k minh minh giai cho

a) ( 5 - 2x )( 2x + 7 ) - 4x² + 25 = 0

<=> ( 5 - 2x )( 2x + 7 ) + ( 5 - 2x )( 5 + 2x ) = 0

<=> ( 5 - 2x )( 2x + 7 + 5 + 2x ) = 0

<=> ( 5 - 2x )( 4x + 12 ) = 0

<=> \(\orbr{\begin{cases}5-2x=0\\4x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)

b) ( 5x² + 3x - 2 )² - ( 4x² - x - 5 )² = 0 ( như này chứ nhỉ ? )

<=> [ ( 5x² + 3x - 2 ) - ( 4x² - x - 5 ) ][ ( 5x² + 3x - 2 ) + ( 4x² - x - 5 ) ] = 0

<=> ( 5x² + 3x - 2 - 4x² + x + 5 )( 5x² + 3x - 2 + 4x² - x - 5 ) = 0

<=> ( x² + 4x + 3 )( 9x² + 2x - 7 ) = 0

<=> ( x² + x + 3x + 3 )( 9x² + 9x - 7x - 7 ) = 0

<=> [ x( x + 1 ) + 3( x + 1 ) ][ 9x( x + 1 ) - 7( x + 1 ) ] = 0

<=> ( x + 1 )( x + 3 )( x + 1 )( 9x - 7 ) = 0

<=> ( x + 1 )²( x + 3 )( 9x - 7 ) = 0

<=> x + 1 = 0 hoặc x + 3 = 0 hoặc 9x - 7 = 0

<=> x = -1 hoặc x = -3 hoặc x = 7/9

c) 15x⁴ - 8x³ - 14x² - 8x + 15 = 0

<=> 15x⁴ + 22x³ - 30x³ + 15x² + 15x² - 44x² - 30x + 22x + 15 = 0

<=> ( 15x⁴ + 22x³ + 15x² ) - ( 30x³ + 44x² + 30x ) + ( 15x² + 22x + 15 ) = 0

<=> x²( 15x² + 22x + 15 ) - 2x( 15x² + 22x + 15 ) + ( 15x² + 22x + 15 ) = 0

<=> ( 15x² + 22x + 15 )( x² - 2x + 1 ) = 0

<=> ( 15x² + 22x + 15 )( x - 1 )² = 0

<=> \(\orbr{\begin{cases}15x^2+22x+15=0\\\left(x-1\right)^2=0\end{cases}}\)

+) ( x - 1 )² = 0 <=> x = 1

+) 15x² + 22x + 15 = 15( x² + 22/15x + 121/225 ) + 104/15 = 15( x + 11/25 )² + 104/15 ≥ 104/15 > 0 ∀ x

Vậy phương trình có nghiệm duy nhất là x = 1

Cảm ơn bạn câu b thiếu cái mũ 2 sorry :))

Lời giải:

a) \(x^3+5x^2-5=15x-32\)

Bạn xem lại xem có sai đề không

b)

\(8x^2+2x-15=0\)

\(\Leftrightarrow 16x^2+4x-30=0\)

\(\Leftrightarrow (4x+\frac{1}{2})^2-\frac{121}{4}=0\)

\(\Rightarrow \left[\begin{matrix} 4x+\frac{1}{2}=\sqrt{\frac{121}{4}}=\frac{11}{2}\\ 4x+\frac{1}{2}=-\sqrt{\frac{121}{4}}=\frac{-11}{2}\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=\frac{5}{4}\\ x=\frac{-3}{2}\end{matrix}\right.\)

1.

\(\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

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

2.

a)  \(27x^4-8x=x\left(27x^3-8\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b)  \(16x^2y-4xy^2-4x^3+x^2y\)

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

\(=x\left(4x-y\right)\left(4y-x\right)\)

c) \(x^2-2x-5+2\sqrt{5}\)

\(=\left(x-1\right)^2-6+2\sqrt{5}\)

\(=\left(x-1\right)^2-\left(6-2\sqrt{5}\right)=\left(x-1\right)^2-\left(\sqrt{5}-1\right)^2\)

\(=\left(x-\sqrt{5}\right)\left(x-2+\sqrt{5}\right)\)

Bài 1:

 \(\left(25x^4y^3-15x^3y^5+20x^2y^4\right):\left(5x^2y^3\right)\)

\(=\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

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

Bài 2: 

a) \(27x^4-8x\)

\(=x\left(3x-2\right)\left(3^2x^2+2.3x+2^2\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b) \(16x^2y-4xy^2-4x^3+x^2y\)

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

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

a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)

\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)

\(A=9x\)

Thay x = 15 vào, ta có: 

\(A=9.15=135\)

b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(B=5x^2-20xy-4y^2+20xy\)

\(B=5x^2-4y\)

Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có: 

\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)

c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)

\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(C=9x^2y^2-xy^3-8x^3\)

Thay \(x=\frac{1}{2};y=2\) vào, ta có:

\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)

d) \(D=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)

\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)

\(D=18x^2+12x-7\)

Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)

+) Với x = -2

\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)

+) Với x = 2

\(D=18.2^2+12.2-7=89\)