\(x^2y^2+2x^2+y^2+2\)

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

\(=\left(x^2+1\right)\left(y^2+2\right)\)

\(a^2-b^2+a-b\)

\(=\left(a+b\right)\left(a-b\right)+\left(a-b\right)\)

\(=\left(a+b+1\right)\left(a-b\right)\)

\(a,x^2y^2+2x^2+y^2+2\)

\(=y^2\left(x^2+1\right)+2\left(x^2+1\right)\)

\(=\left(y^2+2\right)\left(x^2+1\right)\)

\(b,a^2-b^2+a-b\)

\(=\left(a+b\right)\left(a-b\right)+\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b+1\right)\)

bạn nào giải đúng mình cho 1 tích

\(A=\left(9xy^2-6x^2y\right):\left(-3xy\right)+\left(6x^2y+2x^4\right):\left(2x^2\right)\)

\(=-3y+2x+3y+2x\)

\(=4x\)

Biểu thức không có GTNN

Câu hỏi của Chi Chi - Toán lớp 8 - Học toán với OnlineMath

\(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)

\(ĐKXĐ:x\ne1;x\ne3\)

\(pt\Leftrightarrow\frac{2x-2}{x^2-4x+3}+\frac{x^2-8x+15}{x^2-4x+3}=1\)

\(\Leftrightarrow\frac{x^2-6x+13}{x^2-4x+3}=1\)

\(\Leftrightarrow x^2-6x+13=x^2-4x+3\)

\(\Leftrightarrow-2x+10=0\Leftrightarrow x=-5\left(t/mđkxđ\right)\)

Vậy pt có 1 nghiệm là -5

2/x - 3 + x - 5/x - 1 = 1

2(x - 1) + (x - 5)(x - 3) = (x - 3)(x - 1)

-6x + 13 + x^2 = x^2 - 4x + 3

-6x + 13 = -4x + 3

13 = -4x + 3 + 6x

13 = 2x + 3

13 - 3 = 3x

10 = 2x

5 = x

=> x = 5

(3x - 5)(5 - 3x) + 9(x + 1)^2 = 30

=> 15x - 9x^2 - 25 + 15x + 9(x^2 + 2x + 1) = 30

=> 30x - 9x^2 - 25 + 9x^2 + 18x + 9 = 30

=> 38x - 16 = 30

=> 38x = 46

=> x = 23/19

\(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30.\)

\(15x-9x^2-25+15x+9\left(x^2+2x+1\right)=30\)

\(15x-9x^2-25+15x+9x^2+18x+9=30\)

\(48x-16=30\)

\(48x=30+16=46\)

\(x=\frac{46}{48}=\frac{23}{24}\)

  \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(x^2+8x+16-\left(x^2-1\right)=16\)

\(x^2+8x+16-x^2+1=16\)

\(8x+17=16\)

\(8x=16-17=-1\)

\(x=-\frac{1}{8}\)

\(M=2x^2-x+1=\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\frac{1}{\sqrt{8}}+\frac{1}{8}+\frac{7}{8}\)

\(=\left(\sqrt{2}x-\frac{1}{\sqrt{8}}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)

Vậy \(M_{min}=\frac{7}{8}\Leftrightarrow\sqrt{2}x-\frac{1}{\sqrt{8}}=0\)

\(\Leftrightarrow\sqrt{2}x=\frac{1}{\sqrt{8}}\Leftrightarrow x=\frac{1}{16}\)

\(x=\frac{1}{\sqrt{16}}=\frac{1}{4}\)nha. Gõ nhầm

2x(5 - x) + (x - 1)(2x + 3) = 8

=> 10x - 2x² + 2x² + 3x - 2x - 3 = 8

=> 11x - 3 = 8

=> 11x = 8 + 3

=> 11x = 11

=> x = 11 : 11 

=> x = 1

Đặt f(x) = x^4 + ax^3 + bx +b 

xét f(-1)=0 và f(1) =0(vì f(x) chia hết cho a khi f(a) =0)

f(-1) = 1 - a -b + b = 1-a =0

+

f(1) = 1+a+b+b = 1+a+2b = 0

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=> 2+2b = 0

=> b= -1

=> 1+a-2 = 0

=> a=1