( x + 1 )2 +3( x - 5)(x+ 5)-( 2x-1)2
giup minh voi cac ban
de bai la rut gon
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\(A=\frac{\left(x-1\right)-5\sqrt{x-1}+6}{\sqrt{x-1}\cdot\left(\sqrt{x-1}-3\right)}=\frac{\left(\sqrt{x-1}-2\right)\cdot\left(\sqrt{x-1}-3\right)}{\sqrt{x-1}\cdot\left(\sqrt{x-1}-3\right)}\) Đk x\(\ne\) 1;10
\(A=\frac{\sqrt{x-1}-2}{\sqrt{x-1}}=1-\frac{2}{\sqrt{x-1}}\)
\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)
\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)
\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)
\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)
a, x.(x-y) +y.(x+y)
=x2-xy+xy+y2
=x2+y2
b, (x2-5).(2x+3)-2x.(x-3)
=2x3+3x2-10x-15-2x2+6x
=2x3-x2-4x-15
c, 8-5x.(x+2) +4 .( x-2) . (x+1) +2.( x+2)+ 2.(x-2)+10
=8-5x2-10x+4.(x2+x-2x-2)+2x+4+2x-4+10
=18-6x-5x2+4x2+4x-8x-8
=10-10x-x2
Bài 1:
a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)
b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x
=>P=5x+4
\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)
=>-x2+x=10x+8
=>x2-x=-10x-8
=>x2+9x+8=0
=>x=-8(nhận) hoặc x=-1(loại)
Bài 1:
\(a,\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^2\left(x^2-1\right)\)
\(b,\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)^3\)
\(=x^6+27-27-27x^2-9x^4-x^6\)
\(=-9x^2\left(3-x^2\right)\)
Bài 5:
\(A=x^2-2x+1\)
\(=\left(x^2-2x+1\right)-2\)
\(=\left(x-1\right)^2-2\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)
Vậy Min A = -2
Để A = -2 thì \(x-1=0\Rightarrow x=1\)
b, \(B=4x^2+4x+5\)
\(=\left(4x^2+4x+1\right)+4\)
\(=\left(2x+1\right)^2+4\)
Với mọi giá trị của x ta có:
\(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+4\ge4\)
Vậy Min B = 4
Để B = 4 thì \(2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
c, \(C=2x-x^2-4\)
\(=-\left(x^2-2x+1\right)-3\)
\(=-\left(x-1\right)^2-3\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-3\le-3\)Vậy Max C = -3
để C = -3 thì \(x-1=0\Rightarrow x=1\)
\(a\text{)}.\:\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\\ =x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)\\ =x^4+4x^2+4-x^4+16\\ =4x^2+20\)
\(b\text{)}.\:\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\\ =\left(x+1+x-1\right)\left(x+1-x+1\right)-3\left(x^2-1\right)\\ =4x-3x^2+3\)
Tinh bang hai cach:
a. (1/3 + 1/5) x 1/2 b. (1/3 - 1/5) x 1/2
giup minh voi, tra loi minh tick cho nha
a.\(=\dfrac{8}{15}\times\dfrac{1}{2}=\dfrac{4}{15}\)
b.\(=\dfrac{1}{3}\times\dfrac{1}{2}-\dfrac{1}{5}\times\dfrac{1}{2}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\)
( x + 1 )2 +3( x - 5)(x+ 5)-( 2x-1)2
=x2 + 2x + 1 + 3(x2 - 25) - 4x2 - 4x + 1
= x2 + 2x + 1 + 3x2 - 75 - 4x2 - 4x + 1
= -2x - 73
k cho mk nhe!!
( x + 1 )2 +3( x - 5)(x+ 5)-( 2x-1)2
=x2+2x+1+3x2-75-4x2+4x-1
=(x2+3x2-4x2)+(2x+4x)-(1-1)-75
=6x-75
Vậy ms đúng bn kia sai r`