\(ĐKXĐ:x\ne\pm1\)

a) \(A=\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)

\(\Leftrightarrow A=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)

\(\Leftrightarrow A=x-1+x+1-3\)

\(\Leftrightarrow A=2x-3\)

b) Thay x = 3 vào A, ta được :

\(A=2.3-3=3\)

Thay x = 0 vào A, ta được :

\(A=2.0-3=-3\)

c) Để A = 2

\(\Leftrightarrow2x-3=2\)

\(\Leftrightarrow2x=5\)

\(\Leftrightarrow x=\frac{5}{2}\)

Vậy để \(A=2\Leftrightarrow x=\frac{5}{2}\)

a) 3x(x - 1) + 7x²(x  - 1) = 0

<=> x(x - 1)(3 + 7x) = 0

<=> x = 0

hoặc : x - 1 = 0

hoặc 3 + 7x = 0

<=> x = 0

hoặc x = 1

hoặc x = -3/7

b) x² - 2018x - 2019 = 0

<=> x² - 2019x + x - 2019 = 0

<=> x(x - 2019) + (x - 2019) = 0

<=> (x + 1)(x - 2019) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-2019=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-1\\x=2019\end{cases}}\)

c) (x + 3)² - x(x - 2) = 13

<=> x² + 6x + 9 - x² + 2x = 13

<=> 8x = 13 - 9

<=> 8x = 6

<=> x=  6/8 = 3/4

a/\(3x\left(x-1\right)+7x^2\left(x-1\right)=0.\)

\(\Leftrightarrow\left(x-1\right)\left(3x+7x^2\right)=0\)

\(\Leftrightarrow\left(x-1\right)x\left(3+7x\right)=0\)

Th1: x - 1 = 0

=> x = 1

Th2: x= 0

Th3: 3 + 7x = 0

=> x= -3/7

\(\Rightarrow x\in\left\{1;0;-\frac{3}{7}\right\}\)

b/ \(x^2-2018x-2019=0\)

\(\Leftrightarrow x^2+x-2019x-2019=0\)

\(\Leftrightarrow\left(x^2+x\right)-\left(2019x+2019\right)=0\)

\(\Leftrightarrow x\left(x+1\right)-2019\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2019\right)\left(x+1\right)=0\)

Th1 : x -2019 = 0

=> x =2019

Th2: x + 1  =0

=> x = -1

\(\Rightarrow x\in\left\{2019;-1\right\}\)

c/ \(\left(x+3\right)^2-x\left(x-2\right)=13\)

\(\Leftrightarrow x^2+6x+9-x^2+2x=13\)

\(\Leftrightarrow8x=4\Rightarrow x=\frac{1}{2}\)

\(3x\left(x-1\right)+7x^2\left(x-1\right)=0\)

\(\Rightarrow\left(3x+7x^2\right)\left(x-1\right)=0\)

\(x\left(3+7x\right)\left(x-1\right)=0\)

\(\hept{\begin{cases}x=0\\3+7x=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-3}{7}\\x=1\end{cases}}\)

\(x\in\left\{0;\frac{-3}{7};1\right\}\)

b) \(x^2-2018x-2019=0\)

\(=x^2+x-2019x-2019=0\)

\(=x\left(x+1\right)-2019\left(x+1\right)=0\)

\(=\left(x-2019\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2019=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2019\end{cases}}\)

\(x\in\left\{-1;2019\right\}\)

c) \(\left(x+3\right)^2-x\left(x-2\right)=13\)

\(\Rightarrow x^2+6x+9-x^2+2x-13=0\)

\(\Rightarrow8x-4=0\)

\(\Rightarrow4\left(2x-1\right)=0\)

\(\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)

a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)

+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)

\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)

+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)

+) \(x+1\ne0\Leftrightarrow x\ne-1\)

+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)

\(\Leftrightarrow x\ne0;x\ne-2\)

+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)

Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)

a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)

\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)

\(=\left[\frac{x^2+2x+1}{x^2-x+1}-\frac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{x+1}\right].\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x^2+2x+1\right)\left(x+1\right)-2x^2-4x+1-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)

\(=\frac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)

\(=\frac{3x}{x-2}=3+\frac{6}{x-2}\)

b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)

Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Vậy ............................

b) \(A=\left(\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right):\frac{x^2-4}{3x^2+6x}\)

\(=\left(\frac{\left(x+1\right)^2}{x^2-x+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{x^2-4}{3x^2+6x}\)

\(=\left(\frac{\left(x+1\right)^3}{x^3+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{x^3+1}\right)\)\(.\frac{3x^2+6x}{x^2-4}\)

\(=\left(\frac{x^3+3x^2+3x+1}{x^3+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{x^3+1}\right)\)\(.\frac{3x^2+6x}{x^2-4}\)

\(=\frac{x^3+1}{x^3+1}\)\(.\frac{3x^2+6x}{x^2-4}\)\(=\frac{3x^2+6x}{x^2-4}\)

\(=\frac{3x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x}{x-2}\)

A nguyên\(\Leftrightarrow3x⋮x-2\)

\(\Leftrightarrow3\left(x-2\right)+6⋮x-2\)

Mà \(\left(x-2\right)⋮x-2\Rightarrow6⋮x-2\)

\(\Rightarrow x-2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Lập bảng:

Vậy\(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

1 M=\(x^2-4xy+4y^2-2x+4y+10\)

=\(\left(x^2-4xy+4y^2\right)+\left(-2x+4y\right)+10\)

\(=\left(x-2y\right)^2-2\left(x-2y\right)+10\)

\(=\left(x-2y\right)\left(x-2y-2\right)+10\)

vì \(\left(x-2y\right)\left(x-2y-2\right)\ge0\)

nên \(\left(x-2y\right)\left(x-2y-2\right)+10\ge10\)

\(\Rightarrow\)A\(\ge13\)

dấu "=" xảy ra khi (x-2y)(x-2y-2)=0

\(\left[{}\begin{matrix}x-2y=0\\x-2y-2=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2y=x\\x-2y=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0;y=0\\x=2;y=1\end{matrix}\right.\)

vậy GTNN của M=10 khi x=0; y=0

x=2;y=1

Khó vl , dẹp mẹ điiii

a)     \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)

\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=4\)

b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)

\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)

\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)

\(\Leftrightarrow B=x^3-20x^2+18x+69\)

c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)

\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)

\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)

\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)

\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)

d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)

\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)

\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)

\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)

Chúc bạn học tốt !

a, \(P\left(x\right)=4x^3+2x-3+2x-2x^2-1\\ =4x^3-2x^2+\left(2x+2x\right)+\left(-3-1\right)\\ =4x^3-2x^2+4x-4\)

Bậc của P(x) là 3

\(Q\left(x\right)=6x^3-3x+5-2x+3x^2\\ =6x^3+3x^2+\left(-3x-2x\right)+5\\ =6x^3+3x^2-5x+5\)

Bậc của Q(x) là 3

b, \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=4x^3-2x^2+4x-4+6x^3+3x^2-5x+5\\ =\left(4x^3+6x^3\right)+\left(-2x^2+3x^2\right)+\left(4x-5x\right)+\left(-4+5\right)\\ =10x^3+x^2-x+1\)

Bị tự tin quá khả năng nhẩm mồm, sai em xin lỗi ...

a, Ta có \(P\left(x\right)=8x^3+2x^2-3x-3x^3+10-x-2x^2-3\)

\(=5x^3-4x-7\)

\(Q\left(x\right)=9x^3-4x^2+2x-3+2x+3x^2+4x^3-2\)

\(=13x^3-x^2+4x-5\)

b, Ta có : \(P\left(-\frac{1}{2}\right)=5.\left(-\frac{1}{2}\right)^3-4.\left(-\frac{1}{2}\right)-7=-\frac{45}{8}\)

c , \(M\left(x\right)=P\left(x\right)+Q\left(x\right)\) 

  \(5x^3-4x-7+13x^3-x^2+4x-5=18x^3-x^2-12\)

\(N\left(x\right)=P\left(x\right)-Q\left(x\right)\)

\(5x^3-4x-7-13x^3+x^2-4x+5=-8x^3-8x-2+x^2\)

d, Đặt \(5x^3-4x-7=0\)( vô nghiệm )

Các bạn ơi giúp mình với !!! mình cần gấp mà HUHU

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