Bài 1.

a) ( 7x - 3 )² - 5x( 9x + 2 ) - 4x² = 18

<=> 49x² - 42x + 9 - 45x² - 10x - 4x² = 18

<=> -52x + 9 = 18

<=> -52x = 9

<=> x = -9/52 

b) ( x - 7 )² - 9( x + 4 )² = 0

<=> x² - 14x + 49 - 9( x² + 8x + 16 ) = 0

<=> x² - 14x + 49 - 9x² - 72x - 144 = 0

<=> -8x² - 86x - 95 = 0 

<=> -8x² - 10x - 76x - 95 = 0

<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0

<=> ( x + 5/4 )( -8x - 76 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)

c) ( 2x + 1 )² + ( 4x - 1 )( x + 5 ) = 36

<=> 4x² + 4x + 1 + 4x² + 19x - 5 = 36

<=> 8x² + 23x - 4 - 36 = 0

<=> 8x² + 23x - 40 = 0

=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))

Bài 2.

a) x² - 12x + 39 = ( x² - 12x + 36 ) + 3 = ( x - 6 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) 17 - 8x + x² = ( x² - 8x + 16 ) + 1 = ( x - 4 )² + 1 ≥ 1 > 0 ∀ x ( đpcm )

c) -x² + 6x - 11 = -( x² - 6x + 9 ) - 2 = -( x - 3 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

d) -x² + 18x - 83 = -( x² - 18x + 81 ) - 2 = -( x - 9 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

a. A= x²-7x+20 = x²-2*\(\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{31}{4}\)=(x-\(\dfrac{7}{2}\))²+\(\dfrac{31}{4}\)>0 \(\forall x\)(đpcm)

b. B= 2x²+5x+14=2(x²+2*\(\dfrac{5}{4}x+\dfrac{25}{16}+\dfrac{87}{16}\))=2(x+\(\dfrac{5}{4}\))²+\(\dfrac{87}{8}\)>0(đpcm)

thanks you !!!

\(A=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\)

\(B=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}\)

\(C=\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\)

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

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

\(M=-\left(x+\frac{7}{2}\right)^2-\frac{11}{4}\)

\(N=-5\left(x-\frac{3}{5}\right)^2-\frac{41}{5}\)

\(C\) đề sai ví dụ \(x=3\Rightarrow C=2>0\)

\(D=-5\left(x-\frac{7}{10}\right)^2-\frac{131}{20}\)

Bài 1:

a) \(x^2-2xy-25+y^2\) (Sửa đề)

\(=x^2-2xy+y^2-25\)

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

\(=\left(x-y-5\right)\left(x-y+5\right)\)

Vậy ...

b) \(x\left(x-1\right)+y\left(1-x\right)\)

\(=x\left(x-1\right)-y\left(x-1\right)\)

\(=\left(x-1\right)\left(x-y\right)\)

Vậy ...

c) \(7x+7y-\left(x+y\right)\) (Sửa đề)

\(=7\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(7-1\right)\)

\(=6\left(x+y\right)\)

Vậy ...

d) \(x^4+y^4\)

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

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

\(=\left(x^2+y^2-\sqrt{2}xy\right)\left(x^2+y^2+\sqrt{2}xy\right)\)

Vậy ...

Bạn xem lại 1 sô câu sai và bài 2 hộ mk

E=4x^​2​+5x+5>0 với mọi x

=(4x^​2 +4x+1)+4

=(2x+1)\(^2\)+4

Với mọi x thuộc R thì (2x+1)\(^2\)>=0

Suy ra(2x+1)\(^2\)+4>=4>0

Hay E>0 với mọi x thuộc R(đpcm)

F=5x²​-6x+7>0 với mọi x

=(5x\(^2\)-6x+\(\dfrac{36}{25}\))+\(\dfrac{139}{25}\)

=5\(\left(x-\dfrac{6}{5}\right)^2\)+\(\dfrac{139}{25}\)

Với mọi x thuộc R thì 5\(\left(x-\dfrac{6}{5}\right)^2\)>=0

Suy ra 5\(\left(x-\dfrac{6}{5}\right)^2\)+\(\dfrac{139}{25}\)>0

Hay F >0 với mọi x(đpcm)

G=-x^​2​​+5x -6<0 với mọi x​

=-(x^​2​​-5x+6,25)+0,25

=-(x-2,5)² +0,25

Với mọi x thuộc R thì -(x-2,5)² <=0

Suy ra -(x-2,5)² +0,25<0

Hay G<0 với mọi x (đpcm)

chúc bạn học tốt ạ

a: \(\Leftrightarrow3x^3-x^2+3x^2-x-6x+2-a-2⋮3x-1\)

=>-a-2=0

hay a=-2

b: \(-x^2+x-1\)

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

\(=-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< 0\forall x\)

c: \(P\left(x\right)=x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi x=5/2

d: \(f\left(x\right)=x^2-4x+4+5=\left(x-2\right)^2+5\ge5\forall x\)

Dấu '=' xảy ra khi x=2

a. \(x^2+3x+5\)

\(=x^2+2.x^2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

=> đpcm

b. \(4x^2+5x+7\)

\(=\left(2x\right)^2-2.2x.\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{87}{16}\)

= \(\left(2x+\dfrac{5}{4}\right)^2\) + \(\dfrac{87}{16}\) \(\ge\dfrac{87}{16}\)

=> đpcm

Câu a phần I sai. đề là :
a) A = -3x(x - 5 ) + 3(x2 - 4x ) - 3x + 10