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\(\frac{a}{x+1}+\frac{b}{x-1}=\frac{5x+1}{x^2-1}\)

\(\Leftrightarrow\frac{a\left(x-1\right)+b\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{5x+1}{\left(x+1\right)\left(x-1\right)}\)

\(\Leftrightarrow ax-a+bx+b=5x+1\)

\(\Leftrightarrow x\left(a+b\right)-a+b=5x+1\)

\(\Rightarrow\hept{\begin{cases}a+b=5\\b-a=1\end{cases}\Rightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}}\)

a) \(2x^2+3x-8=0\)

Ta có: \(\Delta=3^2+4.2.8=73\)

pt có 2 nghiệm

\(x_1=\frac{-3+\sqrt{73}}{4}\);\(x_1=\frac{-3-\sqrt{73}}{4}\)

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

Đặt \(x^2+2x=t\)

\(pt\Leftrightarrow t^2-2t-3=0\)

Ta có: \(\Delta=2^2+4.3=16,\sqrt{\Delta}=4\)

pt trên có 2 nghiệm

\(x_1=\frac{2+4}{2}=3;x_2=\frac{2-4}{2}=-1\)

\(\Rightarrow\orbr{\begin{cases}x^2+2x=3\\x^2+2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)\left(x-1\right)=0\\\left(x+1\right)^2=0\end{cases}}\)

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

c) \(x^4+8x^3+19x^2+12x=0\)

\(\Leftrightarrow x^4+4x^3+4x^3+16x^2+3x^2+12x=0\)

\(\Leftrightarrow\left(x^4+4x^3+3x^2\right)+\left(4x^3+16x^2+12x\right)=0\)

\(\Leftrightarrow x\left(x^3+4x^2+3x\right)+4\left(x^3+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+x^2+3x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2+3x\right)\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow x\in\left\{0;-1;-3;-4\right\}\)

Ta có:

a) 6x²y - 3y² - 2x² + y = (6x²y - 2x²) - (3y² - y) = 2x²(3y - 1) - y(3y - 1) = (2x² - y)(3y - 1)

b)  2x² + x - 4xy - 2y + 2x + 1 = (x² + x) - (4xy + 2y) + (x² + 2x + 1) = x(x + 1) - 2y(2x + 1) + (x + 1)²

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

c) 16x²y - 4xy² - 4x³ + x²y = 4xy(4x - y) - x²(4x - y) = (4xy - x²)(4x - y)

d) 4x² - 20x + 25 - 36y² = (2x  - 5)² - (6y)² = (2x - 5 - 6y)(2x  - 5 + 6y)

e) x² - 4y² + 6x - 4y + 8 = (x² + 6x + 9) - (4y² + 4y + 1) = (x + 3)² - (2y + 1)² = (x + 3 - 2y - 1)(x + 3 + 2y + 1) = (x + 2 - 2y)(x + 4 + 2y)

g) Ta có : x¹⁰ + x⁵ + 1

= (x¹⁰ - x) + (x⁵ - x²) + (x² + x + 1)

= x(x⁹ - 1) + x²(x³ - 1) + (x² + x + 1)

= x(x³ - 1)(x⁶ + x³ + 1) + x²(x³ - 1) + (x² + x + 1)

= (x⁷ + x⁴ + x)(x - 1)(x² + x + 1) + x²(x - 1)(x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x⁸ - x⁷ + x ⁵ - x⁴ + x² - x + x⁴ + x³ + x² + 1)

= (x² + x + 1)(x⁸ - x⁷ + x⁵ + x³ - x + 1)

h) TT trên (dài dòng ktl)

\(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^3=\left(-c\right)^3\)

\(\Leftrightarrow a^3+b^3+3a^2b+3ab^2=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b\right)=0\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)

Câu b) tương tự nha

x.(x-5)+x-5=o

=> x.(x-5)+(x-5)=0

=>(x-5)(x+1)=0

=> x-5=0    =>x=5

     x+1=0        x=-1

x ( x - 5 ) + x -  5 = 0

x ( x - 5 ) + ( x - 5 ) = 0

(x - 5 ) ( x + 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

Vậy...

a/ \(S=\left(5x+1\right)\left(5x-1\right)=25x^2-1\)

b/ \(S=25.2^2-1=99m^2\)

Ta có : x + y = 1 => x = y - 1

=> P = (y - 1).y - 7 = y² - y - 7 = (y² - y - 1/4) - 27/4 = (y - 1/2)² - 27/4 \(\ge\)-27/4 \(\forall\)y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}y-\frac{1}{2}=0\\x=y-1\end{cases}}\) <=> \(\hept{\begin{cases}y=\frac{1}{2}\\x=\frac{1}{2}-1=-\frac{1}{2}\end{cases}}\)

Vậy Min P = -27/4 <=> x = -1/2 và y = 1/2

Edogawa Conan

Cách em làm ko sai. Nhưng em nhầm từ dòng đầu tiên nhé!

x + y = 1 => x = 1- y

Giải: 

Có: \(\left(x-y\right)^2\ge0,\forall x,y\)

<=> \(x^2+2xy+y^2\ge2xy,\forall x,y\)

<=> \(\left(x+y\right)^2\ge4xy,\forall x,y\)

=> \(P=xy-7\le\frac{\left(x+y\right)^2}{4}-7=\frac{1}{4}-7=-\frac{27}{4}\)

"="  xảy ra <=> \(\hept{\begin{cases}\left(x-y\right)^2=0\\x+y=1\end{cases}\Leftrightarrow}x=y=\frac{1}{2}\)

Vậy GTLN của P là -27/4 đạt tại x = y = 1/2.