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!

\(9x^2+42x+49=\left(3x+7\right)^2\)

Thay x=1 ta có 

\(\left(3.1+7\right)^2=10^2=100\)

\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a+2b^2\right)^2\)

Thay a=2;b=-1 ta có 

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

\(\(9x^2+42x+49\)\)tại x = 1

Ta có:\(\(9x^2+42x+49=\left(3x\right)^2+2.3x.7+7^2=\left(3x+7\right)^2\)\)

Thay x = 1 vào \(\(\left(3x+7\right)^2\)\)ta được:

\(\(\left(3.1+7\right)^2=10^2=100\)\)

\(\(\frac{1}{4}a^2+2ab^2+4b^4\)\)tại a = 2 ; b = -1

Ta có: \(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b^2+\left(2b\right)^2=\left(\frac{1}{2}a+2b^2\right)^2\)

Thay a = 2 ; b = -1 vào\(\left(\frac{1}{2}a+2b^2\right)^2\)ta được:

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

1) \(a,\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)=x^2+4x-21-x^2-4x+5=-16\)

\(b,\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2=\left(x+8-x+2\right)^2=\left(10\right)^2=100\)

c, \(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)=x^3+1-x^3+1=2\)

a, \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x=\left(x+1\right)^3-\left(x-1\right)^3-3x\left(x+1\right)+3\left(x+1\right)x=\left(x+1\right)^3-\left(x-1\right)^3=2\left(\right)\)

hình như sai r thui ko làm nữa

1.

a,\(\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)\)

\(=(x^2+7x-3x-21)-\left(x^2-x+5x-5\right)\)

\(=x^2+7x-3x-21-x^2+x-5x+5\)

\(=-16\)

b,\(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)

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

\(=10^2=100\)

c,\(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

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

\(=x^3+1-x^3+1\)

\(=2\)

2.

a,\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\)

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

\(=x^3+3x^2+3x+1-x^3+1-3x+3x^2\)

\(=6x^2+2\)

(Bạn xem lại đề câu này nhé,hoặc là mình làm sai)

b,\(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=\left(x^3+6x^2+12x+8+x^3-6x^2+12x-8\right)-2x^3-24x\)

\(=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x\)

\(=x^3+x^3-2x^3+6x^2-6x^2+12x+12x-24x+8-8\)

\(=0\)

Vậy giá trị của biểu thức trên không phụ thuộc bởi x

(Câu này cũng không chắc lắm,bạn cũng xem lại xem mình đúng chưa nhé!!!)

\(\)

\(\)

b: Ta có: \(x^3-9x^2+27x-27\)

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

\(=\left(-7\right)^3=-343\)

c: Ta có: \(\dfrac{x^3-1}{x^2+1}\)

\(=\dfrac{6^3-1}{6^2+1}=\dfrac{215}{37}\)

\(a,892^2+216.892+108^2=892^2+2.108.892+108^2=\left(892+108\right)^2=1000^2=1000000\)

b, \(9x^2+42x+49=9.1^2+42.1+49=9+42+49=100\)

c,\(\left(x^2-3x\right)+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)=\left(6+1\right)\left(6-3\right)=7. 3=21\)

Câu 1 :

\(a,\left(3x+2\right)^2=9x^2+12x+4.\)

\(b,\left(6a^2-b\right)^2=36a^4-12a^2b-b^2\)

\(c,\left(4x-1\right)\left(4x+1\right)=16x^2-1\)

\(d,\left(1-x\right)\left(1+x\right)\left(1+x^2\right)=\left(1-x^2\right)\left(1+x^2\right)=1-x^4\)

\(e,\left(a^2+b^2\right)\left(a^2-b^2\right)=a^4-b^4\)

\(f,\left(x^3+y^2\right)\left(x^3-y^2\right)=x^6-y^4\)

Bài 2 :

\(a,A=9x^2+42x+49=9+42+49=100.\)

\(b,B=25x^2-2xy+\frac{1}{25}y^2=\left(5x^2\right)-2.5x.\frac{1}{5}y+\left(\frac{1}{5}y\right)^2\)

\(=\left(5x-\frac{1}{5}y\right)^2=\left(-1+1\right)^2=0\)

\(c,C=4x^2-28x+49=4x^2-14x-14x+49\)

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

\(=\left(8-7\right)\left(4-7\right)=-3\)

Bài 3 :

\(a,\left(x-3\right)^2-4=0\)

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

\(\Rightarrow\left(x-1\right)\left(x-5\right)=0\)

\(\Rightarrow x\in\left\{1;5\right\}\)

\(b,x^2-2x=24\)

\(\Rightarrow x^2-2x-24=0\)

\(\Rightarrow x^2+4x-6x-24=0\)

\(\Rightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Rightarrow x\in\left\{-4;6\right\}\)

\(C = \frac{x^{2} + 6x + 9}{x^{3} + 3x^{2} - 9x - 27}\) (ĐKXĐ: \(x \neq \pm 3\))

= \(\frac{(x + 3)^{2}}{x^{3} -27 + 3x^{2} - 9x}\)

\(= \frac{(x + 3)^{2}}{(x - 3)(x^{2} + 3x + 9) + 3x(x - 3)} = \frac{(x + 3)^{2}}{(x - 3)(x^{2} + 6x + 9)}= \frac{(x + 3)^{2}}{(x - 3)(x + 3)^{2}} = \frac{1}{x - 3}\)

Với x = 103 thỏa mãn ĐKXĐ nên giá trị của bt C = \(\frac{1}{103 - 3}= \frac{1}{100}\)

câu còn lại pn lm tương tự nha