1

1 đó

b)x³-2x²-4xy²+x

=x(x²-2x-4y²+1)

=x[(x²-2x+1)-4y²]

=x[(x-1)²-4y²]

=x(x-1-2y)(x-1+2y)

c) (x+2)(x+3)(x+4)(x+5)-8

=[(x+2)(x+5)][(x+3)(x+4)]-8

=(x²+5x+2x+10)(x²+4x+3x+12)-8

=(x²+7x+10)(x²+7x+12)-8

đặt x²+7x+10 =a ta có

a(a+2)-8

=a²+2a-8

=a²+4a-2a-8

=(a²+4a)-(2a+8)

=a(a+4)-2(a+4)

=(a+4)(a-2)

thay a=x²+7x+10 ta đc

(x²+7x+10+4)(x²+7x+10-2)

=(x²+7x+14)(x²+7x+8)

bài 2 x³-x²y+3x-3y

=(x³-x²y)+(3x-3y)

=x²(x-y)+3(x-y)

=(x-y)(x²+3)

1) \(\frac{x-y}{z-y}=-10\Leftrightarrow x-y=10\left(y-z\right)\)

\(\Leftrightarrow x-y=10y-10z\)

\(\Leftrightarrow x=11y-10z\)

Thay x=11y-10z vào biểu thức \(\frac{x-z}{y-z}\), ta có:

\(\frac{11y-10z-z}{y-z}=\frac{11y-11z}{y-z}=\frac{11\left(y-z\right)}{y-z}=11\)

Chá quá, có ghi nhìn không rõ đề

2) \(2x^2=9x-4\)

\(\Leftrightarrow2x^2-9x+4=0\)

\(\Leftrightarrow2x^2-8x-x+4=0\)

\(\Leftrightarrow2x\left(x-4\right)-1\left(x-4\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow2x-1=0\) hoặc x-4=0

1) 2x-1=0<=>x=1/2

2)x-4=0<=>x=4(Loại)

=> x=1/2

Bài 3: 

1: \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(6x+1\right)\left(6x-1\right)\)

\(=\left(6x+1-6x+1\right)^2\)

\(=2^2=4\)

2: \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

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

\(=2^{32}-1\)

3: \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

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

4: \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

=-11x+24

Ta có : \(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3.\sqrt[3]{\frac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right).64}}=\frac{3a}{4}\)

Tương tự : \(\frac{b^3}{\left(1+a\right)\left(1+c\right)}\ge\frac{3b}{4}\) ; \(\frac{c^3}{\left(1+b\right)\left(1+a\right)}\ge\frac{3c}{4}\)

\(\Rightarrow A\ge\frac{3}{4}\left(a+b+c\right)\ge\frac{3}{4}.\sqrt[3]{abc}=\frac{3}{4}\)

=> Max A = 3/4 <=> a = b = c = 1

a/ Theo bài ra: \(x^2+y^2=6;xy=1\)

=> \(x^2+y^2+2xy=8\)

=> \(\left(x+y\right)^2=8\)

=> \(x+y=\sqrt{8}\)

b/ Theo bài ra: \(x^2+y^2=14;xy=1\)

=>\(x^2+y^2-2xy=12\)

=> \(\left(x-y\right)^2=12\)

=> \(x-y=\sqrt{12}\)

c/ Theo bài ra: \(a^2+b^2=116;ab=40\)

=> \(\left(a^2+b^2\right)^2=116^2;a^2b^2=1600\)

=> \(a^4+b^4+2a^2b^2=116^2\)

=> \(a^4-2a^2b^2+b^4+4a^2b^2=13456\)

=> \(a^4-2a^2b^2+b^4=7056\)

Bài 2 :

a ) \(25-20x+4x^2=0\)

\(\Leftrightarrow\left(5-2x\right)^2=0\)

\(\Leftrightarrow5-2x=0\Rightarrow x=\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)

a,\(\left(-2x^2+3x\right)\left(x^2-x+3\right)\\ \Leftrightarrow-2x^4+2x^3-6x^2+3x^3-3x^2+9x\\ \Leftrightarrow-2x^4+5x^3-3x^2+3x\)

\(b,x\left(x-2\right)\left(x+2\right)-\left(x-3\right)\left(x^2+3x+9+6\right)+6\left(x+1\right)^2=15\\ \Leftrightarrow x\left(x^2-4\right)-\left(x^3-27\right)+6\left(x^2+2x+1\right)=15\\ \Leftrightarrow x^3-4x-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow6x^2+8x+18=0\\ \Leftrightarrow6\left(x^2+\dfrac{4}{3}x+3\right)=0\\ \Leftrightarrow\left(x+\dfrac{2}{3}\right)^2+\dfrac{23}{9}=0\)

Với mọi x thì \(\left(x+\dfrac{2}{3}\right)^2\ge0\Rightarrow\left(x+\dfrac{2}{3}\right)^2+\dfrac{23}{9}>0\)

Do đó ko tìm đc giá trị nào của x thỏa mãn đề bài

Vậy..