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ta có ;
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2=\left(x-y+x+y\right)^2-4x^2\)
\(=\left(2x\right)^2-4x^2=0\)
b, 5(x + 2) (x - 2 ) - 1/2 (6-8x)2 + 17
=5x +10 (x - 2) - 1/2 . 6 - 1/2 . 8x +17
=5x + 10x - 20 - 3 - 4x +17
=15x - 17 -4x + 17
=15x - 4x -17 + 17
=11x - 0 =11x
a, (x+1)2 - (x-1)2 - 3(x+1) (x-1)
=(x+1)+(x-1).(x+1)-(x-1) - 3x+3x -3
=2x.0 - 3x
=-3x
Bài 1:
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+1\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3-x+y\right)\)
\(=2\left(x-y\right)\left(2x+3+y\right)\)
Bài 2:
\(P=\left(3x-1\right)^2+2\left(3x-1\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(3x-1-x-1\right)^2\)
\(=\left(2x-2\right)^2\)(1)
b) Thay \(x=\frac{9}{4}\)vào (1) ta được:
\(\left(2.\frac{9}{4}-2\right)^2\)
\(=\frac{25}{4}\)
Vậy giá trị của P \(=\frac{25}{4}\)khi \(x=\frac{9}{4}\)
Bài 3:
Ta có: \(M=x^2+4x+5\)
\(=\left(x+2\right)^2+1\)
Vì \(\left(x+2\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+2\right)^2+1\ge0+1;\forall x\)
Hay \(M\ge1;\forall x\)
Dấu"="xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Leftrightarrow x=-2\)
Vậy \(M_{min}=1\Leftrightarrow x=-2\)
Bài 1 : trên là sai nha mình làm lại
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3y-x+y\right)\)
\(=2\left(x-y\right)\left(2x+4y\right)\)
\(=4\left(x-y\right)\left(x+2y\right)\)
\(\left(x^2-1\right)\left(x+2\right)-\left(x-4\right)\left(x^2+4x+16\right)\)
\(=x^3+2x^2-x-2-\left(x^3-4^3\right)\)
\(=x^3+2x^2-x-2-x^3+64\)
\(=2x^2-x+62\)
\(2x\left(3x-2\right)^2\)
\(=2x\left(9x^2-12x+4\right)\)
\(=18x^3-24x^2+8x\)
\(\left(x-3\right)\left(x^2-3x+9\right)\)
\(=x^3-3x^2+9x-3x^2+9x-27\)
\(=x^3-3x^2+18x-27\)
\(\left(x^2-1\right)\left(x+2\right)-\left(x-4\right)\left(x^2+4x+16\right)\)
\(=\left(x^2-1^2\right)\left(x+2\right)-x^3-4^3\)
\(=\left(x+1\right)\left(x-1\right)\left(x+2\right)-x^3-64\)
A = 5(x + 3)(x - 3) + (2x + 3)3 + (x - 6)2
A = 5(x + 3)(x - 3) + 4x2 + 12x + 9 + x2 - 12x + 36
A = 5x2 - 45x + 4x2 + 12x + 9 + x2 - 12x + 36
A = 10x2 (1)
Thay x = -1/5 vào (1), ta có:
A = 10x2 = 10.(-1/5)2 = 2/5
A = 2/5
Vậy:...
a: \(Q=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x|=1/3 thì x=1/3 hoặc x=-1/3
Khi x=1/3 thì \(Q=\left(\dfrac{1}{3}\right)^2:\left(\dfrac{1}{3}-1\right)=-\dfrac{1}{6}\)
Khi x=-1/3 thì \(Q=\left(-\dfrac{1}{3}\right)^2:\left(-\dfrac{1}{3}-1\right)=-\dfrac{1}{12}\)
c: Để Q là số nguyên thì \(x^2-1+1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>x=2
d: Để Q=4 thì x^2=4x-4
=>x=2
a/ đkxđ: x \(\ne\pm\)2; x≠3
\(P=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)
\(=\left(\dfrac{\left(2+x\right)^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}+\dfrac{4x^2}{x^2-4}\right):\dfrac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)
\(=\dfrac{x^2+4x+4-x^2+4x-4+4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{2-x}{x-3}\)
\(=\dfrac{8x+4x^2}{2+x}\cdot\dfrac{1}{x-3}=\dfrac{4x\left(2+x\right)}{2+x}\cdot\dfrac{1}{x-3}=\dfrac{4x}{x-3}\)
b/ x = \(\dfrac{1}{3}\Leftrightarrow P=\dfrac{4\cdot\dfrac{1}{3}}{\dfrac{1}{3}-3}=\dfrac{4}{3}:\left(-\dfrac{8}{3}\right)=\dfrac{4}{3}\cdot\left(-\dfrac{3}{8}\right)=-\dfrac{4}{8}=-\dfrac{1}{2}\)
c/ \(P\in Z\Rightarrow\dfrac{4x}{x-3}\in Z\)
Ta có: \(\dfrac{4x}{x-3}=\dfrac{4x-12+12}{x-3}=\dfrac{4\left(x-3\right)}{x-3}+\dfrac{12}{x-3}=4+\dfrac{12}{x-3}\)
=> \(x-3\inƯ\left(12\right)\) thì P ∈ Z
=> \(x-3=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
\(\Leftrightarrow x=\left\{-9;-3;-1;0;1;2;4;5;6;7;9;15\right\}\)
mà x>4
=> x = {5;6;7;9;15}
a, Ta có:
\(P=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)
\(=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{4-x^2}\right):\left[\dfrac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\right]\)
\(=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x-3}{2-x}\)
\(=\dfrac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4+4x+x^2-\left(4-4x+x^2\right)+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4+4x+x^2-4+4x-x^2+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x}{x-3}\)
\(=\left(\dfrac{x+1-1}{x+1}\right)\left(\dfrac{x+2-1}{x+2}\right)...\left(\dfrac{x+2014-1}{x+2014}\right)\)
\(=\dfrac{x\left(x+1\right)\left(x+2\right)...\left(x+2013\right)}{\left(x+1\right)\left(x+2\right)...\left(x+2013\right)\left(x+2014\right)}\)
\(=\dfrac{x}{x+2014}\)
Lời giải:
ĐKXĐ: \(x\neq \left\{2;\pm 3\right\}\)
a) Ta có:
\(P=\left(\frac{x^2-3x}{x^2-9}-1\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}\right)\)
\(P=\left(\frac{x(x-3)}{(x-3)(x+3)}-1\right):\left(\frac{(3-x)(3+x)}{(x-2)(x+3)}-\frac{3-x}{x-2}-\frac{x-2}{x+3}\right)\)
\(P=\left(\frac{x}{x+3}-1\right):\left(\frac{3-x}{x-2}-\frac{3-x}{x-2}-\frac{x-2}{x+3}\right)\)
\(P=\frac{x-(x+3)}{x+3}:\left(-\frac{x-2}{x+3}\right)=\frac{-3}{x+3}.\frac{x+3}{-(x-2)}=\frac{3}{x-2}\)
b) \(x^3-3x+2=0\)
\(\Leftrightarrow (x^3-x)-2(x-1)=0\)
\(\Leftrightarrow x(x-1)(x+1)-2(x-1)=0\)
\(\Leftrightarrow (x-1)(x^2+x-2)=0\)
\(\Leftrightarrow (x-1)[(x^2-1)+(x-1)]=0\)
\(\Leftrightarrow (x-1)^2(x+2)=0\) \(\Leftrightarrow \left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.\)
Với \(x=1\Rightarrow P=\frac{3}{1-2}=-3\)
Với \(x=-2\Rightarrow P=\frac{3}{-2-2}=\frac{-3}{4}\)
c)
\(P=\frac{3}{x-2}\in\mathbb{Z}\Leftrightarrow 3\vdots x-2\)
\(\Leftrightarrow x-2\in \text{Ư}(3)\Rightarrow x-2\in\left\{\pm 1; \pm 3\right\}\)
\(\Leftrightarrow x\in \left\{3,1,5,-1\right\}\)
Do \(x\neq 3\Rightarrow x\in \left\{-1,1,5\right\}\)
..
đc ko bạn
cảm ơn bạn nha