(3x+4)² +(4x -1 )² + 2x + 5(2x-5)

= 9x²+24x+16 + 16x²-8x+1 + 2x + 10x - 25

= 25x² + 28x - 8

b. (2x+1)(4x² -2x + 1 ) + ( 2-3x ) ( 4+ 6x + 9x² )

= 8x³ + 1 + 8-27x³

= -19x³ +9

\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

a) \(\left|x\right|=2\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

+) TH1: \(x=2\)

\(A=\left(3\cdot2+5\right)\left(2\cdot2-1\right)+\left(4\cdot2-1\right)\left(3\cdot2+2\right)\)

\(A=89\)

+) TH2: \(x=-2\)

\(A=\left(-2\cdot3+5\right)\left(-2\cdot2-1\right)+\left(-2\cdot4-1\right)\left(-2\cdot3+2\right)\)

\(A=-27\)

Vậy...

b) \(B=9x^2+42x+49\)

\(B=\left(3x+7\right)^2\)

\(B=\left(3\cdot1+7\right)^2\)

\(B=100\)

Vậy...

c) \(C=25x^2-2xy+\frac{1}{25}y^2\)

\(C=\left(5x-\frac{1}{5}y\right)^2\)

\(C=\left(\frac{-1}{5}\cdot5-\frac{1}{5}\cdot\left(-5\right)\right)^2\)

\(C=0\)

Vậy...

b) \(B=9x^2+42x+49\)

Thay \(x=1\) vào biểu thức B, ta được:

\(B=9.1^2+42.1+49\)

\(B=9+42+49\)

\(B=51+49\)

\(B=100\)

Vậy giá trị của biểu thức B tại \(x=1\) là \(100.\)

Chúc bạn học tốt!

a, (4x-3)(3x+2)-(6x+1)(2x-5)+1

=12x²-8x-9x+6-12x²+30x-2x+5+1

=11x+12

b, (3x+4)²+(4x-1)²+(2+5x)(2-5x)

=9x²+24x+16+16x²-8x+1+4-25x²

=16x+21

c, (2x+1)(4x²2x+1)+(2-3x)(4+6x+9x²)-9

=8x³+1+8-27x³-9

=-19x³

swingrock có thể giải thik rõ hơn đc ko ạ

a)  \(x^2+4x+4=\left(x+2\right)^2\)

b)  \(9x^2-12x+4=\left(3x-2\right)^2\)

c)  \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

d)  kiểm tra lại đề nhé

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

f)  \(9y^2-\frac{1}{4}=\left(3y-\frac{1}{2}\right)\left(3y+\frac{1}{2}\right)\)

g)  \(8x^3+8a^3=\left(2x+2a\right)\left(4x^2-4ax+4a^2\right)\)

h)  \(64x^3-27y^3=\left(4x-3y\right)\left(9y^2+12xy+16x^2\right)\)

Cảm ơn bạn nhìu !!

26:

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

=12x^2+4x-5-12x^2+5x+2

=9x-3

Khi x=-2 thì A=-18-3=-21

25:

b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)

=y^3+y^2+y-3y^2-3y-3-y^3+2y

=-2y^2-3

\(A=27x^3-8-27x^3+3=-5\\ B=8x^3+y^3+8x^3-y^3-16x^3=0\)