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1a) ( x - 2 )( x2 + 2x + 4 ) - x( x - 1 )( x + 1 ) + 3
= x3 - 8 - x( x2 - 1 ) + 3
= x3 - 8 - x3 + x + 3
= x - 5
b) Với x = -1/2 => Giá trị của biểu thức = -1/2 - 5 = -11/2
2a) 3x( 5x2 - 2xy2 + y ) = 15x3 - 6x2y2 + 3xy
b) ( x + y )( x2 - xy + y2 ) = x3 + y3
3) 16x2 - ( 4x - 5 )2 = 15
<=> 16x2 - ( 16x2 - 40x + 25 ) = 15
<=> 16x2 - 16x2 + 40x - 25 = 15
<=> 40x - 25 = 15
<=> 40x = 40
<=> x = 1
a) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
a) Rút gọn :
\(ĐKXĐ:x\ne\pm5\)
Ta có : \(P=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}-\frac{2x}{5-x}\)
\(=\left(\frac{x^2-\left(x-5\right)\left(x-5\right)}{x\left(x-5\right)\left(x+5\right)}\right):\frac{\left(2x-5\right)\left(x-5\right)+2x^2\left(x+5\right)}{x\left(x+5\right)\left(x-5\right)}\)
\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\frac{x\left(x+5\right)\left(x-5\right)}{ }\)
Tui đang định làm tiếp đó, nhưng khẳng định đề này hơi sai sai ở vế bị chia. Bạn xem lại đc k ?
a.)Đkxđ bạn tự tìm nha!!!
A=\(\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow\)\(\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow\)\(\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{x^2+x+1}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{x^2+x+1}\)
\(\Leftrightarrow\)\(\frac{2x+1}{\left(x-1\right)\left(x+1\right)}:\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow\)\(\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow\)\(\frac{x+1}{x-1}\left(tm\text{đ}k\right)\)
b.)Thay \(x=\frac{1}{2}\)vào A \(\Rightarrow\)\(A=-3\)
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
( x + 2 )( x2 - 2x + x2 ) - x( x - 5 )( x + 5 ) + 17
= ( x + 2 )( 2x2 - 2x ) - x( x2 - 25 ) + 17
= x( 2x2 - 2x ) + 2( 2x2 - 2x ) - x3 + 25x + 17
= 2x3 - 2x2 + 4x2 - 4x - x3 + 25x + 17
= x3 + 2x2 + 21x + 17
Thế x = -1 ta được :
(-1)3 + 2.(-1)2 + 2.(-21) + 17 = -1 + 2 - 21 + 17 = -3
nhầm dòng tính :v
(-1)3 + 2.(-1)2 + 21.(-1) + 17 nhé .-.