1.

a)

\(-x^2+2x-7\left(1\right)\\ \Leftrightarrow-\left(x^2-2x+7\right)\\ \Leftrightarrow-\left[\left(x^2-2x+1\right)+6\right]\\ \Leftrightarrow-\left[\left(x-1\right)^2+6\right]\le-6\forall x\)

=> BT (1) luôn âm với mọi x

b)

\(-5x^2+20x-49\left(2\right)\\ \Leftrightarrow-\left(5x^2-20x+49\right)\\ \Leftrightarrow-\left(x^2-4x+\dfrac{49}{5}\right)\Leftrightarrow-\left[\left(x^2-4x+4\right)+\dfrac{29}{5}\right]\Leftrightarrow-\left[\left(x-2\right)^2+\dfrac{29}{5}\right]\le\dfrac{29}{5}\forall x\)

=> BT (2) luôn âm với mọi x

Bài 1 :

\(-x^2+2x-7\)

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

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

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

Do \(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-6\le-6< 0\)

Vậy biểu thức luôn âm với mọi giá trị của x .

\(-5x^2+20x-49\)

\(=\left(-5x^2+20x-20\right)-29\)

\(=-5\left(x^2-4x+4\right)-29\)

\(=-5\left(x-2\right)^2-29\)

Do \(\left(x-2\right)^2\ge0\Rightarrow-5\left(x-2\right)^2\le0\Rightarrow-5\left(x-2\right)^2-29\le-29< 0\)

Vậy biểu thức luôn âm với mọi giá trị của x

Bài 2 :

\(x^2+8x=x^2+8x+16-16=\left(x+4\right)^2-16\ge-16\)

\(2x^2+4x+15=2x^2+4x+2+13=2\left(x+1\right)^2+13\ge13\)

1) \(2x^3-8x=0\)

\(\Leftrightarrow2x\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

Vậy \(x\in\left\{0;\pm2\right\}\)

2) \(2x\left(x-15\right)-4\left(x-15\right)=0\)

\(\Leftrightarrow\left(2x-4\right)\left(x-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=15\end{cases}}\)

Vậy \(x\in\left\{2;15\right\}\)

1 

\(2x^3-8x=0\)   

\(2x\left(x^2-4\right)=0\)   

\(\orbr{\begin{cases}2x=0\\x^2-4=0\end{cases}}\)   

\(\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\)    

\(\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)   

2 

\(2x\left(x-15\right)-4\left(x-15\right)=0\)    

\(\left(2x-4\right)\left(x-15\right)=0\)   

\(\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\)    

\(\orbr{\begin{cases}2x=4\\x=0+15\end{cases}}\)   

\(\orbr{\begin{cases}x=2\\x=15\end{cases}}\)

1) 2x³ - 8x = 0

<=> 2x( x² - 4 ) = 0

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

<=> 2x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0

<=> x = 0 hoặc x = ±2

2) 2x( x - 15 ) - 4( x - 15 ) = 0

<=> ( x - 15 )( 2x - 4 ) = 0

<=> x - 15 = 0 hoặc 2x - 4 = 0

<=> x = 15 hoặc x = 2

Bài 2: 

a: \(A=x^2+8x\)

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

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

Dấu '=' xảy ra khi x=-4

b: \(B=-2x^2+8x-15\)

\(=-2\left(x^2-4x+\dfrac{15}{2}\right)\)

\(=-2\left(x^2-4x+4+\dfrac{7}{2}\right)\)

\(=-2\left(x-2\right)^2-7\le-7\)

Dấu '=' xảy ra khi x=2

c: \(C=x^2-4x+7\)

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

\(=\left(x-2\right)^2+3\ge3\)

Dấu '=' xảy ra khi x=2

e: \(E=x^2-6x+y^2-2y+12\)

\(=x^2-6x+9+y^2-2y+1+2\)

\(=\left(x-3\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu '=' xảy ra khi x=3 và y=1

8x² + 2x - 15

= 8x² + 12x - 10x - 15

= 4x (2x + 3 ) - 5 ( 2x + 3 )

= (2x + 3) (4x - 5)

Phân tích đa thức thành nhân tử

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

\(=8x^2+12x+10x-15\)

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

\(=\left(4x-5\right)\left(2x-3\right)\)

8x^2+2x-15

=8x^2+12x-10x-15

=4x(2x+3)-5(2x+3)

=(2x+3)(4x-5)

Các bn giúp mình với mình đang cần gấp

nhiều quá bạn ơi , mk nghĩ bạn nên tách ra rồi hãy đăng lên

Bài 1:

16:

=>15-30x-8+8x=4x-9

=>-22x+7=4x-9

=>-26x=-16

=>x=8/13

15: \(\Leftrightarrow12x-6+12-4x=10-6x\)

=>8x+6=10-6x

=>14x=4

=>x=2/7

14: \(\Leftrightarrow12x-18+9-15x=8x-9\)

=>-3x-9=8x-9

=>x=0

13: \(\Leftrightarrow10x+15x-10=25-10x\)

=>25x-10=25-10x

=>35x=35

=>x=1

12: \(\Leftrightarrow15x-18-4x+10=11x-10\)

=>11x-8=11x-10(loại)

( x + 5 ) . ( x + 6 ) = 0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

Vậy \(x=-5\) hoặc \(x=-6\)

8x - 9x -2x - 15 = 0

\(\Rightarrow8x-9x-2x=0+15\)

\(\Rightarrow-3x=15\)

\(\Rightarrow x=15:\left(-3\right)\)

\(\Rightarrow x=-5\)

a, \(\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

Vậy ......

\(\left(x+5\right)\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)

\(8x-9x-2x-15=0\)

\(\Rightarrow8x-9x-2x=15\)

\(\Rightarrow-3x=15\)

\(\Rightarrow x=-5\)

Bài 1:

a) \(ay-ax-2x+2y\)

\(=-a\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(-a-2\right)\)

b) \(5ax-7by-7ay+5bx\)

\(=5x\left(a+b\right)-7y\left(a+b\right)\)

\(=\left(a+b\right)\left(5x-7y\right)\)

c) \(4x^2-9x+5\)

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

\(=4x\left(x-1\right)-5\left(x-1\right)\)

\(=\left(x-1\right)\left(4x-5\right)\)

d) \(x^2-8x+15\)

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

\(=x\left(x-3\right)-5\left(x-3\right)\)

\(=\left(x-3\right)\left(x-5\right)\)

Bài 2:

a) \(x^2+x+\frac{1}{2}\)

\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{1}{4}>0\forall x\)

b) \(x^2+5x+7\)

\(=x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}+\frac{3}{4}\)

\(=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}>0\forall x\)

c) \(2x^2-3x+9\)

\(=2\left(x^2-\frac{3}{2}x+\frac{9}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}+\frac{63}{16}\right)\)

\(=2\left[\left(x-\frac{3}{4}\right)^2+\frac{63}{16}\right]\)

\(=2\left(x-\frac{3}{4}\right)^2+\frac{63}{8}>0\forall x\)

Bài 1: Phân tích đa thức thành nhân tử.

a, ay - ax - 2x + 2y

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

b, 5ax - 7by - 7ay + 5bx

=5x(a+b)-7y(b+a)=(a+b)(5x-7y)

c, 4x^2 - 9x + 5

=4x²-4x-5x+5=4x(x-1)-5(x-1)=(x-1)(4x-5)

d, x^2 - 8x + 15

=x²-3x-5x+15=x(x-3)-5(x-3)=(x-3)(x-5)

Bài 2: Chứng minh rằng.

a, x^2 + x + 1/2 >0 với mọi x (mình nghĩ phải là >0 )

Ta có :

x²+x+1/2=x²+2x.1/2+1/4 +1/4=(x+1/2)²+1/4

vì (x+1/2)² ≥0 ∀ x nên (x+1/2)²+1/4 ≥ 1/4>0 ∀ x

Vậy x^2 + x + 1/2 >0 với mọi x

a) \(2x^2+8x+15\) \(=2\left(x^2+4x+\frac{15}{2}\right)\) \(=2\left(x^2+4x+4+\frac{7}{2}\right)=2\left(x+2\right)^2+7\ge7>0\)

b) \(-x^2+x-3=-\left(x^2-x+3\right)\) \(=-\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{11}{4}\right)=-\frac{11}{4}-\left(x-\frac{1}{2}\right)^2< 0\)

c) \(-4x^2+8x-11=-\left(4x^2-8x+11\right)\) \(=-\left(4x^2-2\cdot2\cdot2x+4+7\right)=-7-\left(2x-2\right)^2< 0\)

d) \(-9x^2+12x-15=-\left(9x^2-12x+15\right)\) \(=-\left(9x^2-2\cdot3x\cdot2+4+11\right)=-11-\left(3x-2\right)^2< 0\)