
Cho x,y>0,x+y=1.CM:`A=(x+1/x)^2+(y+1/y)^2>=25/2`
`A=x^2+1/x^2+2+y^2+1/y^2+2`
`=x^2+y^2+1/x^2+1/y^2+4`
`=(x^2+1/(16x^2))+(y^2+1/(16y^2))+4+15/16(1/x^2+1/y^2)`
Áp dụng BĐt cosi và `1/a^2+1/b^2>=8/(a+b)^2`
`=>A>=1/2+1/2+4+15/16(8/(x+y)^2)`
`<=>A>=5+15/2=25/2`
Dấu "=" `<=>x=y=1/2`
Không làm theo cách sau:
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 "=" \(x=y=\dfrac{1}{2}\)
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Bài 1: rút gọn
a, (2x-1)^2 +4(x-1) (x+3)-2(5-3x)^2
b, (2a^2+2a+1)(2a^2-2a+1)-(2a^2+1)^2
c, (9x-1)^2+(1-5x)^2+2(9x-1)(1-5x)
d, (x^2+5x-1)^2+2(5x-1)(x^2-5x+1)+(5x-1)^2
e, x(x-1)(x+1)-(x+1)(x^2-x+1)
f, x(x+4)(x-4)-(x^2+1)(x^2-1)
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: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
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1/x-1-x^3-x/x^2+1(x/x^2-2x+1-1/x^2-1)
[2/(x+1)^3.(1/x+1)+1/x^2+2x+1(1/x^2+1)]:x-1/x^3=x/x-1
(x/x^2-36-x-6/x^2+6x):2x-6/x^2+6x+x/6-x
giúp mik với ;-; mik cần gấp
Quy đồng phân thức (mình gấp lắm ạ)
b) x-1 và 1/x+1
c)1/3x-3y và 1/x^2-2xy7^2
d)x/x+3 và 1/3-x và 1/x^2-9
e) 1/x^2 và 1/xy-y^2 và 2/x^2-y^2
g)x^2-4/x^2-4x+4 và 2x/x^2-4
h) 1/x^3+1 và 2/x+1 và 3/x^2-x+1
i)1/x^3-1 và 1/x^2+x+1 và 1/1-x
giúp mình với mình gấp lắm ạ
d: \(\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x+3\right)\left(x-3\right)}\)
\(\dfrac{1}{3-x}=\dfrac{-1}{x-3}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x^2-9}=\dfrac{1}{\left(x+3\right)\left(x-3\right)}\)
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1) \(\dfrac{1}{x^2+6x+9}+\dfrac{1}{6x-x^2+9}+\dfrac{x}{x^2-9}\) 2) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\) 3) \(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\)
thực hiện phép tính
a, 2x-1/2x+1 :(2x-1+2-4x/2x+1)
b,(1/1-x -1) : (x- 1-2x/1-x +1)
c,(1/x + x-2/x^2 -4 -2+x/x^2+2x):(x^2+2x+1)/x^2
d,{1/x^2 + 1/y^2 + 2/ x+y.(1/x + 1/y )} : x^3+y^3/x^2+y^2